Extraordinary events do not always require extraordinary causes. Given enough time, they can happen by chance. Knowing this, Mlodinow says, “we can improve our skill at decision making and tame some of the biases that lead us to make poor judgments and poor choices ... and we can learn to judge decisions by the spectrum of potential outcomes they might have produced rather than by the particular result that actually occurred.” Embrace the random. Find the pattern. Know the difference.
Note: This article was originally printed with the title, "A Random Walk through Middle Land".
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174 Comments
Add CommentI guess we have to be skeptical of the skeptic. Michael Shermers interpretation of Lets Make a Deal ignores the effect of the additional information injected into the situation. With door number two exposed as a goat, the original dynamics are irrevocably changed. The bad, good, bad scenario no longer exists and the contestants decision comes down to choosing to stay with door number one or changing to door number three. Under those conditions, there is no way switching to door number three will generate the two-thirds success rate Shermer contends.
Reply | Report Abuse | Link to thisI second Mr. King's comment of 9/15.
Reply | Report Abuse | Link to thisI agree with David King and Marcus Millet and I disagree with the published solution to the Monty Hall Problem . I wrote this before seeing their comments but I will post it here anyway in support.
Reply | Report Abuse | Link to thisMichael Shermeer says: "You had a one in three chance to start, but now that Monty has shown you one of the losing doors, you have a two-thirds chance of winning by switching. Here is why. There are three possible three-doors configurations: (1) good, bad, bad; (2) bad, good, bad; (3) bad, bad, good. In (1) you lose by switching, but in (2) and (3) you can win by switching."
Our skeptic neglects that the matrix of possible configurations changes when it is learned that the middle door contains a goat.
Configurations before knowing the middle door contains a goat.
(1) good, bad, bad; P=1/3
(2) bad, good, bad; P=1/3
(3) bad, bad, good. P=1/3
Configurations with additional information that the middle door contains a goat.
(1) good, bad, bad; P=1/2
(2) bad, good, bad; P=0
(3) bad, bad, good. P=1/2
Knowing that configuration (2) is impossible changes its probability to zero. Since the probabilities have to add up to 1.0 the probabilities for (1) and (3) become 1/2 each. There is no reason to switch your initial choice of (1).
Sincerely,
Stan Hilliard
Here's why the above comments are wrong and you should change all the time.
Reply | Report Abuse | Link to thisWith 3 doors, the initial probabilities are 1/3 that door #1 is the winner, and 2/3 that all other doors are the winner. The 2/3 probability is initially distributed to 2 doors (1/3 each).
Now Monty unveils a door you didn't choose. The entire original 2/3 probability can now only be allocated to the one remaining door. Since 2/3>1/3 you should switch.
The erroneous argument of 50/50 outcomes after revealing one door is only valid of Monty can relocate of the prize. If Monty does not relocate after revealing, then the original probabilities must remain with the original sets.
Lets now take the 10 door case. You choose door #1. The same argument says you have 10% chance of being the winner, and doors 2-10 are 90% likely to be the winner. As Monty reveals doors 2-9, those original probabilities do not change but the 90% chance of winning with doors 2-10 is allocated to fewer and fewer doors until, as the Skeptic notes, your chance of winning by switching to door #10 is 90%.
If you still don't think so, then have your spouse, child or a friend play Monty using the 10 door case. The 90% chance of winning will be evident much more rapidly than using the 3 doors.
Sincerely,
Chris
I don't buy Soccer PhD's explanation either. Since number 3 also had a 1/3 probability to start, it doesn't automatically inherit #2's probability when number 2 is revealed to be bad.
Reply | Report Abuse | Link to thisdduehren - But its ok for both doors to "inherit" 1/2 - 1/3 automatically?
Reply | Report Abuse | Link to thisThis gets to the heart of what "probability" means. Is the answer 50% for each remaining door because there are only 2 doors left or does probability mean "if I played this game thousands of times how often would door #1 win?"
I agree with King et al; and disagree with SoccerPhd, whom we'd all like to play a game of chance with.
Reply | Report Abuse | Link to thisSoccerPhd, how does your argument NOT lead to the belief that if you flip a fair coin 9 times and get 9 heads, the 10th flip of the coin will have greater than 1/2 probability of coming up tails?
Reply | Report Abuse | Link to thisIn truth, if the coin-flip is fair the probability remains 1/2.
If the 10th flip came up heads again though, I think I would start to suspect that the guy might be using a two-headed coin -- or something.
After cogitating further about SoccerPhD's arguments, I am now convinced that he's right. This is really a very cool problem that 90% of the people (including myself) will initially believe vehemently is always 50-50. I guess that was the Skeptic's whole point.
Reply | Report Abuse | Link to thisSwitch or stay?
Reply | Report Abuse | Link to thisHere is another argument in support of the view that you will NOT improve your odds of winning the new car by switching your choice of doors after seeing the goat in door #2.
Assume that instead of choosing door #1 initially, you had chosen #3. If your strategy is to switch your choice, then you would switch your choice to door #1. The switched-to door is solely determined by which door you chose first.
But your initial choice of doors (#3 in this case) is arbitrary and based on ignorance as to which door contains the new car. Your switched-to door depends only on which door you chose first, which in turn was based on ignorance.
So if you switch, regardless of whether you win the new car or not, the contribution of your thinking to that outcome will have been that you supplied ignorance-squared.
Three possibilities:
Reply | Report Abuse | Link to thisCar behind #1 Car behind #2 Car behind #3
You initially choose #1
To show goat Monty opens #2 Monty opens #3 Monty opens #2
You stick with #1 You Win You Lose You Lose
or
You switch to #3, You Lose #2, You Win #3, You win
Thus you have twice the chance of winning by switching
Three possibilities:
Reply | Report Abuse | Link to thisCar behind #1 Car behind #2 Car behind #3
You initially choose #1
To show goat Monty opens #2 Monty opens #3 Monty opens #2
You stick with #1 You Win You Lose You Lose
or
You switch to #3, You Lose #2, You Win #3, You win
Thus you have twice the chance of winning by switching
I don't know why this works, but it does.
Reply | Report Abuse | Link to thisI wrote a simple program that mimics the scenario. There are 3 doors, one of which has randomly been selected to be the winner. You randomly choose one. You are shown one of the other doors which is a loser. Now I wrote my program to iterate this same contest 10000 times. It has 3 modes, one where you never switch doors, one where you always switch doors, and one where you randomly choose whether or not to switch doors.
To my great surprise, when running the program, I find:
When never switching doors, you win about 1/3 of the time.
When always switching doors, you win about 2/3 of the time.
When randomly choosing whether or not to switch doors, you win about 1/2 of the time.
It was much more dramatic when I changed the door count to 100. Then I got:
When never switching doors, you win about 1/100 of the time.
When always switching doors, you win about 99/100 of the time.
When randomly choosing whether or not to switch doors, you win about 1/2 of the time.
Simply amazing!
Now, I think I've wrapped my head around this.
Reply | Report Abuse | Link to thisLook at the ten door situation. And don't think in terms of the contestant, think in terms of the host.
There are two possibilities:
One, the contestant first chooses the wrong door (90% of the time), then as the host I must show the other 8 wrong doors and leave the correct door for the contestant to possibly switch to.
Two, the contestant first chooses the correct door (10% of the time), then as the host I can show any combination of the remaining doors leaving one loser for the contestant to possibly switch to.
So, 90% of the time the host must "give away" the winner by eliminating all other choices.
As the contestant, by always switching, I will get the first situation 90% of the time where I first chose wrong and now am switching to the correct door, and I will get the second situation 10% of the time where I first chose correctly and now switch to the loser.
Probability is certainly a difficult science, which is why the casinos make money.
If you played the 3-door game 3,000 times selecting the same door each time and the good door is randomly placed, you should win 1,000 times (probably) regardless of which door you choose. Likewise, to win 1,000 times with the 10-door game, you will need to play 10,000 times. The added knowledge does not change these probabilities.
Reply | Report Abuse | Link to thisDavid King an his supporters are definitely right. At the start the prize is equally likely to be behind any one of the three doors. The only fact other fact that matters is that Monty eliminated Door 2. The prize is now equally likely to be behind Door 1 or Door 3. The correct description "equally likely" helps to keep one from getting confused by fractions.
Reply | Report Abuse | Link to thisThe comment about "playing the game a thousand times" is pertinent. It is easy to write a computer program that models this scenario, with appropriately random draws, and run it as many times as desired. I assure you that the result is the two remaining doors are equally likely to contain the prize. This might be called the "Francis Bacon approach," since Bacon reputedly poked fun at this sort of debate by describing a group of philosophers trying to determine the number of teeth a horse has by doing everything except looking in his mouth.
No one so far has pointed out the unstated assumption behind the 2/3 win result: The host must be unbiased! If you don't believe that this is relevant, consider the mathematically identical problem of one bean under three half-walnut shells where you choose the one you think you saw the bean put and then the person who shuffled the shells puts his finger on your choice and "accidentally" exposes an adjacent shell to be empty. By Shermer's (& Marilyn's) logic, you should immediately say, "No, I mean the other (remaining) shell and you will probably win. However, if I were the shuffler, I would have this "accident" only when the chooser has made the right choice. I have never seen this problem discuss when the host might chose to open the third door. On TV, he might want to improve the contestant's odds by only opening a door when the first choice was wrong. At a county fair, the guy with the walnut shells would want to decrease the contestant's odds by only uncovering another shell when the contestant was initially right. So, depending on the motivation of the host, the odds of winning if you switch can range from 0% to 100%!
Reply | Report Abuse | Link to thisI was wrong! I mistakenly assumed that giving the contestant a preliminary choice was irrelevant. In fact, it is crucial. Consider the three cases:
Reply | Report Abuse | Link to this1) Prize behind door 1: Contestant chooses door 1. Monty opens either door 2 or door 3. Contestant switches to the remaining door. Contestant looses.
2) Prize behind door 2: Contestant chooses door 1. Monty MUST open door 3. Contestant switches to the remaining door. Contestant wins.
3) Prize behind door 3: Contestant chooses door 1. Monty MUST open door 2. Contestant switches to the remaining door. Contestant wins.
Thus the contestant wins two times out of three. He gets the advantage by eliminating one of Monty's choices in the beginning.
I now agree with Ed Rush. He has given a clear argument. There are only three possible arrangements as he has shown. If the contestant switches he wins 2 out of 3 times. Conversely, if the contestant sticks to door 1 he wins only 1 out of 3 times.
Reply | Report Abuse | Link to thisIn response to Ed Rush, there are actually possible four cases:
Reply | Report Abuse | Link to this1) Prize behind door 1: Contestant chooses door 1. Monty opens door 2. Contestant switches to door 3. Contestant loses.
2) Prize behind door 1: Contestant chooses door 1. Monty opens door 3. Contestant switches to door 2. Contestant loses.
3) Prize behind door 2: Contestant chooses door 1. Monty MUST open door 3. Contestant switches to the remaining door. Contestant wins.
4) Prize behind door 3: Contestant chooses door 1. Monty MUST open door 2. Contestant switches to the remaining door. Contestant wins.
-The first two cases can't be grouped together- notice the word 'either' which implies two distinct outcomes which each have a separate chance of occurrence. Furthermore, even a computer simulation will fail if it doesn't eliminate former potential outcomes which are later shown to be eliminated from chance. If door one is chosen, and door two is shown to be a loser, the "lose, win, lose" outcome is impossible. The 1/3 chance it had doesn't get magically transferred to the "lose, lose, win" outcome (nor to the "win, lose, lose"). The odds of winning by always switching (or always not switching) are best determined by a probability tree, not a table, which fails to represent decision making.
Ed Rush is right, Gabriel Enck is wrong. There indeed are four possible cases with a switch, but losing cases 1) and 2) each have a probability of 1/6 while winning cases 3) and 4) each have a probability of 1/3.
Reply | Report Abuse | Link to thisHaving read the comments to the article, I find that both sides of the argument appear to have been based on sound reasoning. We are often subject to pre-conceived conviction and influenced by psychological preferences before making a decision.
Reply | Report Abuse | Link to thisPerhaps, the best way to solve the dilemma is to try it out. Just play the 3-door game hundreds if not thousands of times and evaluate the outcome.
Verily, probability is a strange animal. If life itself is a probability, what else is not? (btt1943@yahoo.com)
Reply | Report Abuse | Link to thisTheory must be verified or falsified by experiment – the indisputable golden rule of science (mathematics included). This is exactly what is happening at the HLC center in CERN, though temporary inactive due to a technical glitch.
That said, I would certainly love to know the outcome if the game is indeed played that many times by someone, somewhere, sometime.
I agree with David King and others who agree with him. Put it this way. Suppose you don't choose which door until the goat is revealed behind the middle door. The only choices would then be Doors 1 and 3, each of which would have a 50-50 chance of being right.
Reply | Report Abuse | Link to thisI may be wrong, but it seems to me that all the switching proponents in this discussion, plus Michael Shermer and Marilyn vos Savant themselves I suspect, are smuggling excluded goats in through the back door so to speak rather than sticking to the situations specified.
Reply | Report Abuse | Link to thisRecognizing the apparently inexhaustible range of arguments on behalf of switching, however, as a true scientific skeptic I did do an actual empirical experiment! It's one, moreover, that anyone can replicate since I didn't do it with a computer where the incorrect answer might be built into the programming -- garbage in, garbage out -- or any other recondite equipment. Instead I simply used three playing cards to precisely model the original situation. Two of these I designated as goats, and one as a car. I then shuffled them with my eyes closed and laid them out on the table, the card on my left always being the chosen, again to conform with the original example. Further to conform with our specified conditions I also of course excluded those deals where the car card was in the middle. Of the remaining ten deals five had the car on the left and five on the right.
You may think ten deals too few for a reliable conclusion in this matter, and I would hope some of you with more patience than I have might contribute more. But those ten alone actually have considerable significance. I included the third card merely for verisimilitude. Two cards, one a goat and one a car, is clearly enough to model the choice point of these games however many doors they involve. So instead of the three door game I could just as well have been playing, for example, a million door game with 999,998 goats eliminated leaving one goat and one car. According to Shermer the probability of the card on the right being the car would then have been 999,999 to 1, and that would with a very high probability be reflected even in a sample so small. If someone is still unsatisfied with that, however, remember that we could just as well pretend that it was a billion or trillion door game or however many would be required to give even my results statistical significance by anyone's standards.
