Still not convinced? Google “Monty Hall Problem simulation” and try the various computer simulations yourself and you will see that you double your actual wins by switching doors. One of my correspondents, who was skeptical at first, ran his own simulation over 10,000 trials, concluding that “switching doors yields a 2/3 success rate while running without switching doors yields a 1/3 success rate.” (Go to www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html for the simulation.)
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50 Comments
Add CommentI think the reason why the Monty Hall Problem has been described as maths most contentious problem is that the problem is almost always poses a different question to the one that questioner is expecting the answer from. This is generally because the questioner normally does not understand the subtleties of the problem.
Reply | Report Abuse | Link to thisMost times I have heard this problem posed, the question goes something alone this line (which seems the most common sense way to describe the problem.).
You are in the audience, three boxes are bought on to stage and you are asked by Monty to come down and pick a box. You are told that behind two of these boxes is a goat and one of the boxes has a car behind it. You pick a box and then Monty, to drum up suspense, slowly turns over one of the other boxes that you did not pick and a goat appears. He then gives you the option to switch boxes.
I think most people who a given this problem see the problem as I have described above. I think it is fair to assume that you count yourself as lucky that the car was not turned over and that you now have a 50:50 chance of winning the car and switching boxes will not change your odds of winning. Note: the quantity of information from seeing the goat has raised you odds from 1/3 to 1/2.*
This is the intuitive answer and is correct in this circumstance based on your assumptions.
*(the value of this information increases your chances by 1.66 times).
Some people posing the question will forward additional detail by stating that, you are told that Monty knows what is behind the boxes. They may think this is key to doubling your odds by switching. This information is of little use to you unless you feel that Monty is feeling generous to you and has purposely turned around a goat. If you feel there is this chance then your odds will increase an amount proportional to the odds that you weigh up that Monty is feeling generous , by swapping (perhaps you take into account your good looks and great personality).
A different problem is as follows.
You are in the audience, three boxes are bought on to stage and you are asked by Monty to come down and pick a box. You are told that behind two of these boxes is a goat and one of the boxes has a car behind it. You then pick a box. To drum up suspense Monty then goes up to one of the boxes and says there is a goat behind this box and turns the box around to prove it. Well what information have you gained here? You know that Monty knows what is behind the two boxes that you have not chosen and if you have chosen a box with a goat then by switching you will win the car 100 % of the time.
That it is a goat that Monty turns around is of no information if you stay with the box you have chosen, as a goat had to be behind one of the other two boxes. So you are not using any information to increase your odds if you stay. That is, your odds remain at 1/3. However Monty has sorted the other two boxes for you by eliminating a goat behind one of them. You can only take advantage of this sort by choosing the box that he did not pick. The other two boxes have a 2/3 chance of having a car behind them in total. You choose the remaining box and you claim the benefit of the information offered by Montys sort and these increased odds*. Because the information is restricted only to these two boxes (not three) the value is of the information is concentrated and your odds double by switching boxes.
*(The value of concentrating this information to two boxes increases your chances by 2.0 times, when swapping)
Note:
Reply | Report Abuse | Link to thisThese increase odds are only valid if you have been made aware that Monty has in fact performed a sort..
The second problem above is not completely accurate. If Monty had remember that the Goat was behind one of the ones you did not pick (from before the show say) and turned this box around, and he was unsure about where the car was in the other two boxes then Monty does not perform a sort and your odds are as in the initially posed problem That is 50/50.
So to be accurate Monty should not say to you there is a goat behind this one and turn it around to prove it. He should in fact say I know what is behind these two remaining boxes and this one definitely has a goat behind it &..
So in fact the problem has become quite contrived in order to double your odds.
Johnnyct@optusnet.com.au and johnnymac are among the many who just don't get it. When you pick a door (or box) you have a 1/3 chance of picking a car. When Monty opens a door and reveals a goat, you still have a 1 in 3 chance of having picked the door with a car. It does not matter how Monty picks the door to open; if it shows a goat, you double your chances by switching.
Reply | Report Abuse | Link to thisThe only way to have 50-50 odds is if Monty reveals a goat BEFORE you pick.
While I agree with Mr. Schermer's analysis, I take issue with professor Rosenhouse's third requirement. If the contestant picks the door with the car, Monty can pick from the other two doors at random, or can pick the lowest numbered door, or the door with the fatter goat, or can use any other criterion. There is no need for Monty to pick at random.
ReReply to JHasty
Reply | Report Abuse | Link to thisThere are two different scenarios here either:
You are made aware that Monty has performed a sort on the remaining boxes and you choose accordingly.
Or
You are not made aware that Monty has performed a sort on the remaining boxes and you choose accordingly.
In both of these scenarios the probabilities calculations are as per school. There is no magic going on here.
In the first scenario you calculate the probabilities of a car on all three boxes.
In the second scenario you calculate the probabilities of a car on only the two boxes you did not pick. Obviously the process of the sort will magnify the cars probabilities in these remaining boxes.
In the first scenario there is no advantage to switching boxes. In the second scenario there is.
If you write a computer program you must be very careful to make sure you are replicating the real life problem that you consider the program is duplicating.
JHasty does not state which scenario above he is commenting on.
If it is the first scenario then by virtue of randomly turning over a box with a goat or if Monty knows what is behind the boxes and randomly chooses to select a goat in the remaining boxes then there is no advantage to swap.
In this scenario just by virtue that a car could have been revealed increases the odds of the remaining two boxes to 50:50. The odds do not remain at 33% in this scenario as JHasty seems to be suggesting.
I agree with JHasty that you have a 50/50 chance if the Goat is revealed before you choose. But you also have a 50:50 chance after you choose if this box is revealed at random and does it does not reveal a car.