I've given this more thought and I've changed my mind. Ddevader's explanation convinced me. With 10 doors, you only have a 1 in 10 chance of choosing the right door and the right door has a 9 in 10 chance of being among the others. When the host eliminates 8 wrong doors, there is a 9 in 10 chance that the remaining one is right == so you should change your choice to that door! Amazing!!!?
Reply | Report Abuse | Link to thisWhat if Bill and Tom play the game simultaneously? Bill chooses Door 1 and Tom Chooses Door 3. Monty opens Door 2, revealing a goat. Would Bill and Tom BOTH be better off switching?
Reply | Report Abuse | Link to thisHugh de Mann's refutation is brilliant. I wish I'd thought of that.
Reply | Report Abuse | Link to thisIt's certainly tempting to think: "Damn it, we had only a one in three chance of selecting the correct door. Why should that change?" But that's just like thinking that, since the odds of getting three heads in a row on three tosses of a fair coin are only one in eight, the chances of your also getting a head on your third toss after two heads must also be only one in eight since that would make three in a row.
The whole point of Shermer's article was that improbable things happen. He was right about that, but wrong, apparently, in not fully realizing that when they do happen we need to recalculate our probabilities.
Much of the confusion here arises because Michael Shermer did not specify the rules of the game. In detail, after the contestant has chosen a door, Monty Hall is constrained to open a door that does not end the game, i.e., not the one the contestant has chosen and not the one with the automobile behind it (both might be the same door). Phil's 9/25 comment provides a concise example of why we needn't be skeptical about Michael Shermer's conclusion.
Reply | Report Abuse | Link to thisThe Hugh de Mann "refutation" is not applicable because there is a 1/3 chance that the automobile is behind the one door neither contestant chose, so Monty Hall cannot follow the rules: he would have to either end the game by showing the position of the automobile or eliminate one of the contestants. To continue the game, he would have to eliminate one of the contestants, and, oddly enough, with this situation the remaining contestant has a 100 percent probability of winning by switching because Monty Hall would not have broken the rules by eliminating a contestant unless he was forced to by the automobile being behind the unchosen door..
The only thing clear about this game is that it has confused many of us. Michael Shermer gives a clear explanation of why the contestent should switch to the other remaining door not exposed. No matter how manner doors are used there will only be two doors closed (the chosen door and the door to switch to) after doors not containing the car are exposed.
Reply | Report Abuse | Link to thisEveryone agrees there is an equal probability that the randomly placed car may be behind any of the 3 (or n) doors provided by this game. If the contestent initially picks the door with the car behind it he will lose the car when he switches. However, if he initially picks a door with a goat behind it he is guaranteed to win the car. We only need to consider his probability of initially picking a door with a goat behind it. Therefore, the 2 out of 3 (or n-1 out of n) probability applies to his chances of winning the car if he swiches.
That was me! On TV with Monty Hall! I remember (except I can't remember what happened two minutes ago. I have that kind of short-term amnesia. What's it called? I forget!). Anyway, there I was with the audience urging me to 'keep it' or 'change' and I couldn't remember what was going on. So Monty said the prize was behind one of those two doors (nevermind that there was a third door opened with a goat standing in it and I LIKE GOATS!). So I figured I had a fifty-fifty chance of picking the prize. I didn't know I had already picked one of the doors but since I got to pick again, it made no difference. And since I couldn't remember which I had already chosen, "change" was irrelevant. So, frustrated and confused, I pleaded, "can't I just have that goat?!"
Reply | Report Abuse | Link to thisThis is counterintuituve, like the birthday paradox. But I must also be honest that the wording of the original problem is intentionally misleading to increase entertainment value. Technically, the instance of Monty revealing the goat behind door number 2 is correct, but it is misleading when presented almost like a generality - or like a given. SoccerPhD and others had to add crucial information back into the problem - that Monty doesn't always reveal door number 2.
Reply | Report Abuse | Link to thisHere is how the deception occurs, both in the trick and in the problem as stated.
Take 3 coins, one different from the others. Let the "good" coin visit all 3 positions (1, 2, and 3). With the good coin "chosen" at 1 by placing a finger on it, then "reveal" either bad coin at 2 or 3 by pointing at it. Switching to the remaining coin will lose each time (2/6). But now move the good coin to position 2. Choose 1 again. Since the Monty finger will not reveal position 2, it must reveal position 3 in both instances of random assignment. Switching to the remaining position (2) actually wins both times (2/6). Ditto if the good coin is in position 3, in which case the Monty finger always reveals position 2, and the remaining postion (3) wins both
times (2/6). The two paths in the universe where the Monty finger acted randomly and told the truth ("OMG - look at that! It's the car!") do not occur. They are converted to deceptions ("Look - no car! And you may have even picked it already!")
In a strange way, the "mark" doesn't correctly factor in the possibilities that Monty steered away from the car when showing the goat. We tell ourselves that Monty is not so stupid as to give away information. We do a rough calculation on the spot - "folk numeracy" - and get 50/50. Monty actually brazens out the situation by riskily showing bad positions, inducing the mark to "stay" instead of switching. Fascinating. And a moneymaker, just like the birthday paradox. But also a loser for Monty if the mark is wise.
Shermer also deceives by hiding information. Not allowed in a journal ("You were not clear about picking any bad door. Fix it."), but perfectly alright for entertainment. And I have to admit, this was entertaining, and extremely enlightening.
I also started out agreeing with David King, but on further consideration must agree with SoccerPhD (and Michael Shermer) that switching is a winning strategy.
Reply | Report Abuse | Link to thisThe important consideration here has been made various ways by others who've commented, but maybe another way of stating it will help clarify it.
Shermer explained that there are three possible 3-door combinations, but he failed to make clear how in two of them you win by switching. It appears at first glance that there are only two left after Hall opens a door, so you ought to have a 50/50 chance whether or not you switch.
What Shermer failed to make clear was why the original count of three remains significant. It is because Hall does not always open the middle door; he always opens a door with a goat behind it. So, going with Shermer's numbering and the assumption that you always first choose the left door, Hall will open either the middle or right door.
Case 1: Hall can open either door to show you a goat, and you will lose when you switch.
Case 2: Hall must open the right door, and you will win when you switch.
Case 3: Hall must open the middle door, and you will win when you switch.
So, indeed, your chances of winning if you do not switch are 1/3, and of winning if you do switch are 2/3.
The problem with the example, like a poorly told joke is in the telling. In the article he said that Monty shows "shows you that a goat is behind number two". Calling out the door number is too specific, everybody gets hung up on door number two. As a result, the example with the three cases doesn't make sense because Monty can't show what's behind door 2 in the second case. It just should have said that "after you pick a door Monty opens one of the other two doors to reveal a goat". Shermer should definitely not go on stand-up statisticians circuit.
Reply | Report Abuse | Link to thisWhy not simply make a "truth-table"? Like this:
Reply | Report Abuse | Link to thisThe car can be behind any of three doors.
The contestant can select any door.
The host opens one of two doors - sometimes HAS to (making p = 1.0) sometimes may choose (making p = 0.5)
#1: Car = Door #1. Cont sel #1. Host opens #2 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#2: Car = Door #1. Cont sel #1. Host opens #3 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#3: Car = Door #1. Cont sel #2. Host MUST open #3 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#4: Car = Door #1. Cont sel #3. Host MUST open #2 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#5: Car = Door #2. Cont sel #1. Host MUST open #3 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#6: Car = Door #2. Cont sel #2. Host opens #1 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#7: Car = Door #2. Cont sel #2. Host opens #3 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#8: Car = Door #2. Cont sel #3. Host MUST open #1 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#9: Car = Door #3. Cont sel #1. Host MUST open #2 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#10: Car = Door #3. Cont sel #2. Host MUST open #1 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#11: Car = Door #3. Cont sel #3. Host opens #1 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#12: Car = Door #3. Cont sel #3. Host opens #2 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
Total NOT SWITCH and win = 1/3
Total SWITCH and win = 2/3
Regards,
KFS
Why not make a "truth-table"?
Reply | Report Abuse | Link to thisSomething like this:
The car can be behind any of three doors.
The contestant can select any door.
The host opens one of two doors - sometimes HAS to (making p = 1.0) sometimes may choose (making p = 0.5)
#1: Car = Door #1. Cont sel #1. Host opens #2 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#2: Car = Door #1. Cont sel #1. Host opens #3 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#3: Car = Door #1. Cont sel #2. Host MUST open #3 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#4: Car = Door #1. Cont sel #3. Host MUST open #2 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#5: Car = Door #2. Cont sel #1. Host MUST open #3 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#6: Car = Door #2. Cont sel #2. Host opens #1 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#7: Car = Door #2. Cont sel #2. Host opens #3 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#8: Car = Door #2. Cont sel #3. Host MUST open #1 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#9: Car = Door #3. Cont sel #1. Host MUST open #2 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#10: Car = Door #3. Cont sel #2. Host MUST open #1 = (1/3 x 1/3 x 1/1) 2/18
Sw & win = 2/18
#11: Car = Door #3. Cont sel #3. Host opens #1 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
#12: Car = Door #3. Cont sel #3. Host opens #2 = (1/3 x 1/3 x 1/2) 1/18
Not sw & win = 1/18
Total NOT SWITCH and win = 1/3
Total SWITCH and win = 2/3
/KFS
It is funny that people don't see that you have an original 90% chance of having been wrong in the 10 door example, and since you are given the assurance that 1 of the 2 doors left are the correct door, you have a higher chance of getting the prize if you choose door 10. The comments are all proving the postulation of the article. Brilliant!!
Reply | Report Abuse | Link to thisWith 3 doors you have a 2 out of 3 chance of being WRONG! Given the choice to change you now have a 1 out of 3 chance of being wrong. This is better. Over a string of 1000 tries, you would win 33.3333% of the time by never switching and 66.66666% of the time by switching every time.
In the 10 door example it gets even more lopsided. If you never switch in a random selection of a random placement, you win only 10% of the time over a long trial. Yet if you have 9 of the 10 remaining doors eliminated, you win 90% of the time by switching every time.
The key here is that the original opportunity to be WRONG is more important than the new level of opportunity to be RIGHT.
The known information includes 1 correct and multiple wrong answers. Yes, you have a final chance of one of the two doors being correct at your last decision. BUT you first had to choose which door to start with. This decision had a lower than 50% chance of being right. So no matter how many of the other choices you eliminate, you still are better off changing if the remaining choices are reduced to 1 other door.
This kind of observation is very entertaining in a geeky way. Great article.
For those who still can't believe this (like I couldn't), I have created an experiment on my blog: http://rand1-365.blogspot.com/2008/10/doors-experiment.html
Reply | Report Abuse | Link to thisYou only gain an advantage is switching if a door with the goat MUST be shown after you select.
Reply | Report Abuse | Link to thisWe assume that things cannot be swapped between doors or the game is not worth playing.
We know that we will never be shown the car behind one of the two doors not yet picked because that would end the game.
If it is manditory to show a goat then switching is best.
If it is optional to show the goat then the game becomes one of bluff instead of math. Even being told at this point that a goat must be shown will not override our instincts that it is a trick.
So I disagree with the article that we never evolved a proability network, we evolved an instinct to run counter to proability when another intelligence is at work as opposed to always going with proability or ignoring it altogether.
A predator when seeing a bird pretending to be wounded would be distracted. A human would have a feeling that the prize.
It is the instinct of "good luck" should be viewed with suspicion which is why we use science to help prove that some good things are not luck.
I have to disagree with Mr. Shermer. The only way that you have a 2/3 probability of winning is if you maintain a scenario which is no longer an option. When you make your first choice, Mr. Hall reveals a goat. This revelation eliminates one of the three possibilities ( in this case; loss, win, loss) which continues to be included in Mr. Shermer's calculation. There are at this point only two possibilities left and a 50/50 shot at the prize.
Reply | Report Abuse | Link to thisNo matter which losing door Mr. Hall reveals, it eliminates one of the possibilities which would be counted towards the 2/3 win. I'm afraid that this is a scenario where instinct seems to be more evolved than intellect.
The Monty Hall Problem was settled LONG ago, you people are just making yourselves look foolish by arguing against the correct solution. There is a 2/3 chance of winning if you always switch. See wikipedia - http://en.wikipedia.org/wiki/Monty_Hall_problem
Reply | Report Abuse | Link to thisStop this madness. This problem is a solved problem! See http://semanticvector.blogspot.com/2008/10/lets-make-deal-let-monty-rest.html
Reply | Report Abuse | Link to thisI did have to read through some of the comments before I was convinced! Nicely done, commenters!
Reply | Report Abuse | Link to this"It doesnt matter, because the roulette wheel has no memory, yet gamblers notoriously employ both the hot streak fallacy and the dueness fallacy, much to the delight of casino owners."
- of course, if it doesn't matter, than you might as well employ either fallacy - unless of course you could choose just to stop while you're ahead.
Actually, though, if applied to judging the rationality of choices made by actual people, this has the potential to be erroneous - by assuming that winning money is the only value involved. On a case by case basis, there could be all sorts of other factors. A person might rationally choose to employ a fallacy in the same way a person chooses to enjoy a superhero movie (which also costs money - and that isn't made up for by the food!). Few actual games happen in isolation - they are all part of bigger games.
Let's look at the possibilities, we'll assume that we know which door holds the car behind it, and that the contestant will ALWAYS choose door #1, and will ALWAYS switch when given the chance. Note that this will still apply even if the contestant chooses any other door.
Reply | Report Abuse | Link to this1. The car is behind door #1.
- The contestant chooses door #1 (33.3% chance of being correct)
- The host opens door #2 to reveal the goat
- The contestant switches
Result: Contestant LOSES
2. The car is behind door #2
- The contestant choose door #1 (33.3% chance of being correct)
- The host opens door #3 to reveal the goat
- The contestant switches
Result: The contestant WINS
3. The car is behind door #3
- The contestant chooses door #1 (33.3% chance of being correct)
- The host opens door #2 to reveal the goat
- The contestant switches
Result: The contestant WINS
I was skeptical at first too; but probability is not random as shown above. Every time the contestant chooses a door, she only has 33.3% of being correct; by switching, most (not all) of the time, the chances of being correct are bumped up to 66.6%.
The reason why it's not 50/50 is because when the contestant has their initial choice at the beginning of the game, she is only guaranteed to be 33.3% correct. It would only be a 50/50 chance if there were only two doors. When the number of options increase it seems, the likely hood of switching and being correct also increases.
I think here is talking a paradox.
Reply | Report Abuse | Link to thisThis discussion is somewhat interesting, but I'm surprised nobody thus far has referred to the Wikipedia article on this subject. It does a very nice job explaining the history of the problem as well as the accepted solution. Plus, it illustrates why the counter-intuitive answer is the correct answer.
Reply | Report Abuse | Link to thishttp://en.wikipedia.org/wiki/Monty_hall_problem#Solution
If the host opened door number 2 because it was door number 2 and there just happened to be a goat behind it the odds of winning by changing doors is 1/2
Reply | Report Abuse | Link to thisIf the host opened door number 2 because it had a goat behind it and would have opened door number three if door number 2 had had the prize behind it then the odds of winning by swapping doors is 2/3
The question in this article does not make it clear which situation it is. So you are all right and you are all wrong :-). if the question is asked properly as it is on wiki then the answer is always 2/3
The confusion arises here hrough the mistating of the poblem. The host does NOT show you door two - he shows you a goat - which may be from door two, OR door three. He therefore effectively"collapses" the combined probability of the car being in door two and three into the one remaining unshown door - hence switching now gives you 66.6% winning chance. The whole thing depends on the fact that the host has to show you a goat - and is therefore giving you information you did not have before.
Reply | Report Abuse | Link to thisI'm afraid you do not understand how statistics works. The fact that the middle option was removed does not change the outcome: there are still originally three possible outcomes, and only one of them has door 1 as the winning option. Having exposed the second door as having a "bad" outcome doesn't change the fact that there are still two doors with possible good outcomes, and that switching from the one you have chosen can have two thirds chance of picking the correct door.
Reply | Report Abuse | Link to thisI do not agree with the 2/3 probability of a particular player winning if he or she switches. A single player only plays the game once. And once Monty reveals the goat, it's a new game for that player, one with only two doors now (unless the player doesn't care about winning the car). The car is either behind the door the player picked or the other one, so for that player, the probability is 50/50.
Reply | Report Abuse | Link to thisDoctorja, think of it like this. The contestant intialy picked randomly from 1 of 3 doors, so the chance of getting it right must be 1 in 3. This can't change whatever happens subsequently. So where is the remaining 2 in 3 chance? Initially it resided in the other two doors, but a empty one has been chosen (not seected at random, but chosen by the host), and then shown to you. There is only one remaining unopened door, so the probability of it having the car must be 2 in 3 - the remaining probability. Another way of looking at it: If the car is not in the door you initially choose (which will be the case 2 in 3 times), and is in one of the other two, then the host has kindly eliminated the goat out of the two for you, so you don't have to pick between two, but have just one door, which must contain he car.
Reply | Report Abuse | Link to thisThe door problem is obviously correct and it dsnt matter if you do this or that or have a goat or whatever.
Reply | Report Abuse | Link to thisI'm not going to repeat everything people have already said, I will just say this one more time: It's correct. But I love the fact that some are questioning it, question everything!
"With 3 doors, the initial probabilities are 1/3 that door #1 is the winner, and 2/3 that all other doors are the winner. The 2/3 probability is initially distributed to 2 doors (1/3 each).
Reply | Report Abuse | Link to thisNow Monty unveils a door you didn't choose. The entire original 2/3 probability can now only be allocated to the one remaining door. Since 2/3>1/3 you should switch."
as you said, the 2/3 probability is initially distributed to 2 doors (1/3 each).
after Monty unveils the door2, the 1/3 probability of door 2 should be distributed evenly to the other two doors, instead of reallocate the 2/3 all to the last door.
Afraid I have to agree with San. Here's another way of looking at it: Suppose you had a die with the amazing property that it lost a face each time you rolled it. In other words, after the first foll the hexahedron becomes a five-sided solid (imagine that its possible). After the second roll, it becomes a tetrahedron, etc. Again, imagine that it's possible and forget about changing dimensionality. Each time you lose a face, the number on that face is correspondingly removed from play.
Reply | Report Abuse | Link to thisNow, the "Let's Make A Deal" problem is analagous to starting with a three-sided die. Each face represents one of the three possibilities outlined in the article. After the first "roll", the side with the number 2 is removed, leaving a one dimensional die (a line segment, or rod) - each end representing possiblities 1 and 3. Now, remember we chose the number 1 to start with. The crucial (and I believe completely analagous) question is: "When you roll your 2-sided die, do your chances of picking the right face (or "end" in this case) increase by switching your choice from 1 to 3?" I say "no".
Note this is really the same argument that Stan is making. I agree with him. Someone has to convince me that the probability of choosing correctly doesn't change in this way. One commentor mentions the additional requirement that Monty can't change the prizes behind the doors as the game progresses. I claim that's irrelevant . From the contestant's point of view, an unknown is still an unknown with probabilties determined only by the number of possiblities, which change as doors open, but must always sum to 1 (as Stan points out). The notion that your probability goes up to 2/3 by switching after door number 2 is opened is analogous to suggesting that switching your die pick from 1 to 3 in my example increases your chances because the die already landed on face 2. The crucial notion here is that probablilty number 2 is now "out of play" and the remaining possiblities must still sum to 1. Its interesting to note that the author concedes this in his roulette wheel example, but the logic completely changes for the "Let's Make a Deal" example.
Actually, here's perhaps a better (although again completely analagous) way of looking at the "Let's Make A Deal" dilemma. The author points out the three possibilities (good,bad,bad), (bad,good,bad), and (bad,bad,good). Imagine that these are arranged in a matrix with each possibility representing a row. Now imagine a mask or barrier of some kind that obstructs an entire row and has three little doors arranged from left-to-right such that, when one is opened, the element in the currently blocked row is revealed (either a "good" or "bad"). This mask may moved up and down the matrix, but we do not know what row its on. The only way we can know which row we're on is by opening two doors of the mask. So the host moves the mask up and down randomly, finally settling on a particular row. He then asks us to guess which row (or which door contains the "good" element, which is analagous). Let's assume we pick row 1.
Reply | Report Abuse | Link to thisNow the host opens the middle door (door number 2) of the mask to reveal a "bad" element, and asks us if we want to change our guess based on this new information. The author of this article argues that we are better off by changing our guess. In other words, your odds increase from 1/3 (the original guess) to 2/3 simply by knowing that we are NOT on row 2 (or equivalently, door number 2 is not "good"). I claim, to the contrary, that your odds at this point are 50/50, regardless of whether you change an original guess from 1 to 3 or 3 to 1. You are either on row 1 or row 3. Period.
The key to the correct solution to this problem (2/3 not 1/2) lies in the statement that the host "knows what is behind all three doors." Implicit in this is that he does not choose randomly among the remaining doors but must choose a door with a goat. (A couple of others have noted this.) Where "folk numeracy" goes wrong is in assuming that the host chooses randomly among the remaining doors. Note that in actual instances the host gives the impression he is picking randomly, and the contestant does not really have the opportunity to question this. (If the host really picked randomly, he would find a car one third of the time and say "Sorry, you lose.") In probability theory the first thing you have to do is define a sample space, i.e., explicitly state the conditions of the problem. This problem is a rather suble application of conditional probability that should have been stated and explained more clearly. The Wikipedia article referred to by michael_teter is good, but a shorter explanation could make the point.
Reply | Report Abuse | Link to thisI would also like to comment on the casino owners "delight." If the roulette wheel is honest the actual outcome is independent of any theory the gambler might have (with a fixed or neutral take by the house). The only reason the owners have to be delighted is if gamblers with fallacious theories tend to keep playing longer than gamblers without such theories.
Reply | Report Abuse | Link to thisIf that's the case why does all the extra probability adds to the door that you haven't chosen. Common sense tells me after he shows you the door(s) with the goat the probability should be added to both remaining doors in an equal amount 0.5/3 each (in the first case), so the probability will still remain 50-50 no mater how many doors he show you.
Reply | Report Abuse | Link to thisI don't know what DWRoss means by "remaining" doors. The only opportunity for the host to misdirect the player is when he opens the first door. So let's say the host knows there's a shiny new car behind door 1, and that happens to be the door the player picked. The player doesn't yet know he's right. At this point, the host opens door two to reveal the goat and asks if the player wants to change his pick. He didn't open door 1 because then the player wins immediately (and they have to go to commercial).
Reply | Report Abuse | Link to thisI claim that at this stage, the player is in exactly the same position as if he had chosen door 3. Whether he changes his choice at this point, his odds are the same (unless the host replaces the car with a goat, in which case we have a completely different situation).
Another point about DWRoss's first comment: I don't believe the question of whether the host is choosing randomly has anything to do with the player's chances (that's why I offered my second example, where all three possibilities are laid out. The host is simply asking you to guess which of the three possibilities he is concealing). If we trust that the host isn't free to shift goats and cars around behind the scenes, he is very limited in what he can do to misdirect the player. In particular, the worst he can do is not open the door the player picked the first time. At that point, the players's chances are 50/50, regardless. Of course, the next thing he could do is ask tauntingly "are you sure....?" - after he's opened the door with a goat, simply to falsely hint that maybe the player's chosen the wrong door and he's giving him a chance to change his choice. Note that this is immaterial, and doesn't change the player's odds.
Reply | Report Abuse | Link to thisOk, never mind. I finally get it. It evolves due to a combination of two facts:
Reply | Report Abuse | Link to this1. The host must always open a door with a goat
2. The host never opens the player's door first, regardless of his choice
This means that the probability of the car being behind door 1 + the player selecting door 1 + the host opening door 2 is different than the probability of the car being behind door 1 + the player selecting door 2 + the host opening door 3. I finally got it once I saw the table provided by KFS.
I feel silly.
Alright, here's a stupid simple explanation, done with brute force:
Reply | Report Abuse | Link to thisConfigurations:
good, bad, bad
bad, good, bad
bad, bad, good
Do this next part with me so you can see that I'm right.
If you pick the first door, a goat is revealed, then you switch:
Lose, Win, Win
If you pick the second:
Win, Lose, Win
If you pick the third:
Win, Win, Lose
You see, you have a 1/3 chance of picking the car and a 2/3 chance of picking a goat on your first try. Notice a pattern in the wins and losses? You only lose by switching if you picked the car first, which we know you have a 1/3 chance of doing. This means that by switching, you have a 2/3 chance of winning. You should switch every time.
I have only a moment, but consider this thought in favor of Shermer's example: What if you take the "House" position as a test to see whether the house has better odds than the "Player". Since the Player is selecting only one card, this means the house has a 2/3 chance of winning on every initially dealt "hand". If the house subsequently reveals one card on every round played as being a non-winner, who will then (at that point in the round) have the better odds of winning on every hand? If you could switch hands with the dealer at that point, would you? Remember that the House's hand always starts out with a 2/3 chance of winning versus the player's hand at 1/3. Just a thought.
Reply | Report Abuse | Link to thisI have only a moment, but consider this thought in favor of Shermer's example: What if you take the "House" position as a test to see whether the house has better odds than the "Player". Since the Player is selecting only one card, this means the house has a 2/3 chance of winning on every initially dealt "hand". If the house subsequently reveals one card on every round played as being a non-winner, who will then (at that point in the round) have the better odds of winning on every hand? If you could switch hands with the dealer at that point, would you? Remember that the House's hand always starts out with a 2/3 chance of winning versus the player's hand at 1/3. Just a thought.
Reply | Report Abuse | Link to thisHere's another way to look at it:
Reply | Report Abuse | Link to thisInstead of telling Monty which door you choose, you tell him it's either 1 or 3, but definitely not 2. You have a 1/3 chance of being wrong about 2, but Monty reveals the goat behind two so you're good to go.
Now you have to make a choice between 1 and 3. Since you didn't pick either before, you have a 50/50 chance with either door you pick now. Each door has exactly the same odds of being correct under this scenario.
Now, this scenario as stated is in not significantly different from the original scenario. In both, you start with three doors and you make a decision, which leaves you with two doors to chose from. When there are three doors, any one of them has a 1/3 chance of being the correct door. It doesn't matter how one of the choices is eliminated, because now that it is there are two choices, either of which has 1/2 chance of being the correct one. You still have to chose between the two that are left, so regardless of what the percentages or odds were originally, you are left with a 50/50 decision to make.
Increasing the number to 10 doesn't matter, either. Without knowing which door is correct, eliminating 8 choices doesn't magically assign the remaining odds to the one you didn't chose but not the one you did. There's no logic involved in that.
Okay, one last scenario to try to make my point. You're in the original Monty problem, you pick door a door, Monty shows you door 2 and it's a goat. If the door I didn't pick now has the 2/3 chance of being correct and the one I did pick has only 1/3, I change my pick. Now Monty asks me if I'm sure, and do I want to change my mind again. What are now the odds that my original door is right? 1/3 or 1/2? Why not 2/3?
There's no getting around it: once one choice is eliminated, the odds for all of the remaining choices get reshuffled, not just the ones you didn't chose.
If probability means "if I play this game 1000 times, how often will door #1 win", probability also means "if I play this game 1000 times how often will door #3 win". If "this game" means that we know door #2 has a goat every time, how can the answer to either question be less than 500?
Reply | Report Abuse | Link to thisAhh. I am no mathematician but I see SoccerPhD's point. The flipping coin system has no 'memory'. The Monty Hall problem does have memory, in the form of the prize which can stay in one of three locations. As long as it is not relocated, the memory stays and the probability of the single door remains 2/3. How mind boggling !
Reply | Report Abuse | Link to thisKeep in mind that the phrasing of the Monty Hall problem is crucial to its outcome (see en.wikipedia.org/wiki/Monty_Hall_problem). The assumption here is that the Host knows where the car and two goats are and always opens one of the two doors containing a goat. The hosts knowledge is directly influencing the odds. If the host did not know where the car and the goats were and randomly opened a door revealing a goat or the car then the contestants odds would be 50/50 on the switch instead of 2/3s.