The second criteria in Professor Rosenhouses criteria - Monty always opens a door concealing a goat in the remaining boxes - is only possibly in the second scenario. That is Monty performs a sort (he must have information on all remaining boxes to do so) and you need to be made aware of this sort (be told that this in fact is what has being performed) in order to take advantage of the increase odds by choosing to switch.
TO DECIDE ON THIS PROBLEM YOU MUST BE CLEAR WHICH SENARIO YOU HAVE BEEN ASKED.
The two senarios have the wrong order for comment. Please read my comments in the order as below.
Reply | Report Abuse | Link to thisReply to JHasty
There are two different scenarios here either:
You are not made aware that Monty has performed a sort on the remaining boxes and you choose accordingly.
Or
You are made aware that Monty has performed a sort on the remaining boxes and you choose accordingly.
In both of these scenarios the probabilities calculations are as per school. There is no magic going on here.
In the first scenario you calculate the probabilities of a car on all three boxes.
In the second scenario you calculate the probabilities of a car on only the two boxes you did not pick. Obviously the process of the sort will magnify the cars probabilities in these remaining boxes.
In the first scenario there is no advantage to switching boxes. In the second scenario there is.
If you write a computer program you must be very careful to make sure you are replicating the real life problem that you consider the program is duplicating.
JHasty does not state which scenario above he is commenting on.
If it is the first scenario then by virtue of randomly turning over a box with a goat or if Monty knows what is behind the boxes and randomly chooses to select a goat in the remaining boxes then there is no advantage to swap.
In this scenario just by virtue that a car could have been revealed increases the odds of the remaining two boxes to 50:50. The odds do not remain at 33% in this scenario as JHasty seems to be suggesting.
I agree with JHasty that you have a 50/50 chance if the Goat is revealed before you choose. But you also have a 50:50 chance after you choose if this box is revealed at random and does it does not reveal a car.
The second criteria in Professor Rosenhouse’s criteria - Monty always opens a door concealing a goat in the remaining boxes - is only possibly in the second scenario. That is Monty performs a sort (he must have information on all remaining boxes to do so) and you need to be made aware of this sort (be told that this in fact is what has being performed) in order to take advantage of the increase odds by choosing to switch.
TO DECIDE ON THIS PROBLEM YOU MUST BE CLEAR WHICH SENARIO YOU HAVE BEEN ASKED.
Sorry I have commented on the senarios in the wrong order.
Reply | Report Abuse | Link to thisMy reply should read as follows:
Reply to JHasty
There are two different scenarios here either:
You are not made aware that Monty has performed a sort on the remaining boxes and you choose accordingly.
Or
You are made aware that Monty has performed a sort on the remaining boxes and you choose accordingly.
In both of these scenarios the probabilities calculations are as per school. There is no magic going on here.
In the first scenario you calculate the probabilities of a car on all three boxes.
In the second scenario you calculate the probabilities of a car on only the two boxes you did not pick. Obviously the process of the sort will magnify the cars probabilities in these remaining boxes.
In the first scenario there is no advantage to switching boxes. In the second scenario there is.
If you write a computer program you must be very careful to make sure you are replicating the real life problem that you consider the program is duplicating.
JHasty does not state which scenario above he is commenting on.
If it is the first scenario then by virtue of randomly turning over a box with a goat or if Monty knows what is behind the boxes and randomly chooses to select a goat in the remaining boxes then there is no advantage to swap.
In this scenario just by virtue that a car could have been revealed increases the odds of the remaining two boxes to 50:50. The odds do not remain at 33% in this scenario as JHasty seems to be suggesting.
I agree with JHasty that you have a 50/50 chance if the Goat is revealed before you choose. But you also have a 50:50 chance after you choose if this box is revealed at random and does it does not reveal a car.
The second criteria in Professor Rosenhouse’s criteria - Monty always opens a door concealing a goat in the remaining boxes - is only possibly in the second scenario. That is Monty performs a sort (he must have information on all remaining boxes to do so) and you need to be made aware of this sort (be told that this in fact is what has being performed) in order to take advantage of the increase odds by choosing to switch.
TO DECIDE ON THIS PROBLEM YOU MUST BE CLEAR WHICH SENARIO YOU HAVE BEEN ASKED.
Lets define the problem explicitly. There are three doors, with a prize behind each. Behind one door is a car, and behind each of the two others is a goat. During the game, the prizes are not moved. The contestant has absolutely no knowledge regarding which door hides which prize. The contestant picks a door. Then Monty, the host, opens one of the two other doors and reveals a goat. The contestant is then given the option of sticking with the door he initially chose, or switching to the other still-closed door. Once the contestant makes his final choice, that door is opened. If a goat is revealed, he loses. If a car is revealed, he wins.
Reply | Report Abuse | Link to thisIt is important to know how Monty makes his choice. In the classic Monty Hall Problem, Monty knows which door hides the car, and deliberately opens a door that he knows will reveal a goat. This is the scenario discussed in Schermers column, and in this case, switching doubles the contestants chance of winning.
If, on the other hand, Monty either does not know or does not care where the car is hidden, and he makes his choice between the two remaining doors at random, this changes things. Since there is a chance that Monty could have opened the door with the car, but happened not to, the odds that the contestant originally picked the door with the car increase. It turns out they increase to 50%, so there is, in this second scenario, no advantage to switching.
The original points I wished to make were, 1) I agree that in the scenario in the original column, switching doubles the chance of winning, and 2) I disagree with professor Rosenhouses third condition: When & Monty [has] a choice of doors to open (which happens in those cases where your initial choice was correct) he makes his choice at random. I just dont see that this makes any difference at all. Perhaps I need to buy his book and see if he can convince me.