Reply | Report Abuse | Link to thisNow suppose there are 2 players A & B; A chooses door 1, B chooses door 3; Monty reveals that door 2 contains a goat. Should both players then switch their choice ?!
Reply | Report Abuse | Link to thisA simple thought experiment will make the answer more clear. Suppose there are a billion doors with goats except 1 with a car. You select one of the doors and have a 1 in a billion chance of winning. The host reveals 999,999,998 of the 999,999,999 doors with goats behind them. The only doors that remain are the one you selected initially and one of the 999,999,999 you did not select.
Reply | Report Abuse | Link to thisThe probability that you selected the car on the first try is 1 in a billion. The probability that the car is behind the other remaining closed door is 999,999,999 out of 1,000,000,000. In 999,999,999 out of a billion possible scenarios, the remaining door that you did not pick has the car. With your first pick, you had worse than lottery chances of winning. By switching, you are virtually gauranteed a new car. By revealing 999,999,998 doors with goats, the host narrows down 999,999,999 possible winning doors to only 1 door.
Try it with a 52 card deck. Lay out all of the cards face down and place an ace of spades anywhere. Ask one of the PHD's who doesn't understand the solution to pick a card that they believe is the Ace of Spades. Then, after they pick, if they selected the Ace of Spades on the first try (1 in 52), randomly select another card to remain unturned and turn over 50 of the 51 unselected cards to reveal that they are not the Ace of Spades. If the PHD does not pick the Ace of Spades (51 in 52), turn over every unselected card except the Ace of Spades. In either case ask the increduluos PHD if he/she wants to switch his selection. If he/she does not switch, first pray that he/she is not a tenured professor, and then start offering them betting odds of 10 to one 1 in their favor. Eventually, they'll catch on, and even though they've just been ripped off, they'll thank you!
Suppose that after Door 2 is shown to be a goat, another participant is brought into the game. The information available to the new participant is that there are two doors, only one of which has the car. For the new participant, the probability is 50/50.
Reply | Report Abuse | Link to thisThe author and some participants are claiming that the probability of success is different, depending on the history of the particular participant.
Which by definition can't be true.
Ralph
There is another way of convincing oneself of that I thought of:
Reply | Report Abuse | Link to thisImagine there are three contestants, you and two others, each having a different door assigned to him at random. So at the beginning, each of the contestants, including you, has a 1/3 chance of having been assigned the winning door. Simple so far. Now, by revealing one of the losing doors, Monty is eliminating one of the other contestants (but always keeping you around). Since each of your 2 opponents faces a 50% chance of being eliminated by Monty, each is expected to miss 50% of the games played.
Now imagine also that you are all familiar with this door switching strategy and you all use it all the time. That means your opponent will always choose the door that you are abandoning by pursuing exactly the same strategy as you are. And each of you believes he's got a 66% chance of winning by switching.
So you may ask: how can you both be facing the same odds of success (66%) that add up to more than 100% by making a choice which is exactly the opposite of one another? At first glance this looks like a paradox that makes some feel they'be found a proof why vos Savant's claim would be wrong.
But what you need to remember is that each of the other contestants stood only a 50% chance of even making this far in the game and not being eliminated by Monty in the first place. So while the particular opponent you are now facing may (rightfully) think he's got a 66% chance of beating you by taking the door you're abandoning, it's 66% of the times he gets to play the game, not 66% of the times YOU play the game.
You're facing this particular opponent only 50% of the time. So from your point of view, the odds of this particular opponent winning by switching to your door when this game is played many times are really 50% times 2/3, which is obviously 1/3. But that means that if this particular opponent's chances of winning by taking your door
are 1/3 (of the times YOU play the game), then it must be the case that your chances of winning by taking your opponent's door (whichever of the two it may be) are 2/3.
(This line of reasoning extends elegantly into the 10 door scenario, I'll spare you from the math, but it's simple.)
It is interesting to note that from your opponent's point of view, he's still got a 66% chance of beating you whenever he is in the game, it's just that you make up for it by being in the game 2 times more often than he is (or the other contestant). So even if from his point
of view, your chance of beating him should be 33% whenever he plays, your chances of beating either of these other 2 contestants winds up adding up to 66%.
In fact, laws of probability (if not logic) would be clearly violated if, after Monty eliminates one of the opponents,
you stood a 50% chance of winning by not switching. That's because it'd mean that your remaining opponent, whom you would see only 50% of the time, would also have to stand a 50% chance of beating you by choosing your door. But that means his probability of beating you whenever he's playing against you (i.e. 50% of the times the game is played that he is not eliminated by Monty) would have to be 100% over time. That obviously makes no sense, because there is nothing guaranteeing that once one of your opponents stays in the game, he's 100% guaranteed to win.
It may seem convoluted, but this is the line of reasoning finally convinced me.
Here is another way of looking at it: Suppose I have already decided to switch, i.e. I choose my first door on the assumption that it does NOT lead to the car. Then if my assumption was right (2/3 of the time) the host is forced to reveal the car by opening the other non-winning door and I alway win. If the assumption was wrong (1/3 of the time) I always loose.
Reply | Report Abuse | Link to thisThink about the information conveyed by Monte. Since he always, in the one person game, has a door with a goat to open after the pick, there is no new information with which to change your initial probabilty of 1/3. There remains a 2/3 chance of being wrong, but Monte has now shown which door has that 2/3 probability. Switch! Same logic as the 999,999,999/100 billion probability in the billion door game.
Reply | Report Abuse | Link to thisIf there are 3 doors and Monte always has to show you one goat, then your chances of winning are 1/2 going into the game, not 1/3. If there are a billion doors and Monte must always show you 999,999,998 goats, then your chances of winning the game are 1/2 and not 1/billion. Switching doors will not improve your initial chances of 1/2 no matter how many doors there are.
Reply | Report Abuse | Link to thisPeople are confusing the chances of winning the game with the chances of picking the correct door initially. They are not the same.
If there are 3 doors and Monte must always show you a goat, then your chances of WINNING are 1/2 going in, not 1/3. If there are a billion doors and Monte must always show you 999,999,998 goats, then your chances of WINNING are 1/2 going in and not 1/billion. Switching doors will not increase your initial odds of winning no matter how many doors there are. What has been done here is data on the chances of initially picking the right door has been used to argue chances of WINNING THE GAME. They are two seperate and unrelated probabilities given the way the game is played. I like Shermer, but he is wrong here.
Reply | Report Abuse | Link to thisImagine that you choose #1 (doesn't matter does it 1/3 of a chance the car is there, two thirds it isn't just like #2 & #3)
Reply | Report Abuse | Link to this#1 #2 #3
O X X car at one - two opens: ta-da: goat: you switch:you lose
X O X car is at two - three opens: goat: you switch: you win
X X O car is at three - two opens: goat: you switch: you win
Oops - hadn't noticed the four pages of new comments – sorry about that
Reply | Report Abuse | Link to thisI'd really rather win a goat though, so i'll stay at one
Reply | Report Abuse | Link to thisThe main problem with probability theory is that the logic is completely maddening.
Reply | Report Abuse | Link to thisThis is correct.
Reply | Report Abuse | Link to thisHowever, if monty is allowed to force the contestant to switch (by opening door one if a goat is under it) then the odds are always 50/50.
This is correct. However, if monty is allowed to open door one (assuming door one holds a goat) and thus FORCE the contestant to switch, the odds return to 50/50 for each switch.
Reply | Report Abuse | Link to thisWhen you attach probability to intuition, it makes a whacky combination. Probability at one hand is itself uncertainty in future while intuition is somewhat of the other nature. So, when you choose to choose the door you're right but when you start calculating probability you're chances of getting it wrong increases.
Reply | Report Abuse | Link to thisThe main problem with probability theory is that the logic is completely maddening. The main problem with probability theory is that you cannot
Reply | Report Abuse | Link to thishave your cake and eat it too. Prove me wrong (please!). Everything hangs on whether hypothetical choices are counted as real choices (they are like
apples and oranges). Suppose you switch again after the second (2/3)
choice. Assume an observer just arriving. From their point of view, the
process is just starting (ie. they are not aware of the first choice.
) They can equally assume that switching (actually for the second time ,
but to them for the first time) wil produce a 2/3 certainty. So, how do 2/3 + 2/3 = 1? Either the observer and the chooser come from two separate and
parallel universes or, there is a glitch in our formulation of logic. Furthermore, reasonable and logical thinking is a largely inadequate tool
for investigating the universe. Look how long it took to overturn the
assumption that the sun revolves around the earth. I suspect we are just beginning to understand the universe.
It's all in how you define the problem. Let's carefully define the possibilities:
Reply | Report Abuse | Link to this1. You choose the prize door, Monty randomly opens one of the other two.
2. You DON'T choose the prize door, Monty is forced to open the only other door without a prize.
People who believe in the switch argument look at the possible combinations of which door you chose and where the prize is. So they come up with the following combinations where the first number is the location of the prize and the second number is the door you chose originally: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3). In this analysis, you do win 2/3 of the time.
Unfortunately, this is a flawed analysis because it only recognizes one option when you have chosen the prize door to begin. In fact, there are two possibilities: if the prize is behind door 1, and you chose door 1, Monty can open either door 2 or door 3.
So the full set of combinations includes where the prize is, what door your choose, and what door Monty opens. Here are the 4 combinations if the prize is behind door 1: (1,1,2), (1,1,3), (1,2,3), (1,3,2). Switching loses in the first 2, and wins in the second 2. 50/50.
Rather than running a few trials with a friend where there is no assurance of randomness, I created a spread sheet with a random assigned prize door, a randomly chosen door, and Monty randomly choosing one of the non-prize doors when the prize and the guess match.
The result ... 50% for switching.
Here's another way to verify that the 2/3 idea doesn't work. Let's put the prize behind door 1 all the time. Let's choose door 1 all the time. Switching loses 100% of the time. Now leave the prize behind door 1 and choose door 2 all the time. Switching wins 100% of the time. This illustrates the potential problem with non-random experimentation of the kind you get with a neighbor.
iggles14: the couting argument is only valid if the different outcomes have the same probability. (1,1) is just as likely as (1,2) or (1,3), each being 1/3.
Reply | Report Abuse | Link to thisBut (using '=' to denote equal probabilities) (1,1) = (1,1,2) + (1,1,3) (Monty has a choice: 2 or 3), while (1,2) = (1,2,3) (Month has no choice). So (1,1,2) and (1,1,3) together have the same probability as (1,2,3), i.e. 1/6 each.
In your final example your forgot to include 'choose door 3 all the time'.
All this is beginning to bring me to one thought only: how small are angels,
Reply | Report Abuse | Link to thisreally?
I thought of something that may support SoccerPHD's argument. Monty has inserted knowledge into the system that was not available before, and more than simply the information that there was a goat behind door 2. For you to have the option of a switch, he had to choose a goat door instead of a good door.
Reply | Report Abuse | Link to thisHis choice was not random, but informed, and constrained, and changes your probabilities. Even if Monty had guessed it, like another gambler, his choosing the good door would have ended the game. So the knowledge that skews the probability is not just the goat behind door 2, but the fact that you are still in the game.
I thought of something that may support SoccerPHD's argument. Monty has inserted knowledge into the system that was not available before, and more than simply the information that there was a goat behind door 2. For you to have the option of a switch, he had to choose a goat door instead of a good door.
Reply | Report Abuse | Link to thisHis choice was not random, but informed, and constrained, and changes your probabilities. Even if Monty had guessed it, like another gambler, his choosing the good door would have ended the game. So the knowledge that skews the probability is not just the goat behind door 2, but the fact that you are still in the game.
I thought of something that may support SoccerPHD's argument. Monty has inserted knowledge into the system that was not available before, and more than simply the information that there was a goat behind door 2. For you to have the option of a switch, he had to choose a goat door instead of a good door.
Reply | Report Abuse | Link to thisHis choice was not random, but informed, and constrained, and changes your probabilities. Even if Monty had guessed it, like another gambler, his choosing the good door would have ended the game. So the knowledge that skews the probability is not just the goat behind door 2, but the fact that you are still in the game.
Originally I was skeptical but now I have to admit it...SoccerPhD is dead right. It is obvious when you do it with ten doors, (or peas under cups in my case).
Reply | Report Abuse | Link to thisThe error lies in thinking of the final state as an isolated 1/2 chance. It's not. The chance that you picked the correct door originally is 1/3 and that probability stays with that choice. The chance that the prize is behind the other doors was 2/3 originally and when one of those doors is shown to be the goat, the remaining door takes on the 2/3 probability. Therefore, you should always switch to the other door.
Don't take my word for it though, try it! The ten door version makes it obvious. You'll soon see that switching gives you a much greater hit rate.
Nathaniel is also perfectly correct. Very clear explanation too!
Reply | Report Abuse | Link to thisSuppose there was another contestant who originally chose door number 3. The same argument would prove the second contestant wins two thirds of the time by switching to door number 1. Adding the two probabilities we get 4/3, meaning that 1 time in 3 the prize is in two places at once: Schroedinger's door?
Reply | Report Abuse | Link to thisBecause there is only 1 good prize out of 3 doors, there will always be at least one bad prize left behind the other two doors.
Reply | Report Abuse | Link to thisIf you chose the winning door, the act of revealing a goat behind another door has no affect. Odds are still 1/3 that you have the winning door if you don't switch.
If you chose a losing door, then revealing a bad door gives 100% odds that the final door is the winner. The extra step is not a random event because the host has prior knowledge. This scenario will happen 2/3rds of the time because your original chance of picking a losing door was 2 out of 3.
Summary: by not switching, your odds are 1/3. By switching, your odds are 2/3.
The "Game show paradox" was written about by Marilyn Vos Savant in her book "The Power of Logical Thinking". The answer to switch is absolutely correct. She also gives an example with a larger number of doors. If you picked door #1 and the host opened every other door to show you a goat, except door 777,777. You'd switch pretty quickly wouldn't you?
Reply | Report Abuse | Link to thisOk, I'm not PhD, but this is what doesn't make sense to me.
Reply | Report Abuse | Link to thisIf I pick door #1 and then I am shown that door #2 contains a goat, then, according to some, the chances of the goat being behind door #3 increase from 1/3 to 2/3. What if I originally chose #3 instead of #1? Would the chances of it being behind door #1 then increase to 2/3?
If the chance of winning by picking door #1 is 2/3, and the chance of winning by picking door #3 is 2/3, then the chance of winning by picking either #1 or #3 becomes 4/3.
How can that be?
Oops, I misspoke above. The first sentence of the second paragraph should read "If I pick door #1 and then I am shown that door #2 contains a goat, then, according to some, the chances of the PRIZE being behind door #3 increase from 1/3 to 2/3."
Reply | Report Abuse | Link to thisDang it, no Edit function here.
My entire post, corrected, is below.
Ok, I'm not PhD, but this is what doesn't make sense to me.
If I pick door #1 and then I am shown that door #2 contains a goat, then, according to some, the chance of the prize being behind door #3 increases from 1/3 to 2/3. What if I originally chose #3 instead of #1? Would the chance of it being behind door #1 then increase to 2/3?
If the chance of winning by picking door #1 is 2/3, and the chance of winning by picking door #3 is 2/3, then the chance of winning by picking either #1 or #3 becomes 4/3.
How can that be?
I understand the statistical point. Initially there is a 1/3 chance of choosing the correct door, so there is a 2/3 chance that the car is behind one of the other two doors. Thus if no doors are revealed after one is chosen, then for a single player in a single game, there is a 2/3 chance of winning by switching. And even if the goat is revealed behind one of the other doors, it might well be that over many repetitions of the game, the probability of winning by switching would be 2/3. But on Let's Make a Deal, you don't get to play the game over and over, so what really matters is the probability of winning by switching for a single player playing a single game. Consider this: Suppose the game were played from the start with only two doors, with the car behind one of them. The player chooses one of them, and then has the opportunity to switch. First, please tell me what the probability of winning by switching is in this case. Then please explain to me how, for a single player in a single game, this is any different from the situation with three doors after the goat is revealed behind one of the doors not chosen.
Reply | Report Abuse | Link to thisYou can fool all of the people some of the time, and all the people some of the time, but you can fool all of the people all of the time,
Reply | Report Abuse | Link to thisYou are give two times to select therefor two "games". The first you selected door # 1 of three doors. The conclusion of that choice (game) was not determined.
In the new game you are asked to select again. This time from two choices. Each has an equal chance.
You can fool all of the people some of the time, and some the people all of the time, but you can not fool all of the people all of the time,
Reply | Report Abuse | Link to thisYou are give two times to select therefor two "games". The first you selected door # 1 of three doors. The conclusion of that choice (game) was not determined.
In the new game you are asked to select again. This time from two choices. Each has an equal chance
Your original selection has no bearing! The host has eliminated one of the choices and in this case one of the goats.
Reply | Report Abuse | Link to thisThe new game becomes 1 car, 1 goat, two doors.
The misinforamtion in the article explantaion is this
Reply | Report Abuse | Link to this" You had a one in three chance to start, but now that Monty has shown you one of the losing doors, you have a two-thirds chance of winning by switching. Here is why. There are three possible three-doors configurations: (1) good, bad, bad; (2) BAD, GOOD, BAD; (3) bad, bad, good. In (1) you lose by switching, but in (2) and (3) you can win by switching."
You can't win by switching to configuration # 2 because it has been eliminated by showing the goat behind door # 2
It does not matter how paradoxical it is but Michael Shermer is correct. The open door did not add any knowledge. We knew from very beginning that one of two other doors would reveal a goat.
Reply | Report Abuse | Link to thisOf course we know that one of the other doors will reveal a goat. But when Monty reveals the goat, we now know which one, so we have a 50/50 chance of winning by switching to the door he did *not* open.
Reply | Report Abuse | Link to thisI think the confusion is between the frequency of winning by switching when measured over the entire history of all the games like this ever played, which might well be 2/3, and the probability of winning by switching for a single player playing a single game. I agree with adamsmith36: when the goat is revealed it becomes in effect a new game, one with only two doors. But maybe adamsmith36 and I are wrong. Please see my comment from 1:14 PM on 10/05/08 and explain to me how, once the goat is revealed for a single player playing a single game, the Monty Hall Problem is any different from a game starting with only two doors, one with a goat behind it and one with a car.
imagine it this way.
Reply | Report Abuse | Link to thisTwo people play the game where 1 person gets to chose 1 door and the other gets the other 2 doors. You'd always want to be the person getting two doors even if you don't get to choose them yourself, correct? Even if one of your choices was shown to be bad, it doesn't matter because by getting to chose 2 doors to start, you started with a 2/3 probability of winning. When the host asks player 1 to switch, he's really asking if player 1 wants to switch starting positions.
I'm looking at it this way: 10 doors. The car is behind door #10. Our
Reply | Report Abuse | Link to thiscontestant chooses door 1. The contestent doesn't know where the car is,
so the probability situation is:
Car is behind door 1: 10%
Car is behind door 2-10: 90%
Sum of probability = 100%
Monty shows that doors 2-9 have goats. At no time is the car moved.
The new situation is:
Car is behind door 1: 10%
Car is behind door 2-10: 90%
Sum of probability = 100%
Before doors 2-9 were opened, there was a 10% chance the car is
behind door 1.
After doors 2-9 were opened, the is STILL a 10% chance the car is behind
door 1.
(Only if the car is moved randomly between doors 1 and 10 after 2-9 are
opened, can we say the car has a 50/50 chance of being behind 1 or 10.)
Now, with 2-9 open, there is a 90% chance the car is behind door 10.
The 10%/90% partition of probability stays. The number of places the car can hide in doors 2-10 is reduced to a single door, 10.
So, there is a 10% chance the car is behind door 1 and a 90% chance it's
behind door 10. (90% chance of being behind doors 2-10 with 2-9 now known to contain goats.)
The problem for me was realizing the probabilities aren't redistributed:
Car is behind door 1: 10%
Car is behind door 2-10: 90%
Sum of probability = 100%
Does NOT become:
Car is behind door 1: 50%
Car is behind door 10: 50%
... after doors 2-8 are shown to contain goats. In this (false) scenario, the
probability the car is behind door 1 grows by 40%. There's no justification
for this.
Put another way, the car has a 10% chance of being behind door #1 and
a 90% chance of NOT being behind door #1.
Therefore, from this analysis I'd say it's best to switch.
David.
I'm looking at it this way: 10 doors. The car is behind door #10. Our
Reply | Report Abuse | Link to thiscontestant chooses door 1. The contestent doesn't know where the car is,
so the probability situation is:
Car is behind door 1: 10%
Car is behind door 2-10: 90%
Sum of probability = 100%
Monty shows that doors 2-9 have goats. At no time is the car moved.
The new situation is:
Car is behind door 1: 10%
Car is behind door 2-10: 90%
Sum of probability = 100%
Before doors 2-9 were opened, there was a 10% chance the car is
behind door 1.
After doors 2-9 were opened, the is STILL a 10% chance the car is behind
door 1.
(Only if the car is moved randomly between doors 1 and 10 after 2-9 are
opened, can we say the car has a 50/50 chance of being behind 1 or 10.)
Now, with 2-9 open, there is a 90% chance the car is behind door 10.
The 10%/90% partition of probability stays. The number of places the car can hide in doors 2-10 is reduced to a single door, 10.
So, there is a 10% chance the car is behind door 1 and a 90% chance it's
behind door 10. (90% chance of being behind doors 2-10 with 2-9 now known to contain goats.)
The problem for me was realizing the probabilities aren't redistributed:
Car is behind door 1: 10%
Car is behind door 2-10: 90%
Sum of probability = 100%
Does NOT become:
Car is behind door 1: 50%
Car is behind door 10: 50%
... after doors 2-8 are shown to contain goats. In this (false) scenario, the
probability the car is behind door 1 grows by 40%. There's no justification
for this.
Put another way, the car has a 10% chance of being behind door #1 and
a 90% chance of NOT being behind door #1.
Therefore, from this analysis I'd say it's best to switch.
David.
Please excuse my double post.
Reply | Report Abuse | Link to thisDavid.
I also stand in support of "folk numeracy". Let's reverse Soccer PhD's argument. Since initially all three doors have 1/3 probability, then it is just as reasonable to assert that there is 2/3 probability that the prize lies in the first two doors as it is to assert that there is a 2/3 probability that it lies in the second two. Following his line of reasoning, one could just as reasonably claim (erroneously), that the revealing of the second door leaves the first door with a 2/3 probability.
Reply | Report Abuse | Link to thistest
Reply | Report Abuse | Link to thisAt first I thought this article was wrong, but I was looking at the "picks" as independent events instead of dependent events, I was ignoring effect of Monty's "pick". If Monty chose the goat door at random, your odds on the second pick would be 50/50. But Monty knows, and he knowledge changes the odds.
Reply | Report Abuse | Link to thisSoccerPhD's argument becomes the intutive one as you increase the number of doors, as he mentions to 10 doors. Let's use 1000 doors to make it even more obvious. You pick door #1. You knows your chances are 1 in 1000. If you have choose the car (1 chance in 1000) Monty then can pick any 998 of the remaining 999 doors, all have goats. If you have choosen a goat (999 chances in 1000) Monty must then pick the remaining 998 doors with goats, leaving the one with the car. Under this scenario, would anyone think that he has a 50-50 chance by staying with door #1? It is now intuitive to switch to the other door. Monty has improved your odds if you switch, because his pick is not random.
I address this comment to Messrs. King, Millet and Hilliard. The problem curiously causes very strong emotional responses by people who have a opinion one way or another. Most people fall into the trap that psychologists call a "cognitive tunnel". The only way you guys are going to prove what is reality is to try the experiment over a series of runs, 27 runs should be sufficient.
Reply | Report Abuse | Link to thisI had to do this myself in a bar in Sydney with a brilliant professor of electronic engineering who still didn't accept it. We used three beer mats and marked one with a cross. I took the role of Monty Hall and dealt the beer mats ace down, a girl kept the score. The professor consistently changed his mind after one mat was revealed. At 27 runs the professor became so exasperated with how the results were coming out that he tried to cheat by covertly marking the beer mat by bending it, at which point the girl invigilator, herself a genius who misses nothing, reprimanded the professor and the game stopped. However the professor tried to rationalise that 27 runs were "not statistically significant", and remains unconvinced.
Nevertheless you are in good company, one of the worlds most eminent mathematicians, Paul Erdos, also never understood it. In his biography "The Man Who Loved only Numbers", it relates how the problem was explained to Paul Erdös by a mathematician friend, Vászonyi. But to his surprise Erdös didn't get it! Vászonyi was then sorry he brought up the problem because it was also his experience with his own students that people get excited and emotional about the answer, and Erdös
was naturally temperamental anyway. The most successful approach Vászonyi tried to convince Erdös was by performing the problem with cards to show that experimentally the odds were not 50:50 but 1:2, just as I did with the professor using beer mats. This showed Erdös the truth of the answer but he still never appeared to be able to understand why, he remained in the 'cognitive tunnel'. In Erdös's defence this was an area of mathematics outside his own field.
I explained the problem to another engineer who had to write a program in Matlab to experimentally test the results to prove it to himself. After the he had confirmed the results he said he enjoyed making a nuisance of himself on campus proving the results to anyone he could capture to listen, he won a few beers out of it.
The human brain is wired wrongly to understand probability, I suffered for a while during trying to understand this problem. Have fun with the experiment.
Regards,
Julian Russel
The problem cannot be solved experimentally. As with mathematical problems in general, it can only be solved by rigorous reasoning.
Reply | Report Abuse | Link to thisThe Monty Hall puzzle is an elementary one IF one understands that the host opens a goat door intentionally rather than by chance, and IF one understands how to reason correctly and rigorously about probability – a very big if. It’s not so much intuition as insufficiently rigorous (OK, sloppy) reasoning that leads otherwise intelligent and mathematically educated people astray with this problem. Those who claim that it doesn’t matter if the contestant switches simply lack an adequate understanding of how to solve probability problems of this nature and overestimate the understanding they possess; cases in point can be found in previous posts in this thread. This problem would be an excellent one to use for teaching purposes in Probability/Statistics 101 courses, and probably (!) has been used in this way.
Wikipedia has a good article on the Monty Hall problem that thoroughly explains the correct solution.
Here's a similar problem: What is the probability that a two-child family that has a boy also has a girl?
Reply | Report Abuse | Link to thisAnswer: 2/3. If you answered 1/2, the problem is again sloppy reasoning, not the perils of "intuition."
I agree with the article & SoccerPhD. Here's the trick, I believe. Each door has a 1/3 chance as agreed earlier. However, Monty Hall selects the one of the unchosen doors that is not a car (& reveals it). This is the key. He has more information & thus moves the 2/3 probability to one door by revealing a goat from one of those unchosen doors. Basically, he has gone into the deck of cards & removed a bad card that you could not see. It is confusing because someone with more information (Monty Hall) is manipulating the deck (doors in this case).
Reply | Report Abuse | Link to thisI am not willing to say for certain that Monty Hall is manipulating the deck. The problem statement does not tell us that.
Reply | Report Abuse | Link to thisWhat reason does one have for assuming that Monty would have opened door #3 if door #2 had contained the car. To make such an assumption would be to unnecessarily assume something to be true that we do not know.
The two different interpretations of the 3-door problem are: (1) Monty opens door #2, and (2) Monty opens a door containing a goat. That interpretation determines which probability is correct: p=1/2 or p=2/3. That interpretation is non-probabilistic. It comes from your belief about the nature of the game being played. This either/or situation makes the calculated probability a conditional probability -- conditional on your belief about Monty's behavior.
Prior to Monty opening the second door:
door1=car, door2=goat, door3=goat, p=1/3
door1=goat, door2=car, door3=goat, p=1/3
door1=goat, door2=goat, door3=car, p=1/3
Your initial choice is door #1. Before seeing behind door #2, there are three equally probable original configurations: p=1/3.