I agree with jhasty analysis.
Reply | Report Abuse | Link to thisHowever
Jhasty states that the Classic Monty Hall Problem is - In the classic Monty Hall Problem, Monty knows which door hides the car, and deliberately opens a door that he knows will reveal a goat.
What I was trying to stress in my comments was that this is not the problem most people think they are answering. When most people present this problem they do not explicitly state that Monty does a sort on the remaining boxes (as you state above) and this is essential information if you hope to double your odds.
The classic Monty Hall Problem you state is not the Monty Hall problem Shermer states which I quote below and I think Schemers way is the most common way the problem is expressed.
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?
As I tried to express knowing that Monty knows what is behind the doors does not by itself double your odds if you switch. You must be given the information that he knows what is behind the boxes and has sorted them so that he will always reveal a goat. These two pieces of information are not the same.
So the answer to Shermers problem is that you have a 50:50 chance of winning the car whether you swap or not.
The large debate and confusion does not centre on the nature of the problem but is caused by people incorrectly stating the problem.
Also imagine the real life scenario. You are in the game show and the host says I am now going to look behind these two remaining boxes and I will tell you which one has a goat behind it. This certainly does not have the same appeal as the way Schemer has described the problem&
I think Professor Rosenhouses conditions were setout for those wishing to make a computer program of the problem. JHasty is of course correct in that it does not matter how a goat is selected in the remaining boxes if the box chosen contains the car.
Schermer is right, of course.
Reply | Report Abuse | Link to thisIt is true that in the world of mathematics, it is necessary to present the tightest definition of the problem and explicitly state the assumptions that are made. Jhasty and johnnymac have worked hard to do this.
However, I think the issue is better viewed in the light of the purpose of Schermer's column. In his column he is illustrating the innumeracy of the typical human, and how people in the know take advantage of that. (Casino owners, et. al.)
It is easy to nit-pick Schermer's point on the Monty Hall problem if one wants to argue that the problem is not stated as well as it could be. But in the real world Schermer is commenting on, people make ill-informed assumptions all the time. e.g. Travel by car is safer than travel by plane.
Given that, whether stated or not, I think the following are assumptions that most typical humans would make:
1) Monty knows exactly what is behind each door at any point.
2) Monty will not open an unopened door that has a car behind it.
It is a game show, and the audience and players are not typically mathematicians!
So it seems reasonable to make assumptions that an average person would make. If one operates on the assumptions and interpretation that a reasonable, average person makes, then most people don't realize their chance of winning increases from 1/3 to 2/3.
Assuming that, Schermer's point is right on.
Why is there so much To-Do about this simple logical problem. When you pick a door, you will win about 1/3 of the time, you will loose 2/3 of the time. It follows then, if you could pick both of the other doors, you would win 2/3 of the time. When Monty gives you one of the other two doors and you switch your selection to the second of the other two doors, you have effectively picked both of the other two doors and will win 2/3 of the time.
Reply | Report Abuse | Link to thisThe puzzle becomes more clear when you use more extreme cases.
Reply | Report Abuse | Link to thisSuppose Monte randomly places 1 firetruck and 101 dalmations behind closed doors. He then asks you to pick a door and then opens up 100 doors, each showing a dalmation. The 2 remaining unopened doors conceal 1 firetruck and 1 dalmation. Do you switch?
You initially picked the door with the firetruck with probability 1/102 or with a dalmation with probability 101/102. The odds against you picking a firetruck is then 101:1. You would only keep your door if you thought that your initial selection was correct but the odds heavily favor the fact that your initial selection was a dalmation. So if you want a firetruck, you are much better off by switching.
It's not perfect because you will regret not switching 1 in 102 tries on average.
I mean you will regret switching 1 in 102 times.
Reply | Report Abuse | Link to thisThis is a tempest in a tea pot. Statistics don't apply to single events. If you are on Monty Hall's show once, you either pick the right door or you don't . Switching doesn't help. If you were on every week for a year, you should switch because you would win more often.
Reply | Report Abuse | Link to thisIf, after you pick a door, Monty allows you to switch your door for both of the other two, statistically, you can raise your chanches of winning from 1 in 3 to 2 in 3. Therefore, you should switch, even though you fully know that at least one of the other two doors has a goat behind it. Don't let Monty confuse you be showing you what you already know.
Reply | Report Abuse | Link to this"Statistics don't apply to single events."
Reply | Report Abuse | Link to thisWhen you need a big base hit, do you want your best hitter up or your worst? Of course it applies since you improve your chances of winning. You can't guaranty it but if you can improve your chances of winning, you should.
Two variations make the always-switch strategy more intuitive.
Reply | Report Abuse | Link to thisFirst, imagine you're a contestant and there are a million doors. You pick one and it stays shut. Other contestants open doors as fast as possible, where they get to keep the prize if they find it. None finds it by the time there's only one door left aside from yours. Should you switch? It doesn't matter, since the other contestants are as clueless as you.
Second case: You pick your door and it stays shut. At the outset, there was only one in a million chance you were right on the first pick. There is a 999,999/million chance the prize door is among the rest. Monty opens doors fast and deliberately avoids the door with the prize. It's down to your door and one other. If you switch, you'd increase your chance of winning from 1/million to 999,999/million.
The puzzle resolves by involving the state of mind of the other participant, like the three-paint-spots-on-foreheads puzzle. The examples also show the increase in odds. If there are n doors, a policy of switching raises the odds from 1/n, to (n-1)/n, which means in the case of 3 doors, the odds go from 1/3 without switching to 2/3 with switching.