Case A)
Suppose that Monty's intention was to open door #2 regardless.
door1=car, door2=goat, door3=goat, #2=goat, switch, you win the goat.
door1=goat, door2=car, door3=goat, #2=car, this option is out because problem states 2=goat.
door1=goat, door2=goat, door3=car, #2=goat, switch, you win the car
In this case, there is nothing to gain by switching, p=1/2
Case B)
Suppose that Monty's intention was to pick/open a door containing a goat.
door1=car, door2=goat, door3=goat, switch, you win the goat
door1=goat, door2=car, door3=goat, switch, you win the car
door1=goat, door2=goat, door3=car, switch, you win the car
In this case, switching raises your probability of winning the car from p=1/2 to p=2/3.
Cases A and B both occur in various games. Think of the variations of blackjack, poker, etc. We are dealing with conditional probabilities and cannot know which one is correct because we don't know which rule Monty follows.
To Julian Russell: Perhaps that "cognitive tunnel" boils down to whether a person is inclined to read into the problem an assumption that is not contained in the problem statement, and the truth of which he/she does not know.
Stan Hilliard
I agree that Shermer’s statement of the problem is ambiguous and his explanation is confusing, making one wonder whether he really understands the problem. Shermer wrote:
Reply | Report Abuse | Link to this“Host Monty Hall, who knows what is behind all three doors, shows you that a goat is behind number two, then inquires: Would you like to keep the door you chose or switch?”
Shermer omits the crucial information that the host is *required*, by the conditions of the Monty Hall problem, to open a goat door among the unselected doors. Shermer merely states that the host knows where the prize is and shows you that a goat is behind #2 – he doesn’t specify *why* he does so. Without the specified constraint on the host’s actions, the probability of winning by switching is 1/2.
Consider this scenario: door #2 accidentally swings open, revealing a goat. Then the probability that the prize is behind door #3 is 1/2. Those who insist that switching doesn’t matter are really solving this problem.
Or consider the following variant of the ten-door problem: You choose door #1, and Monty opens doors 2 through 8 just because he won the lottery with 23-45-67-89, or because he always opens doors 2-8 no matter where the prize is, or because a TV viewer phoned in and suggested it, and lo and behold doors 2-8 all show goats. Then the probability that the prize is behind door #10 is 1/2. But if the host is required, by the conditions of the problem, to open 8 goat doors among the 9 unselected doors, then the probability that the prize is behind the your initial door choice is still 1/10 and you should switch to the remaining door.
Sorry, I meant to say "doors 2 through *9*" in the last paragraph of my previous post.
Reply | Report Abuse | Link to thisI am throughly amused that all responders (that I've seen) to Schermer's brief article are so myopic... completely blinded to the larger question. Is our Universe the result of completely random processes... which would essentially render scientific "truths" or laws moot... or based on causality and thus, Determined. Everthing I have learned as both a physician and in reading about neurobiology, the Natural world and the "observable" universe have convinced me that the latter (deterministic) is true. This particular universe has unfolded inexorably in timespace... and if there could possibly be some agency beyond our ken that "created" it (highly improbable to say the least), then once "created", there was no changing it.
Reply | Report Abuse | Link to thisRandomness clearly does not rule. IMO, those who continue to cling to it have ulterior motives: to save the bogus concept of contra-causal "free will"... and some ethereral supernatural "soul" agency somehow supervening on our evolved physical brains (mind-brain dualists). These folks once defended their positions on stale religious grounds... but the "intellectuals" among them are betting on QM to somehow justify what empirical science denies. However, even heavy-hitters in the QM world, like Gerard t'Hooft (Spinoza Institute) have written of determinism/causality operating at Planck scales.
There's "probability".... then again, there's "reality".
The former a game of "what if's", the latter a universe of "what is".
David King et al omit one factor; the intention of the knowing host.
Reply | Report Abuse | Link to thisDoes the host wants me to win? If so, opening a loosing door seems to suggest I about to loose. So I should switch and win.
But maybe he want me to loose? Knowing I am about to win he opens the loosing door to get me to switch..... mmm...
Or maybe he does not care and just want to prolong the show by giving me another decision to make? This is the scenario David and others have addressed. And as they have said, switching does not improve my chances.
For any skeptics of the skeptic, try writing out each of the 3 scenarios, as follows...
Reply | Report Abuse | Link to thisThere are 3 possible scenarios (X = win, O = lose). Assume door #1 is chosen, so separate it from the rest:
X:OO
O:XO
O:OX
Now remove one of the losing doors from either door #2 or #3. You are left with the following results for each of the initial scenarios:
X:O
O:X
O:X
For 1/3 of scenarios, the winning door will be #1, and switching will be the incorrect choice. But for the other 2/3, switching will result in a win.
Hope this helps some.
I disagree with desheffer and the skeptic. First of all let me say that in math, finding 100 examples that work doesn't prove something right; finding one example that doesn't work, proves it wrong. I will first respond to desheffer's example and all the other posts that reason along the same lines, and then I'll give the one example that doesn't work.
Reply | Report Abuse | Link to thisOriginally there are three doors and the possible configurations are as follows:
XOO
OXO
OOX
with X being the right and O being a wrong door. After the host reveals a goat, we have two scenarios:
a) The goat behind door 2 is revealed. This fact (emphasis on fact) makes the second original configuration (OXO) impossible. Therefore we are left with doors 1 and 3 and the following configurations
XGO
OGX
with G being the goat. Clearly the probabilities are 50/50.
b) The goat behind door 3 is revealed. This fact makes the third of the original configurations (OOX) impossible. Se we are left with doors 1 and 2 and the following possible configurations:
XOG
OXG
which is again 50/50.
Merging doors, scenarios and probabilities is sketchy math and that's why so many people get it wrong. Here is the example that I promised at the beginning:
Two people play the game over the phone and neither knows anything about the other player, or the other player's choices.
Player A chooses door 1, player B chooses door 2.
The host shows the goat behind door 3.
If the "change and your chances go up" philosophy is correct, who must change? both? what sense does that make?
In other words, which door has 1/3 chance now and which one has 2/3. The answer is neither. Both doors have a 50% chance to win.
Anthony
You are assuming that the host randomly chooses the door to reveal, but he doesn't he reveals the one that doesn't contain the car. Therefore, if you made the correct choice the first time you guessed (which I think no one will argue the probability being 1/3) then the host can choose to show either door 2 or 3. If you didn't choose correctly your first guess (probability being 2/3) then the host has only one choice of which door to show you since the remaining door has the car. Therefore, your chances that you choose correctly your first time do not go up because it is not random. As for your second example, you are changing the problem because you are taking away the choice of the host. He can only show you one door otherwise the contestants will know which door the prize is in. Randomness is the key to this problem, and the fact that its not is why there is a 2/3 chance that the contestant will win by switching.
Reply | Report Abuse | Link to thisWell I for one have changed my mind and now believe the sceptic (and Wikipedia) are right.
Reply | Report Abuse | Link to thisWhat finally convinced me was that I decided to set up a test with 300 games to played in parallel with the car behind door 1 in 100 of them, door 2 in 100 and door 3 in 100. Now the host opens door 2 or 3 in every game, no goat in every case. I have chosen door 1 in every game. In 100 cases I am right. In the other 200, the goat is behind the remaining closed door. So of course I switch.
Spin-oza, see also "chance" on Rational Thought.
Reply | Report Abuse | Link to thisKeep Your Eye On The Prize: designating a door or two is not the same as winning the car.
Reply | Report Abuse | Link to thisFor the GIVEN scenario may I suggest that:-
By switching from designated Door x to designate Door z the competitor has, over the course of the game, increased her chances of having designated A winning door from 1 in 3 to 2 in 3.
As for this competitor's chances of choosing THE winning door, these change from 3 in 9 in the first round to 2 in 4 in the second because of the information revealed by the opening of Door y (i.e. run Goat Car Goat is no longer available and there is now only one goat left in play - thus,Car Goat or Goat Car).
I saw the light!! I was wrong, and now I know why. If you switch, you gain indeed.
Reply | Report Abuse | Link to thisAfter the host has opened a goat door, the two remaining doors are NOT equal. The player has more information about the door she didn't choose. That is, if the player chose A and the host opened C, then B is "the other" door from a loosing door (C).
The argument that made me realize that something is wrong with the 50/50 view was the following. Assume that a huge number of players play the game independently and neither of them ever changes doors. What percentage of them do you expect to win if huge approaches infinity? The answer is 1/3. So what would have happened if they had all changed doors? Well, since there is no other case (you either change, or not) the probabilities must add up to 1, so it is 2/3.
An other way to formulate the problem might make it easier to see why the doors are not equal after the (informed, NOT random) opening performed by the host. Assume there are three pockets, and the game will throw a ball that can go into either pocket with 1/3 chance. The player is called to pick a pocket in advance. Say she picks A. The host then reveals a wrong pocket and the player is asked if she wants the remaining pocket instead of her original choice. Now here is the subtle point. The remaining pocket is the "other one" from a loosing one. This means that if the ball went into ANY of B or C, it will be in the one the player is offered to change to, since the other one was shown to be wrong. This is equivalent to connecting pockets B and C into pocket D and asking the player if she wants to drop A for D. In the connected pocket scenario it is easy to see that the probabilities are 1/3 vs 2/3 for A and D respectively. The leap of faith is to realize that when the host opens a wrong door (and NOT one between B and C in RANDOM), he is giving you the extra information that if ANY of B, or C was a winning door, it is going to be the door that is left.
Take this Colbert, wikipedia actually helped me get smarter :-)
"I saw the light!! I was wrong, and now I know why. If you switch, you gain indeed.
Reply | Report Abuse | Link to thisAfter the host has opened a goat door, the two remaining doors are NOT equal. The player has more information about the door she didn't choose. That is, if the player chose A and the host opened C, then B is "the other" door from a loosing door (C)."
I am still not getting it, although the preponderance of adherents to the "switch" argument makes me think I'm just dense. The issue I see in the argument as stated above is that you also gain (or not) the same amount of information about the door you chose, don't you?
Clear case of 50-50-90.
Reply | Report Abuse | Link to thisThe odds may be 50-50, but with my lick, 90% of the time I'll pick the wrong one!
No, you don't. That's what makes it not be 50/50. Consider this: You play the game and choose door A. Now, the host tells you that some other player will also play with you but because you chose first, the other player gets BOTH doors B and C and she reserves the right to throw away one of them, if it's wrong (i.e., if either B or C is right, she wins). At this point you know that you have 1/3 chance of winning and she has 2/3. Then the host looks what's behind HER doors, and throws away one of her bad doors (by revealing it) for her. This is the same with her "practicing her right" to throw away one of the two original doors.
Reply | Report Abuse | Link to thisThat's why the two remaining doors are not equal. Yours is still the one you chose at the beginning, when all you knew was 1/3,1/3,1/3. The door the other player has now is the one that is left from two original doors minus a wrong one.
What confused me and probably is confusing you too is the following thought: "What if someone just walks in the room after the host reveals a goat? There are only two doors, so they have to be 50/50" Indeed that new player would have to assume that because he doesn't have the extra information about the doors that you and that other player have. In my example with the pockets, a player looking at the two resulting pockets would thing that they have a 50/50 chance of winning. This is not correct though, because this conclusion does not use the information that one of the two pockets is the result of connecting two original (and equally probable) ones. Similarly, the door you are offered to change to, is a door that has "inherited" the car from EITHER one of the original two you didn't choose, since the door you are offered is the right one among the original two, if there was a right one among the original two (which will be the case 2/3 of the time).
I hope this helps,
Anthony
Many of you have unnecessarily inferred or added assumptions beyond what is stated in the problem:
Reply | Report Abuse | Link to thisA - (On 10/03/08 at 02:25 PM, RBG said: "The assumption here is that the Host knows where the car and two goats are and ALWAYS OPENS ONE OF THE TWO DOORS CONTAINING A GOAT.")
B - (On 10/14/08 at 04:46 PM, Permacultura said: "... the host ... reveals THE ONE THAT DOESN'T CONTAIN THE CAR.")
C - (On 10/14/08 at 07:55 PM. BillAustralia said: "(experimentally with 300 trials) ... the host opens door 2 or 3 in every game, NO GOAT IN EVERY CASE."
D - (Several other people read that assumption into the problem.)
My question is, what authorizes one to add an unnecessary assumption -- an assumption not contained in a problem statement? Is that part of the scientific method? I think not.
The problem with such add-ons is that the answer depends on the add-on. That way of thinking seems to me to be more like the field of Christian Apologetics than to scientific reasoning:
"Apologetics.com exists to remove intellectual impediments to Christian faith, thereby enhancing believers' confidence in, and weakening skeptics' objections to, the gospel message." http://www.apologetics.com/index.php?option=com_content&view=article&id=48&Itemid=55
I think that by adding an unnecessary assumption one redefines the problem to his/her liking. The add-on changes the answer. In the car/goat example, the unnecessary add-on changes the correct answer from 50/50 to 2/3.
If one believes that such add-on assumptions are good thinking, imagine that I add an unnecessary assumption that differs from your unnecessary assumption? In that case the only way to resolve our differences and reach agreement is for one of us to change our assumptions. And if one believes that his/her own assumptions are true, then we are using faith-thinking -- not science.
The assumption regarding the host's information comes from the statement "Host Monty Hall, who knows what is behind all three doors..." in the original argument.
Reply | Report Abuse | Link to thisThe assumption that he always opens a door with a goat comes from common sense. If he opened the door with the car the game would be over, he wouldn't be asking you if you want to change.
If you go through the wikipedia article you'll see that there are many variations of the problem based on the behavior of the host. However, if the host always open a door and asks you to choose again, and if the door he opens has always a goat behind it, then switching would take your winning chances from 1/3 to 2/3.
By the way, religion is about faith, not assumptions. Making assumptions is what science is all about:
Euclidean geometry is correct assuming the world is flat.
Classical mechanics is correct assuming your subject is not too small, too massive, or too fast.
Most analyses of Internet traffic assume that packets arrive in some distribution, or other.
Medical diagnosis assumes that you don't have the one in a billion disease, but a common one.
Economics assumes that people will behave a certain way under certain conditions.
...
I agree with you about the assumption regarding the host's information coming from the statement "Host Monty Hall, who knows what is behind all three doors.." I didn't notice this the first time I read it, and so assumed that the host was picking a random door, though upon reflection it should be obvious that he would only pick a door with a goat behind it, lest he just give away the prize. So I think this is a just as much of a word problem that easily decieves as it is a math problem. Before realizing the host knowledge/manipulation going on I was prepared to make the following counter argument, which I think would be valid if the host were picking a random door;
Reply | Report Abuse | Link to thisCase 1: This case is simply the one which was presented in the article, with the contestant first choosing door number one. It is suggested after being shown the goat behind door number two that switching to door number three increases the chances of winning.