David Barry, San Francisco
To sbjynyc:For that single event, he will either get a hit or he wont.Obviously you pick your best hitter, because that is where statistics apply and over a season you will do better but for that single event, he will either get a hit or not.If you flip a coin10 times and get 10 heads,that has nothing to do with the next flip. It could be a head or tail because it is a single event and statistics don't apply. Just because the previous ones were heads doesn't make it any more likely that the next will be a tail. If you flip 1000 time about 500 will be heads but that doesn't give you any more prediction information about any single flip. The choosing of a door at Monte Hall is essentially a single flip.
Reply | Report Abuse | Link to thisOther comments here insist that in order to answer this question you first have to know the game and know that Monty is omnescent..????
Reply | Report Abuse | Link to thisWhat type of question would that make this problem?????
I agree with Rodcris.
However:
Switching would help (on this single event) if Monty said to you "OK you have picked this door. I am now going to look behind these two remaining doors and reveal a door that definitely has a goat behind it. Then I am going to offer you the choice to switch to the remaining door which I do not reveal."
I think most peoples' intuition would tell them that that if a car was behind one of the remaining doors that they did not pick, then they will win it 100% if they switch.
ie they have a better chance than remaining with the door they originally picked.
I think this problem illustrates how good our inate intuition is when it comes to the approach of such problems.
Those who assume that when Monty revealed a goat that they were lucky that the car wasn't revealed (else they would have no chance of winning) are right in assuming that it makes no difference whether you switch or not. You have now a 50:50 chance of winning the car.
Sorry to disagree but it is quite clear that when one door is opened to show a goat, the three door game is over and a new two door game begins in which your original choice has a one in two, not one in three, chance of being right and there is no hope of improving your odds by changing your choice. The online game does not seem to reflect this.
Reply | Report Abuse | Link to thisWD Howells
Oregonians for Science and Reason, a non-profit educational group, has proved the advantage of switching in a Manty Hall game, done every year for ~10 years in its puzzles and games booth at Da vinci Days on the OSU campus in Corvallis Oregon.
Reply | Report Abuse | Link to thisPeople in the crowd play the game and a filbert nut is put in one of 4 clear, upright plastic tubes for each result; Stayed & lost, Stayed and won, Switched and lost, Switched and won. After 2 days, and ~3-400 trials there is always approximately twice as many wins to losses for those who switch and twice as many losses to wins for those who stayed.
This graphic display of results is definitive. The filberts don't lie!
In my previous post, I forgot the disclaimer that I am speaking for myself and not reprsenting Oregonians for Science and Reason (O4SR). Sorry.
Reply | Report Abuse | Link to thisIf you set this up using Bayes's equation the unintuitive answer of winning 2/3 of the time by switching falls out quite simply. This has probably been noted before, but I haven't been following this in the past or even read all of the above.
Reply | Report Abuse | Link to thisYou only gain an advantage in 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.
This was a comment on the orginal article that was not remotely addressed.
Reply | Report Abuse | Link to thisIt's not a comment on the math behind the problem, it addresses the problem is set up to hit our instinct to run counter to proability.
The author claims that we didn't develop proability. I assert that we developed an instinct to run counter to probability.
The math problem does not make the biological assertion correct.
"If you are on Monty Hall's show once, you either pick the right door or you don't . Switching doesn't help."
Reply | Report Abuse | Link to thisThat is naive, probabilities still apply to single events. Switching does help because the probably for winning when you switch is higher. Odds are you picked the wrong one so odds are if you switch you win.
That is like saying I roll a dice and don't show you, then you guess whether it falls within 1-5 or is a 6. If there was a prize you would definitely pick the 1-5 range.. this is no different (except the probabilities differ).
What happens if you change the scenario slightly? Monty calls Bob and Mike up to the stage along and tells them that behind the 3 doors are 2 goats and a car. Monty tells the 2 contestants to write down the number of the door they choose and submit them to him. As it turns out, Bob chooses Door 1 and Mike chooses Door 3. Monty then makes things interesting by opening up Door 2 and revealing a goat, and tells each contestant that he may change his choice. According to the theory set forth in Mike Shermer's article, it would appear that each contestant should change his choice and pick the door chosen by the other contestant, since the article states that each contestant would improve his chance of winning if he changed his choice. Although there are now 2 contestants, each one initially made an independent choice, and given the fact that Door 2 contained a goat, each contestant would be foolish not to change, if the mathmatics set out in the problem are correct. In fact, to elminate human indecision, let's say each contestant was provided with a computer to help him decide whether to change his choice, and each contestant has agreed to follow the advice of the computer. Given the scenario, assuming Mr. Shermer's analysis is correct, it would seem that once Monty reveals a goat to be behind Door 2, the computer will always tell each contestant to switch his choice, even though one of the contestants is most definitely going to lose the car by doing so. How does the above situation affect the analysis of the Monty Hall Problem (if at all)?
Reply | Report Abuse | Link to thisIf monty does not know what door you hae chosen thrn if he opens a different door your chance of having chosen the correct door is 50%
Reply | Report Abuse | Link to thisMichael Shermer may have hoped to lay to rest the two goats and a car problem, but he has perpetuated the debate by misconstruing the problem. As he stated in A Random Walk , you are asked to imagine you are a contestant in a game show given the choice of three doors, two concealing goats and the third a new car (the real prize). You choose a door, and the host then opens a second door, revealing a goat, and asks whether you wish to switch to the one remaining door. Should you switch? Shermer says yes , claiming your probability (p) of winning the prize to be 2/3.
Reply | Report Abuse | Link to thisIll refer to this problem as the Savant Puzzle , because it was originally publicized by Marilyn vos Savant in Parade Magazine years ago. Shermers error is to confuse the Savant Puzzle with a mathematical problem known as the Monty Hall Puzzle . The correct p for the Monty Hall Puzzle is 2/3, but the correct p for the Savant Puzzle is a rather vague 1/2. Many respondents have correctly cited the value of 1/2, although often for the wrong reasons.
The difference between the Savant and Monty Hall puzzles lies in the initial conditions. In Savant, all you know is that the host has chosen on this one occasion to reveal a goat. In contrast, in Monty Hall, revealing a goat is a precondition you know in advance that this will always happen. The difference is critical. If you know the host will always open a goat door , p for switching becomes 2/3 for the reasons Shermer describes. If you dont know (the circumstance described by Savant, and in Random Walk ), the only information provided by opening the goat door is that the car is in one of the two remaining doors, hence the p value of 1/2. Ironically, the real world operation of Monty Halls show Lets Make A Deal was closer to the Savant conditions than the Monty Hall puzzle conditions. If it hadnt been, contestants could easily have come to the show prepared always to make a switch.
Its my impression that those of us who do science have an easier time distinguishing the two puzzles than do some mathematicians, because one principle quickly learned in the practice of science is that we cant generalize from a sample size of one. In this case, the fact that the host opened a goat door on this one occasion tells us nothing as to why he opened it or whether he would have done the same thing on other occasions if we had made a different initial choice. If I recall, Monty Hall made the same point in an interview some years ago.
If, after you pick a door, Monte allows you to switch your door for both of the other two, you should switch, even though you know full well that at least one of the other two doors has a goat behind it. Don't let Monte confuse you by showing you what you already know.
Reply | Report Abuse | Link to thisrtissot
if monty does not know your choice there is no advantage in switching if he does not open the door of your choice to reveal a goat
Reply | Report Abuse | Link to thisReply to foxgrover1
Reply | Report Abuse | Link to thisYes this is a perfect example of why this problem is almost always asked wrongly and why most people think it makes no difference if you switch.
You are not including the SORT procedure that Monty performs on two doors as per the Monty Hall Problem&
The only way you can double your odds is if you take part in Montys SORT procedure. In order to do so you must be informed that this is what Monty will perform on the two remaining doors. That is, he will look behind both doors and reveal a goat. This is a SORT operation.
If you choose the door that he does not reveal then you inherit the odds of BOTH these doors. You make use of the information Monty gives you from his SORT procedure.
If you stick with your original choice (and not switch) then you gain no information as Monty always reveals a goat in the remaining doors and you already knew that there was a goat in either or both of these other doors anyway. Hence your odds remain the same.
Foxgrover1 problem is a good example how to see the two scenarios.
SENARIO 1 - Classic Monty Hall Problem.
Classic Monty Hall problem. The problem is divided into two. The door you picked and Montys SORT operation on the other two doors.
Note in this Problem the car can never be revealed.
SENARIO 2 Monty reveals a goat which had the possibility of being the car.
If Monty opens a door which is a goat but could have been a car then this is a different problem. No SORT procedure is done on two doors. By virtue that a car could have been revealed increases the odds of the remaining doors by eliminating a door that could have had the car.
So over many trials the problem that foxgrover1 describes is scenario 2. Monty could have revealed the car when both Bob and Mike choose goat doors. In this case their odds immediately reduce to zero. If a car is not revealed (as per foxgrover1s example) their odds increase to 50:50.
A SCENARIO 2 version of foxgrovers1 problem which includes a SORT by Monty is as follows.
Bob is asked to choose a door.
Monty then says to Mike you can do either of two things. You can simply choose another door or I can look behind both these doors and I will tell you which one definitely has a goat. Mike would be crazy not to take Montys offer to tell him which definitely has a goat and choose the other. Mike must know that if the car is behind one of these two doors and if he chooses the door that is not the goat then he will win the car. That is 2/3s of the time.
Bob does not get the chance to take part in Montys SORT offer. He must stick to his chosen door with its odds of 1/3.
Who would you rather be Bob or Mike in this situation?
Let's look at it this way: If we change the rules a little so that Monty has to show the worst of the three prizes - and the worst is obvious as it usually is - then we really do get new information when he reveals the goat (presumably always the worst prize). In this case, if he doesn't show you the goat, it must be because he can't. That means you picked it, so switching wins 100% of the time. However, this only occurs 1/3 of the time.
Reply | Report Abuse | Link to thisThe other two thirds of the time you picked either the car or the second-worst prize. In this case switching works half the time - one third of the time you picked the car and one third of the time you picked the second worst prize, so it could go either way.
So in this case switching works two times out of three - if you picked the goat or if you picked the second worst prize. Agreed?
But in real play Monty could show you either prize but not the car. Does this change the odds? Advocates of the 50% chance cite the added information from Monty revealing a losing door. However, when you get real information - when he doesn't show you the goat in the above example - it actually made switching work two times out three.
So let's look again at the random bad prize scenario. If you picked the car, he can show you either remaining door. Each one gets picked 1/3 x 1/2 of the time. If you picked the worst prize, he must show you the second worst prize. This happens 1/3 of the time. And if you picked the second worst prize, he must show you the worst prize. Again, this happens one time out of three.
Now if Monty is really picking randomly as above then he reveals the worst prize half the time and the second worst prize half the time. SInce the prize that he shows doesn't affect your decision, no new information has been added.
However, let's look back at the odds again. Of the 50% of the times he shows you the worst prize, 1/6 of the time is when you picked the car and 1/3 of the time is when you picked the second worst prize. Similarly, of the 50% of the times he'll show you the second worst prize, 1/6 is when you picked the car and 1/3 is when you picked the worst prize.