Case 2: The contestant first chooses door number three. The rest is much the same - Monty shows that a goat is behind door number two. It is suggested that after being shown the goat behind door number two that switching to door number one increases the chance of winning.
The symmetry of the problem would suggest there should be no difference in outcomes between case 1 and case 2, otherwise they contradict each other. Therefore there would be no advantage to switching in either case.
Michael Shermer suggests in his article A Random Walk through Middle Land that common sense is wrong in assessing risk in his Let's Make a Deal game. He is the one that is confused.
Reply | Report Abuse | Link to thisIn his scenario, 10 options of 'good' and 'bad' are offered, only one of which has been predetermined to be good. Option 1 is skipped over and options 2 through 9 are shown to be bad. At this point he concludes that option 10 is nine times more likely to be 'good' than is option 1. How droll.
Schermer, usually so opposed to metaphysics, seems to believe that the player's choice of option 1 on the mental level has power over the Newtonian physical level. I don't think so.
If Shermer's interpretation of contemporary probability theory is correct, it's no wonder Wall Street is in so much trouble.
Paul sage
A simple way to see the 2 to 1 advantage of switching your choice in the Let's Make a Deal game is to see switching as getting to choose two doors instead of one.
Reply | Report Abuse | Link to thisYour original tentative choice randomly divides the 3 doors into a group of one and a group of two. If you were told that you could now open the one tentatively selected door, or open both of the other two doors and get the prize if it's behind either of these two doors, then you'd choose the two doors.
Switching your 'choice' after one of the group of two doors is opened, is equivalent to getting the prize if it is behind either of two randomly chosen doors.
My reasoning in the Let's Make a Deal Game (JamesG at 04:45 PM on 10/22/08) is so close to that of SoccerPhD that I wouldn't have posted, if I had first read SoccerPhd's post.
Reply | Report Abuse | Link to thisSome posters are criticizing the always switch rule by covertly appealing to specific cases where it loses. The point is that it wins 2/3 of the time and loses 1/3 of the time.
I should have further clarified that in actuallity case 1 and case 2 are not really symmetric because of the way the host intentionally chooses a door with a goat behind it after the contestant makes a first choice (as opposed to the host making a random choice). The host's door selection will not necessarily be the same in both case 1 and case 2, therefore the symmetry between the two cases does not exist and there is no longer a contradiction presented by the two cases.
Reply | Report Abuse | Link to thisPaul, you are making the same mistake I used to make. The mental choice of the player does not alter the universe, the INFORMED choice of the game host does. As many people (such as JamesG), have pointed out, when the host shows you a goat behind one of the two OTHER doors and asks you to change your choice, he asks you to choose between your original guess which you know nothing new about (and therefore has 1/3 winning chance) and an other door which is as good as both the two doors you didn't originally choose combined. This is true because (since you've seen a goat) if the car was behind ANY of the two doors you didn't choose, it will be behind the only door left for you to switch to.
Reply | Report Abuse | Link to thisIf you still don't buy it, think of the argument that got me thinking that I'm wrong with the 50/50 theory:
Consider a million players playing the game one after the other and consider that no one ever changes. What percentage of them do you expect to win? The answer is (around) 1/3, not half, since only (around) 1/3 will choose the right door out of three that they are shown. So what would have happened, if they had ALL switched AFTER the host opened a door with a goat? Well, the 1/3 that would have won if they had all stayed will now loose, and the 2/3 that would have lost, will now win. Q.E.D.
anthonyG
Reply | Report Abuse | Link to thisWhile I'm not a math guy, let's look at this. If the starting total probability is equal to the sum of it's parts, and each has a one third chance, the total function then equals one.
When door 2 is exposed as a goat, the value of that term in the function drops to zero, making the remaining function equal to 2/3, not 1.
If you insist on keeping the function total equal to one, by what logic do you allocate the value of door 2 only to door 3, rather than distributing anoung the remaining choices?
It seems that in the logic defending Shermer, there is a refusal to see that things are changed by the revelation of door 2. The original situation no longer exists.
Paul
anthonyG
Reply | Report Abuse | Link to thisRegarding your second argument, a million players. Yes, door 1 wins 1/3 of the time. So does door 3. Mental exercises amongst the players does not change the result.
Your last line
"Well, the 1/3 that would have won if they had all stayed will now loose, and the 2/3 that would have lost, will now win. Q.E.D."
In fact, 100% of the people switching will win 50% of the time.
Paul
anthonyG
Reply | Report Abuse | Link to thisIn reply to your second argument above, the million player case.
You state:
"Well, the 1/3 that would have won if they had all stayed will now loose, and the 2/3 that would have lost, will now win. Q.E.D.
Actually, since 100% of the players switch, the correct solution is that 100% of the players either win or lose, depending on the location of the of the 'good'choice. There is no reason in the statement of the problem that the 'good' is twice as likely to be in the 3 position.
Paul
Once you make a preliminary selction the moderator (who knows where the car is and the goats are) opens one of the two other doors that has a goat. The moderator does not open a particular door.
Reply | Report Abuse | Link to thisOne third of the time the moderator can open either one of the two non-selected doors, but 2/3 of the time the moderator must be sure to open the one door where the goat is, because he knows the car is behind the other.
So if you always switch in a large number of trials, then that 2/3 of the time you will get the car, and 1/3 of the time you will get a goat.
jamesG
Reply | Report Abuse | Link to thisAll the moderator's choice proves is that the car is not behind the door he opens. It says nothing about the other two possibilities.
I think your analysis is correct if the moderator only opens door three when the car is behind door two.
There are four possible scenarios
1) Car behind 1, two is exposed
2) Car behind 3, two is exposed
3) Car behind 1, three is exposed
4) Car behind 2, three is exposed
No matter which card the moderator turns over, two possibilities of equal value remain.
Paul
There are only three scenarios in any one trial, and not 4. In any given trial when the car is behind door 1 the moderator can open either door 2 or door 3, but not both on the same trial.
Reply | Report Abuse | Link to thisThere are three equally probable cases:
1) Car behind 1, either 2 or 3 is exposed, but not both.
2) Car behind 2, 3 is exposed.
3) Car behind 3, 2 is exposed.
In a large number of trials, if you always switch, you get the car in cases 2 and 3, or two thirds of the time.
Even though I have taken several courses in statistics and probabiliy and did statistics analyses for myself and my coworkers in an analytical chemistry laboratory, when this originally came out in Marilyn Vos Savant's column in Parade Magazine I got it wrong. I said there was no advantage to switching. And for some time I disputed Marilyn's opinion that there was.
Reply | Report Abuse | Link to thisThis is confirmation of Schermer's point that the configuration of the human brain causes us to misjudge relative probabilities in many cases. If you want some more challenges there are some at "Paradoxes in Probability": http://www.quantdec.com/envstats/homework/class_03/paradox.htm
Alright, try this approach on for size.
Reply | Report Abuse | Link to thisContestant chooses door 1. There are now two possibilities, moderator chooses door 2, or, chooses door 3.
Moderator chooses door 2 and it is opened. Contestant switches to door 3. If he switches and wins 2/3s of the time, this means that the car is behind door 3 twice as often as behind door 1. This is not a random result.
Exactly the same argument applies if moderator discloses door 3. Both cases yield a non random result.
Paul
P. Contestant chooses door 1. There are now two possibilities, moderator chooses door 2, or, chooses door 3.
Reply | Report Abuse | Link to thisJ. Misleading statement. There are two possibilities for choice of the moderator only if the car is behind door 1, which occurs in 1/3 of a large number of trials. Your reasoning from this point in your argument only applies if the car is behind door 1. You are not allowing for the car to be behind the other two doors.
If the car is behind either doors 2 or 3 (which occurs in 2/3 of the trials), then the moderator has no choice in which door to open. And in this subset of the trials the moderator will be forced to open door 2 and door 3 an equal number of times.
P. Moderator chooses door 2 and it is opened. Contestant switches to door 3. If he switches and wins 2/3s of the time, this means that the car is behind door 3 twice as often as behind door 1. This is not a random result.
J. When you say "moderator chooses door 2" you are employing your implicit assumption that the car is behind door 1. Your assertion that the contestant who switches wins in 2/3 of trials where the moderator has a choice is false. When the moderator has a choice in which door to open, the switching contestant always loses. So your subsequent arrival at a contradiction was not obtained by correct reasoning.
OK James, you got me on that one, quote:
Reply | Report Abuse | Link to thisYour assertion that the contestant who switches wins in 2/3 of trials where the moderator has a choice is false.
It is false. That's your assertion.
Paul
Shermer has given an authoritative and readily understandable argument showing that the the switch strategy wins the car in 2/3 of trials, and others have repeated this argument in virtually identical or in equivalent forms. The job now is to explain exactly how specific arguments to the contrary are wrong.
Reply | Report Abuse | Link to thisFor example, the error in Stan Hillard's (Stan Hilliard at 12:46 PM on 09/19/08) argument is that he states without any proof that in a specific trial where the contestant randomly chooses door 1, and door 2 has been opened by the moderator to reveal a goat, that the probability of the car being behind doors 1 and 3 remain equally probable, and so now are each equal to 1/2.
It is certainly true that these two probabilities add up to 1, but not true that they are equal. And, in fact, the probability that the car is behind the door first selected remains 1/3, but that the probability of the car being behind the other unopened door changes from is 2/3.
This does not mean that the probability that the initial probability before any trial that the car is behind door 3 is changed by the actions of the contestant and the moderator.
No. What has change is the success rate of the contestant employing a switching strategy after the moderator opens a non-selected door showing a goat.
SoccerPHd,
Reply | Report Abuse | Link to thisThe fallacy of the 2/3rds switch rests on this being the *same* game.
The three door game finished when door 2 was revealed.
Now it's a two door game, and the odds are changed.
The dice have no memory, as statisticians are fond of saying, therefore you have two doors and only two possible states:
door1: Car door3:Goat
or
door1:Goat door3:Car.
You gain or lose no advantage at all by switching.
(In fact, the whole Monty Hall game was ALWAYS a two door game with three doors in it...Monty knew which door to reveal all the time.)
Bruce,
Reply | Report Abuse | Link to thisThere are three possible (and equally probable) configurations and not two. You left out Door 1: goat, Door 2: car, Door 3: opened to reveal a goat.
The way you describe it a goat is always behind door 2 so that really would be a two door puzzle and there would be no value in switching.
Jim
... Quoting anthonyG from 10/20/08 at 11:21 AM:
Reply | Report Abuse | Link to this... The assumption that he always opens a door with a goat comes from common sense.
... Quoting Paul SageOn from 10/21/08 at 11:53 PM:
... If Shermer's interpretation of contemporary probability theory is correct,
... it's no wonder Wall Street is in so much trouble.
The fact that the host knows what is behind each door does not determine how the host will use his knowledge. Nobody knows the host's intent. Why did he open door 2? A game show host has many factors to consider -- one important factor being to keep up the shows Nielsen rating to make more money. The hosts intuition will tell him what will draw the most audience interest. Another important factor guiding the host is that he must fit the show into a fixed time window.
If you could ask the host why he opened door 2, can you be absolutely sure of what his answer would be? Even when you don't know how to host a game show yourself? If not, you are bringing your own unnecessary assumption to the problem. Because the problem statement did not say WHY he opened the second door. And the host is most certainly guided by his purpose.
If you believe that you are free to add your "common sense" to the problem you can make the probabilities come out any way you want, as the following examples show.
(1) If you can be certain that he only opens a goat-door (your add-in), and if he only opens another door if your door is a goat (my add-in), then your probability of getting the car is 1.0 if you switch.
(2) On the other hand, if the host knows that your strategy is to always switch, and the host can choose to open or not-open any door, and if his intent is that you lose, then the probability that you will win is 0.0.
The only way that you can limit the problem to having a unique "objective" solution is that people not add their own "common sense" assumptions/beliefs to it.
Oh dear! Mr Shermer is quite wrong. Once one of the three doors - with a goat - selected by the show - has been opened the game starts over. The situation is now quite that which holds if the game had started with only 2 doors. The contestant is faced with two possibilities: either the door chosen has the car, or it hasn't. The contestant has no idea which is which. The probability for each is 0.5. Unhappily the matter is confused by the tortuous semantic gymnastics. (Back to 'Why People Believe Weird Things). The only 'reason' the contestant would have to change would be given by an attack of itching thumbs or similar. But then the individual doesn't know if the itching is + or - (Again, back to 'Why People Believe Weird Things). . A sad business I fear .
Reply | Report Abuse | Link to thisI think the misunderstanding of this problem comes with the rules of the gameshow:
Reply | Report Abuse | Link to this1) The host ("Monty") always opens one door after the first round.
2) You always have the option to switch after the first round.
The confusing part is that the scenario changes depending on the situation. I know this has been re-worded many times, but let me give it a try:
Let's suppose that we will ALWAYS switch doors after the first round. What is the best way to win? As long as you choose a door with a goat (2/3 chance) then switching will be the winning choice (because Monty will ALWAYS open the other "goat door", leaving the car as the only other door). The ONLY SCENARIO where switching will be the losing choise is if you choose the car on the first round and switch (1/3 chance). You really only have 2 ways to win:
1) Pick a goat (2/3 chance) and switch.
2) Pick the Car (1/3 chance) and don't switch.
If we choose to never switch after the first round, then our chances remain 1/3, since nothing changes after he opens the other door.
The rules of the gameshow make this more confusing and counterintuitive, but the game is changed by Monty opening a door. Again, this has been explained many different ways on this message board, but you have to be careful that you are not using false logic with your analysis since information being added after round 1 is what changes the scenario.
One more rule to the game:
Reply | Report Abuse | Link to this3) When Monty opens a door, it is ALWAYS a goat.
rekscma, your three assumptions made your calculations support your beliefs. But they are not justified. So your conclusion is incorrect. You said that your three assumptions are:
Reply | Report Abuse | Link to this1) The host ("Monty") always opens one door after the first round.