No matter which prize he shows, you are better off switching.
It appears that the presence or absence of that tidbit of additional information doesn't change the odds. This makes sense because the odds of you having picked the car initially remain the same and Monty's choice of doors is still constrained.
Whether Monty knows what's behind the doors doesn't really change anything. If he doesn't know, then he will reveal the car 1/3 of the time. Clearly this will put a damper on the proceedings, so we're still only interested in the 2/3 of the time when he doesn't show you the car. This reduces the two scenarios to one.
Reply | Report Abuse | Link to thisStop invoking mysticism to solve this problem! The game doesn't change once Monty reveals a goat. There is no quantum collapse of probabilities. This follows a straightforward decision graph. You make a choice that has 1/3 chance of being right. Monty reveals a losing prize. Since your initial guess was likely wrong, it makes sense to switch because2/3 of the time Monty has no choice which prize he reveals. You switch betting that Monty was forced to reveal a particular door, backed up with the knowledge that your original pick was probably wrong.
Reply | Report Abuse | Link to thisIf Monty or Savant open the car door, which happens 1/3 of the time, then the rest of the game loses interest. Therefore the two games become one. If Monty actually did reveal the car 1/3 of the time, then contestants would get upset because they can't make a choice to switch hoping to win it. The reason most contestants don't switch on Let's Make a Deal is because they are no more able to wrap their minds around this problem than a lot of the posters here.
Reply | Report Abuse | Link to thisNote that in the Savant game, if the car is not revealed you still win more often by switching for the same reasons as in the Monty puzzle.
Johnnymac is correct in stating that if the contestant is not informed the host will always open an unchosen door to reveal a goat and then offer a switch, the probability of winning by switching is 1/2 (the Savant and Shermer scenarios), whereas if that information is provided, the probability is 2/3 (the Monty Hall puzzle scenario). Although I don't remember the details of Monty Hall's game show, it's my impression that one of its features was unpredictability - hence more reminiscent of Savant/Shermer (p = 1/2), than the 2/3 probability of the mathematical Monty Hall puzzle.
Reply | Report Abuse | Link to thisIn the real world, there would be innumerable ways to fashion the odds into almost any desired value. If the notion of randomness independent of host decision-making appeals to you, as well as a 50/50 proposition, here is one scenario consistent with a Savant/Shermer outcome of p = 1/2. The three doors are considered to be goat 1, goat 2, and car. Before each show, the host flips a coin - heads then requires him to open goat 1, and tails goat 2 after the contestant makes an initial choice. If the door he opens is not the initial choice, we end up with the Savant/Shermer, "would you like to switch?" scenario, with a 1/2 p value for winning. This would occur 2/3 of the time. On the other 1/3 occasions, he will open the contestant's initial choice to reveal a goat, and will then tell the contestant, "you now get to choose one of the remaining doors". Once more, the probability is 1/2, suspense is maintained, and everyone is entertained.
Just to add to my previous comment, it's important to note that the probabilities we discuss are those that a contestant can estimate from information available to that contestant. For the game show host, the probability that a contestant will win a car by switching to a given door is always exactly 1 or zero, because he has the information needed for that result. The reason why for a contestant, p = 2/3 in the Monty Hall puzzle is well explained by Michael Shermer and can be reviewed for that analysis (or simply plotted out on a grid). For the Savant/Shermer variation, the p value of 1/2 simply reflects the fact that contestant's only relevant knowledge is that the car is behind one of the two remaining doors. The contestant could conjecture that he/she is facing the Monty Hall scenario, or a scenario where the host offers a choice to lure the contestant away from a winning first choice, or a scenario where the host knows the contestant will lose without a switch (respective p values of 2/3, 0, and 1) - or any other scenarios including the one I suggested above - but these would simply be guesses and would not be useful enough to modify a reasonable estimate of 1/2.
Reply | Report Abuse | Link to thisIf the host always opens a losing door a la Savant/Shermer, but the contestant doesn't know this, for whom is the probability 2/3? Certainly not for the host (p = 1 or 0), nor the contestant (p = 1/2). The value of 2/3 would be valid only for those who know what the host always does, but don't know which door wins on this particular occasion. Again, the overall point is that probability is not a fixed, inherent property of an event, but a function of one's knowledge about that event.
In the last paragraph I think Fred Moolten means the mathematical Monty Hall problem - that is 'always opens a losing door'.
Reply | Report Abuse | Link to thisAll Shermer reveals is that Monty knows what is behind the doors and on this occasion reveals a goat.
Fred Moolten describes two different problems.
Reply | Report Abuse | Link to this1) The Savant/Shermer problem which has the solution p=1/2 if you switch
And
2) The mathematical Monty Hall Problem which states that Monty always reveals a goat and which as the solution p=2/3 if you switch.
I would like to take issue with the description of the mathematical Monty Hall problem. In particular the phrase always reveals a goat .
Always reveals a goat is not a mathematical term and I think the phrase makes the problem rather ambiguous for the following reasons.
1) Most people consider this problem to be a problem where they put themselves in the contestants position on the day. The phrase always reveals a goat is not really applicable to such a situation and only confuses the problem. How is the contestant possibly to know that Monty always reveals a goat when they have walked up on stage to make their decision?
2) Always is a vague description when it comes to mathematics. What does it actually mean.?
Does it mean?
That for the entire show which could be 2, 5, 9 or any other number of episodes Monty has managed to reveal a goat.
Now if this is the case two things could be happening:
i) Monty has been revealing a door during each episode and by pure coincidence has revealed a goat each time. In this case the solution is p=1/2 if you switch.
ii) Monty knows what is behind the doors and purposely only reveals a door with a goat in which case the solution is p=2/3 if you switch.
Always reveals a goat is not a mathematical term.
In any infinite number of tosses of a true coin there will always exist a finite sequence of heads or tails or any other combination you can image, of any size.
So to remove all ambiguity in this problem the problem should be describe by saying, MONTY TELLS YOU THAT HE KNOWS WHAT IS BEHIND BOTH OF THE REMAINING BOXES AND THAT THIS ONE DEFINITELY HAS A GOAT BEHIND IT
I can not think of a more simple and precise way to state the problem that removes all ambiguity.
In my encounters with the Monty Hall puzzle, most presenters have taken pains to point out that for the puzzle to clearly yield the 2/3 answer, the contestant must be informed in advance of making a choice that Monty will open a second door to reveal a goat. In that sense, the term "always" means "in every circumstance regardless of initial choice". In fact, the term "always" is often used explicitly, as in the wikipedia explanation - "The standard analysis of the problem assumes that the host is indeed constrained always to open a door revealing a goat," It doesn't mean "in every show", and so I hope I've succeeded in clarifying this uncertainty. Unless the intention to open a goat door is is stated in advance, the contestant cannot know whether the host's decision to open the goat door might have been determined by the initial choice, and the 2/3 answer would no longer be correct.
Reply | Report Abuse | Link to thisThere is no 2/3 advantage in this problem. Your odds of winning are in fact 50/50. The reason the illusion of the 2/3 chance occurs is because mathematically you are evaluating the entire game: two separate rounds. The conditions set forth for the statement can be simplified: Regardless of what you choose, Monty will always reveal a goat, and will never reveal the door you chose. This forces the game to go to the second round. This is where the illusion stems from. The illusion is that your first choice does not matter at all, because the game always goes to a second round. In this second round, you have two choices, car or goat. Two choices of action, stay, or switch. The first round gives you absolutely no information because if you chose a goat, he will reveal a goat, and if you chose the car, he will reveal a goat. Since all of your choices result in the same outcome, the choice has no impact whatsoever. So even if your odds of randomly selecting a goat are 2/3 in the first round, since you are given no indication if your original selection and the resulting goat revelation was because of car selection or goat selection. So in the second round, the only one which choice matters, you have two possible choices, with two possible outcomes, car or goat, stay or switch, win or lose.
Reply | Report Abuse | Link to thisSimulations naturally will show 2/3 success because the simulation is programmed based off of the first round; it uses irrelevant information to calculate the outcome, therefore skewing it.
Dear Dr. West:
Reply | Report Abuse | Link to thisJust come across the "Monty Hall Problem" in the latest issue of Scientific American--poked around a bit--and your name came up. As with most people, my initial take is to say its 50/50 for a change. I am overwhelmed by the folks, learned yet, that believe otherwise. And I am a newby to the MHP--my view is that the counter-intuitive solution is either a hoax (not likely) or a bad application of the implicit assumptions that go into statistical analysis.
My current take is that the situation is not probabilistic--its deterministic. And statistical analysis is not appropriate for this situation, it pops up undependable artifacts because of is structure. I do not have time to read all the stuff (book(s) yet) that have been written on the subject. And I am inviting folks to try and change my view.
There is NOT a probability that the car is behind any given door--its behind one of them, and not behind two of them (or just one of them of two, after Monty, always, opens a goat door, specifically 1,0,0 or the like, not 1/3, 1/3,1/3--and then after Monty opens its 1,0--and take your 50-50 pick. Maybe the process of selection in the mind of the contestant has some sort of randomness associated with the process...but not the placement of the car and goats--that's a done deal before the show. Completely determined.
Try this: otherwise standard scenario--then Monty asks if the contestant wants to change--but he dies before he can answer. Not wanting to to sued, they find the next of kin to stand in--one having no idea what went on before--got two doors--car behind one, goat behind the other. Kin picks--does it matter to him/her what the now dead contestant picked ? Does he/she ask Monty and then switch because of the tripe that has been published on the history--or just pick one or the other. Actually it seems to me it does not matter what the selection history was, he could have been lying to Monty as to which door he "believed" was the right to pick.
At that juncture its is my view that its 50/50 no matter what, (even a resurrection of the original contestant)--of course. Common sense prevails here--correct me if I am wrong.
And if you have any time for me (I am a 64 year old physicist) I have some musings on the "properties of randomness" (oxymoron?) and if it really exists--- and to what extent do we, on faith, or otherwise, implicitly incorporate the concept of randomness (and its properties), rightly or wrongly, in our scientific striving.
Mike Conaway
Reply to Fred Moolten
Reply | Report Abuse | Link to thisI believe this problem is intuitive when put to someone in an explicit way, where they can put themselves in the contestants position to decide what choice is best.
I think we have an innate intuition when we are presented problems and that this intuition can generally be relied on.
I believe that the controversy is generally caused by the way people (almost deceptively - if not for overdue complication) present this problem. If it is presented abstractly as a mathematical problem with a set of rules and constraints then people are not able to put themselves in a position of the contestant and find it difficult to use intuition.
There are two very similar problems here. It is as though the person presenting the problem wants to describe the problem as close as they can to the Savant/Shermer problem without actually doing so, in order to enjoy watching the confusion of the problem solver.
Fred Moolten states that for the puzzle to clearly yield the 2/3 answer; the contestant must be informed in advance of making a choice that Monty will open a second door to reveal a goat.
I know what he is getting at here, but I also think he is presenting the problem with undue complication, further clouding the problem and making it ever more difficult to solve the problem in an intuitive manner. It is adding one more layer of complication.