2) You always have the option to switch after the first round.
3) When Monty opens a door, it is ALWAYS a goat.
I offer two reasons that all three of those assumptions are wrong.
1) There is no mention of any of them in Michael Shermer's description of the situation. I pointed this out earlier.
2) They do not match the actual television game show "Lets Make a Deal". (I watched a rerun of one episode of that game show on the GSN cable channel today.)
In today's episode, Monty told the contestant that behind one door was the big prize of $20,000. Behind the other two doors was $2,000 and an initially undisclosed amount, which was later revealed to be $5,000. The contestant chose door #3. It was the correct door. The contestant won $20,000.
Monty did not open another door. (Your assumption #1 is wrong.)
Monty did not offer the option to switch. (Your assumption #2 is wrong.)
Monty did not open the losing door. (Your assumption #3 is wrong.)
It doesn't matter if Google finds 126,000 web pages that disagree with my analysis. It doesn't matter if that faulty thinking is supported by computer simulations that have the subjective assumptions coded into them -- to code the false assumptions into to Basic and Java doesn't "prove" anything.
The answer to the original sciam.com problem remains p'=0.50.
Well, the problem initially stated by Mr. Shermer stated there are three doors (one is a car, the other 2 are goats) and you can choose one door. Then, one door (a goat) is revealed and you are asked if you want to switch. Regardless of the format of the television show (which changed many times as far as I can tell), we don't really have to make any assumptions. Mr Shermer explicitly states in the first paragraph that Monty opens a door to reveal a goat, and you are offered the chance to switch after the goat is revealed. I believe that my assumptions stated previously would always support this scenario (and could be applied to generate other identical scenarios), and the odds for switching doors are in your favor.
Reply | Report Abuse | Link to thisI guess you could argue that they could be switching what is behind the doors after you make your decisions, but for the sake of argument we would assume that is not happening (otherwise probablity would not apply to the situation at all).
Assumptions are never 'wrong' - they are assumed. When the contestant initially picks a door she has a 2/3 chance of being wrong. Period. Full stop. Nothing happens to change that. Those readers who are unconvinced by simulations do not understand what probability means. If you play this game at home and it is right to change doors two thirds of the time then you have shown that the probability is 2/3.
Reply | Report Abuse | Link to thisThe 'new information' does not effect the probability that we chose a wrong door initially which remains 2/3 , period. Monty Hall shows us a door with a booby prize but so what?- we always new that at least one existed. If our initial choice was wrong (which it will be 2/3 of the time all Monty has done is shown us where the prize is (nice of him really) and we should change. If we chose right initially (1/3 of the time) he has told us nothing and changing doors will have a sad result.
Reply | Report Abuse | Link to thisMichael Shermer correctly answers a scenario for "Let's Make a Deal." The only problem with his answer is the same problem with Marilyn vos Savant's answer to the puzzle posted in Parade in 1990, which is that the problem being answered is not the problem that was presented.
Reply | Report Abuse | Link to thisAny conclusion to this problem must consider the game show host's decision making process. Shermer's and Marilyn vos Savant rests on two assumptions about the thinking of the game show host: 1) that in all circumstance he would have revealed what's behind one of the other doors not chosen and 2) in all circumstances he would have only opened a door that has a goat behind it. The problem never states either of these things. Instead it is assumed as part of the solution. The problem does state that the game show host opens another door to reveal a goat, but this describes the circumstance you are confronted with, not what would have happened in all circumstances, nor the decision making of the game show host that led to the circumstance presented. For example, suppose the game show host planned to simply open one of the 3 doors at random after your first choice regardless of what you picked or where the prize was. If the choice to reveal the goat was the result of a random choice, then the math PhDs who wrote in to disagree with Marilyn vos Savant back in 1990 were absolutely correct: the probability of winning with your original choice is exactly the same as the probability of winning by changing your choice. Suppose the game show host did not want you to win the prize. With this assumption, you should conclude that the only reason he even revealed one of the other choices instead of the door you picked was because you originally picked the correct door – so you should stay with your original choice. Suppose the game show host wants you to win the prize. With this assumption, you should conclude that the only reason for prolonging the game was to give you a chance to pick correctly, since if you had picked correctly the first time, he would have revealed that you were right in order to give you the prize – so you should change your choice (note that the answer in this scenario is the same as Shermer's, but the supporting argument and probability of being correct by switching is much different.) Suppose the game show host doesn't care whether you win or lose but does want to prolong the game by revealing a goat and giving you another choice. In this scenario, you should align with Shermer's thinking and change your choice.
Anyone's conclusion about the game show host's motives is just as valid as any other conclusion because it isn't stated in the problem. Shermer's solution requires assuming a thought process by the game show host that is neither greedy, nor benevolent, nor random. He assumes a more complicated motive and I think it could be argued that his solution is the least plausible solution since there's not even a hint about the game show hosts motives stated in the problem. But he assumes this level of complexity in the host's motives versus simpler ones that you would expect to find in this sort of thought problem like "the agent is greedy" or the "agent is random."
Reply | Report Abuse | Link to thisHere's What's Really Behind the Doors:
In conclusion, if I were actually confronted with this problem in real life, I would choose to assume that the agent is greedy, and I would stick with my original choice.
This problem suddenly interests me as much now as it did when I first read it as a 16 year old, thinking at the time that it was a new math puzzle, which it isn't. The proper understanding of the problem has nothing to do with modeling or statistics. It has to do with carefully separating that which one knows from that which one does not know. The writer, Scott Adams, has suggested that our ability to do this is the definition of awareness. Improving awareness is not related to improving IQ or intelligence, nor is it done by acquiring knowledge or experience, but rather by improving ones ability to separate the known from the unknown. A better awareness might have you conclude that you know why someone on the other side of the office where you work is smiling even though you haven't spoken with that person in days. It might have you conclude that you don't know why someone has said something right after they've told you why they said it. We don't know the motives of the game show host. The best we can do is look for patterns that might be clues to the rules of the game. Historically, these patterns have been found very slowly by proposing solutions and sticking to them – by working through trial and error. But there's a shorter path to understanding that involves listening very carefully and trying diligently to see the world through the eyes of others. If the participants in the controversies over this problem had set aside their proofs and instead focused on discovering the fundamental reason that their opposition had followed a different path than their own, they would have quickly gained a better understanding of the problem for themselves.
What's Really Behind the Doors:
Reply | Report Abuse | Link to thisIn conclusion, if I were actually confronted with this problem in real life, I would choose to assume that the agent is greedy, and I would stick with my original choice.
This problem suddenly interests me as much now as it did when I first read it as a 16 year old, thinking at the time that it was a new math puzzle, which it isn't. The proper understanding of the problem has nothing to do with modeling or statistics. It has to do with carefully separating that which one knows from that which one does not know. The writer, Scott Adams, has suggested that our ability to do this is the definition of awareness. Improving awareness is not related to improving IQ or intelligence, nor is it done by acquiring knowledge or experience, but rather by improving ones ability to separate the known from the unknown. A better awareness might have you conclude that you know why someone on the other side of the office where you work is smiling even though you haven't spoken with that person in days. It might have you conclude that you don't know why someone has said something right after they've told you why they said it. We don't know the motives of the game show host. The best we can do is look for patterns that might be clues to the rules of the game. Historically, these patterns have been found very slowly by proposing solutions and sticking to them – by working through trial and error. But there's a shorter path to understanding that involves listening very carefully and trying diligently to see the world through the eyes of others. If the participants in the controversies over this problem set aside their proofs and instead focus on discovering the fundamental reason that their opposition follows a different path than their own, they would quickly gain a better understanding of the problem for themselves.
There are three equally likely possibilities:
Reply | Report Abuse | Link to this1) If the person had picked the good prize, switching is a losing proposition no matter what other prize Monty reveals.
2) If the person had picked the first not so good prize, then Monty must reveal the second not so good prize. In this case switching wins.
3) If the person had picked the second not so good prize, then Monty must reveal the first not so good prize. Switching is again a winning tactic.
Since there are three equally likely scenarios and switching wins in two of them, then the rational course is to switch. Stated this way there is no paradox.
I am amazed at the silly discussions on this list. Monty Hall has very little choice in his actions. He has no choice if you picked a losing door, and his choice is irrelevant if you picked the winning door. This is pure probability theory, nothing more.
Reply | Report Abuse | Link to thisThere are other potential rules that could have been used that may change the probabilities, but as the game is played, switching is the best option. For those information is added advocates, I can point out that you always know a door that was not chosen contains a goat. Once Monty reveals which one it is, you still know that there was only a one in three chance that the door you picked contained the good prize. However, you now also know that the remaining unselected door must contain the good prize if your door doesn't. Hence you switch.
Note that Monty must pick the goat door at random if you have the car door. If he doesn't then you would know whether to switch or not. For example, if he always picked door #3 unless it contained the car, then if he showed you door #2, you'd know the car must be there. Nor can he rotate through the doors or pick the door next to yours, etc., for similar reasons.
The essence of the game described by Shermer is that:
Reply | Report Abuse | Link to this- a door is opened which contains a not so good prize
- you have a choice to switch or not, and
- one prize is worth significantly more than the others.
If Monty doesn't offer you the choice to switch then there is no game. And if he doesn't open a door then the odds from switching remain 1 in 3 - the same as the odds for sticking. However, once he opens a door, then you know that door doesn't contain the good prize. This gives you the information to rule out that door so switching is now a winning strategy.
From: thomascloomis@comcast.net
Reply | Report Abuse | Link to thisDate: January 26, 2009 4:19:00 PM EST
To: editors@sciam.com
Subject: Michael Shermer's goat problem
OK, I finally decided to think about it. It's simple.
If I simply "choose" a door, and stick (foolishly) with my choice no matter WHAT is done afterwards, I of course have a 1/3 chance of the car and a 2/3 chance of a goat.
But the situation the puzzle describes is not this.
When I "choose" one door, and another person ALWAYS opens one of the two other doors to reveal a goat, this is the same as saying that 2/3 of the time, between me and him, we've found both goats (and 1/3 of the time, a goat and a car).
Therefore 2/3 of the time the other door from the one I first chose, is the car.
So it makes sense to switch.
Simple as that.
TL
Assuming that most agree that switching the door doubles our chances of winning, let me propose a variation of the problem, in order to test further our understanding.
Reply | Report Abuse | Link to thisIn this Modified Problem, Monty does not know (maybe he has forgotten) what is behind the closed doors and he opens one (chosen randomly) that shows a goat. Once asked whether we want to change our initial choice, what would be our decision most likely of winning the car?
This is a good example of a situation where what might appear as a small difference (in one case Monty's action is based on additional information, in the other on a random choice, but both apparently have the same outcome) changes significantly the problem to be solved.
The problem here is that it is mis-stated. For the logic to be true, Monty has to open EITHER door 2 or 3, revealing a goat. In the number two set, door number two number CAN'T be opened to show a goat, because it has the auto. The problem, as stated, eliminates the possibility of opening door number two, produces a 50-50 chance of winning or losing by switching.
Reply | Report Abuse | Link to thisAs previously stated, the problem is that Shermer's problem statement is wrong. The proper description is that Monty opens a door (either 2 or 3) and reveals a goat. It could NOT have been just door number 2 because, in combination two, bad-good-bad, door number 2 doesn't HAVE a goat behind it. In Shermer's described problem, the chances are 50-50.
Reply | Report Abuse | Link to thisScientific American can do a better job of prior review.
Maybe so, but there is also a 2/3 chance of it being door 1 or 2 and 1/3 of it being door 3. Now if Monty reveals that 2 is a goat, the 2/3 chance is entirely allocated to door 1 (at least, using your logic). Since we can't decide which door is which probability, we should simply estimate both probabilities to be 1.5/3, which is really 1/2, like David King, Marcus Millet and Stan Hillard said. The same applies when there are 10 doors and Monty reveals 2-9 to be goats.
Reply | Report Abuse | Link to thisThink about it.
If I were to ever find myself in the improbable role of a game show contestant, I hope that lessons learned in my statistics class would help override my folk numeracy tendencies and increase my odds of winning.
Reply | Report Abuse | Link to thisThat said, how much more boring would our lives be if everyone were to think like a statistician before making a decision?
this article was great. i dont think that average brains work in detail enouugh to grasp the idea around the topic. For exaple if you were on that game show you would want to stay with your origonal pick, unless you were tought about these probabalities.
Reply | Report Abuse | Link to thisI finally get it why SoccerPhD is right. When you select the fist door the odds that you are wrong are 2/3. Then Monty reveals the door with the goat behind it from the remaining 2 doors. The chance that you got it wrong is still 2/3.
Reply | Report Abuse | Link to thisInteresting topic! There paradox of determined coincidence solves the issue of randomness. The logical or gambler’s fallacy can be transcended by means of interacting with chaos through order. Yes! There is order in chaos, and this order can be observed and even predicted. Once you realize that randomness or chaos exists purely in the objective realm then the rest is easy. Quantum mechanics plays a vital role here for the behaviour patterns in randomness are not automatically likened to common sense as we think it should be. The idea of ‘hot’ and ‘cold’ numbers in Roulette play operates in the common sense realm and therefore nothing is solved. We try to capture that which belongs to the objective with the subjective mind, and that is where the confusion sets in. Like Boolean algebra, the logic is different to ordinary algebra, but it works. The same with this: the art is to take the correct perspective.
Reply | Report Abuse | Link to thisRandomness as we see it is infinite. But is it really? No it’s not! It has limits. It is timeless and seemingly without cause and effect. So? Can it be measured? No, it can’t. Just like the Measurement Problem in quantum physics…It doesn’t need to be measured, only observed through interaction. The easiest way to do this is through numbers. The random input is read by a series of numbers; a cyclical repetition of digits that represents the order inside the randomness. In this way order can be seen inside the chaos and predictions can be made on future behaviour.
The above statements has been proven through experimental analysis by running tests on a 128 GB RAM with Xeon 7500 CPU's super server. Numbers were obtained from a TRNG source and fed through the system. Projections and predictions on future behaviour were found to be 100% correct. This breaks all barriers on Marsaglia and Die Hard Tests.
Yes, random can be cracked! It does not have to rule. What happened to mind over matter? ordo ab chao