Freds description.
The Contestant is asked to come on down and take part. Monty says Behind one of these three doors contains a car. You are to pick a door. Once you have chosen I will reveal a goat behind one of the remaining doors. I will then give you the option to switch doors
Apart from this rather unnaturally awkward presentation by Monty, I personally find this one level too far for me to easily use my intuition to see the answer (especially on the moment of the day).
I think it is a fair to assume that the show host is fair and unbiased.
Shows like this, like casinos are aware of their risks, (in fact the car manufacture promoting the car is likely to know the risks and the cost of donating cars (advertising) on the show).
It is just as likely that Monty wants you to win the car to drum up audience enthusiasm as he would perhaps want you to lose the car.
In the contestants mind, weighing up this evidence, I do not think he/she would have information on whether Monty is biased, or if Monty is, in which direction this bias sways. That is, the contestant has no information that they could use to decide whether Monty is bias and if this bias sways for or against winning the car.
So there is no constraints here (There is no use for the contestant to play mind games about Montys intentions) when he reveals a goat door and the offer to switch. The contestant has no information on Montys mindset. Unless you have been given information about Montys mindset, you must assume on this one-off situation that Monty is impartial.
A much more realistic, honest and uncomplicated way to present this problem to someone, without intentionally trying to instil confusion or complication is:
Monty asks you to come on down. Three doors. One has a car, the other two goats. Pick a door.
Monty says:
"OK you have picked this door. I am now going to look behind these two remaining doors and reveal a door that definitely has a goat behind it. Then I am going to offer you the choice to switch to the remaining door which I do not reveal."
I think most peoples' intuition would tell them that that if a car was behind one of the remaining doors that they did not pick, then they will win it 100% if they switch.
An even more honest approach to the same problem is:
Reply | Report Abuse | Link to thisMonty asks you to come on down.
Three doors, one has the car, the two others goats.
Monty says I am going to separate these door. Two on that side of the stage and one on this side.
Pick a side you think the car is.
The contestant intuitively picks the side with the two doors to maximise his odds.
Monty says bring you wife down to look behind the two doors on the side you have chosen and to open one of the doors that definitely has a goat behind it.
Monty then says, Ok we have two closed doors remaining. Would you like to switch your choice to the singe door on other side of the stage.
Of course not......
The professor is wrong.
Reply | Report Abuse | Link to thisLets start by debunking the three rules which seem to be an attempt to mystify the problem.
Rules 1 and 2 are stating the obvious and are implicit in the real world scenario given; the host will not show you what you have chosen before you commit to an answer and the host will not reveal the location of the car.
Rule three is redundant; when your initial choice was correct both remaining doors conceal a goat so it doesn't make a blind bit of difference how Monty chooses.
The notion that you increase your chance of choosing the car by switching ignores the new information provided in round two. In round two you are presented with a simple choice of two doors and the knowledge that one conceals a goat and one a car. It matters not a jot how you came to be in that situation. It is analogous to the chance of throwing a six on a die. The chance is always one sixth no matter how many previous sixes you have thrown. Similarly the chance of picking the car in round two is one half regardless of whether or not you picked correctly, incorrectly or just refused to pick in round one.
By combining the probability of the two rounds the professor is explicitly ignoring the new and useful information provided by Monty for round two. Mathematically you only need round two but that would make for a shorter show.
YOUR ALL WRONG. If your standing out there faced with two doors FORGET MATHS. Even if there is a scientific way of doing it (which I highly doubt), faced with the chance of winning a car you won't even remember it!
Reply | Report Abuse | Link to thisBut writing a book on what would seem such a trivial subject seems a great feat.
let me simplify it for you.
Round one: pick any of the three doors
[Monty narrows it down for you to two doors,making the first round a complete waste of time]
Round two: pick either of the doors
You have now one either a car or a goat. Enjoy!
Wow, lots of controversy here.
Reply | Report Abuse | Link to thisIt really is simple. Three doors. Behind one of them is a new car. Behind each of the other two, a goat. The object is to win a car. Monte knows what is behind each door. You pick a door. Monte reveals a goat behind one of the doors you did not pick. Monte offers you the chance to stay with your original pick, or switch to the other remaining closed door. If you switch, you'll win 2/3 of the time. If you do not, you'll win 1/3 of the time.
Consider that after your initial choice, you've got a 1/3 chance of having picked the car. After Monty opens a door and shows you a goat, the probability of the car being behind the door you chose is 1/3. That does not change by revealing the goat. What DOES change is that the probability of the car being behind the OTHER closed door increases from 1/3 to 2/3. Accordingly, you should take the opportunity to switch to the other closed door.
If you disagree, consider if there were 1 million doors. You pick a door. You have a 1/1,000,000th chance of picking the car. Monte opens 999,998 doors, leaving the door you picked, and one other door. Do you switch? Of course. The probability that the other door has the car is 999,999/1,000,000.
Hope that clears it up,
-Mark
Upon further thought and consideration the [ahem] professor is right, BUT his rules are as I said (or meant to say) pointless and can be summed up by saying the only rule is that monty won't pick a goat and if he did he would probably be fired.
Reply | Report Abuse | Link to this-The Noble Acceptor Of The Truth
no it doesn't mark. the analogy would be for monty to open 333333 doors.
Reply | Report Abuse | Link to thiscould you explain why you think the probability of the car being behind the other closed door increases to 2/3 ?
The easiest way to explain it is, what if there's 100 doors and 99 goats? And after picking one, the host reveals 98 doors of goats. Should you switch?
Reply | Report Abuse | Link to this