Observations and results
Did about 50 percent of the groups of 23 or more people include at least two people with the same birthdays?
When comparing probabilities with birthdays, it can be easier to look at the probability that people do not share a birthday. A person's birthday is one out of 365 possibilities (excluding February 29 birthdays). The probability that a person does not have the same birthday as another person is 364 divided by 365 because there are 364 days that are not a person's birthday. This means that any two people have a 364/365, or 99.726027 percent, chance of not matching birthdays.
As mentioned before, in a group of 23 people, there are 253 comparisons, or combinations, that can be made. So, we're not looking at just one comparison, but at 253 comparisons. Every one of the 253 combinations has the same odds, 99.726027 percent, of not being a match. If you multiply 99.726027 percent by 99.726027 253 times, or calculate (364/365)253, you'll find there's a 49.952 percent chance that all 253 comparisons contain no matches. Consequently, the odds that there is a birthday match in those 253 comparisons is 1 – 49.952 percent = 50.048 percent, or just over half! The more trials you run, the closer the actual probability should approach 50 percent.
More to explore
"Understanding the Birthday Paradox" from BetterExplained
"Probability Central" from Oracle ThinkQuest
"Combinations and Permutations" from MathIsFun
"The Birthday Paradox" from Science Buddies
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9 Comments
Add CommentThis statement is wrong:
Reply | Report Abuse | Link to this"Every one of the 253 combinations has the same odds, 99.726027 percent, of not being a match."
Correct formula for probability of NOT having two persons in a group of 23 to share birthday (excluding 29 Feb):
P = 365! / ((365-23)! * 365^23)
Or, in other words:
P = (365/365)*(364/365)*(363/365)*...*(343*365) ≈ 0.492703 = 49.2703%, which is slightly less than 49.952%.
Simple explanation is that when we do a survey, first person can have any birthday (365/365); second person can have any birthday apart from first person's birthday, so 364 choices, hence (364/365); third person can have any birthday, apart from first and second persons' birthday, (363/365), etc.
With your reasoning nobody would have the same birthday. You are assuming no replacement, like when you remove a card from a deck of cards. Birthdays have replacement.
Reply | Report Abuse | Link to thisActually, il--ya is absolutely correct. The author's formula is close, but not technically correct. The problem is that birthdays are not like coin flips. The chances of individual pairs having the same birthday are not independent (if person 1 = person 3, and person 3= person 5, then person 1 = person 5). The practical difference is not very much as there is not much of a probability of such multiple matches among 23 people, but it exists.
Reply | Report Abuse | Link to thisLook at what happens if use 366 people and the author's formula. The result says there is a non-zero probability of them having different birthdays . . . a non-sensical result.
In problems of this sort, il--ya's approach is usually the correct: calculate the opposite and subtract from 1. This works because the answer you are trying to find and the opposite of that answer combine to equal 100% of all outcomes.
I've never met anyone who has the same birthday as me. Is this unusual? My birthday is 31st. June.Just joking!
Reply | Report Abuse | Link to this"If you use a group of 366 people—the greatest number of days a year can have—the odds that two people have the same birthday are 100 percent."
Reply | Report Abuse | Link to thisNo comment.
The author is correct.
Reply | Report Abuse | Link to thisThink of one person in a roonm and a second person entering...the odds are 365 to one that they have the same birthday. A third person enters and the odds halve...then a fourth...etc. Each added person is a variable for the next person to match. Odds are not one-on-one but collective.
geojellyroll, OK, to keep it simple, let's imagine that we live on a planet where a year lasts for only 3 days. Now let's calculate the probability of the event that three random persons have all different birthdays, using authors method:
Reply | Report Abuse | Link to thisFor 3 persons we need to make 1+2=3 comparisons.
"Every one of the 3 combinations has the same odds". For each person, probability of NOT having the same birthday is (3-1)/3 = 2/3. Probability of not having the same birthday for all three of them, following author's logic, is
(2/3)*(2/3)*(2/3)= 8/27
Now let's look at the sample space. I will list all possible combinations of birthdays for person 1, 2 and 3.
Day 1 - Day 1 - Day 1;
1-1-2
1-1-3
1-2-1
1-2-2
1-2-3 (all different)
1-3-1
1-3-2 (all different)
1-3-3
2-1-1
2-1-2
2-1-3 (all different)
2-2-1
2-2-2
2-2-3
2-3-1 (all different)
2-3-2
2-3-3
3-1-1
3-1-2 (all different)
3-1-3
3-2-1 (all different)
3-2-2
3-2-3
3-3-1
3-3-2
3-3-3
There is 27 possible outcomes in total. All outcomes are equally possible. Of these outcomes, only 6 satisfy our requirement. So the odds are 6/27. Which is different from what we've got following the author's method.
Using the method which I suggested, we get:
(3/3)*(2/3)*(1/3)=6/27, which is the correct answer.
il--ya
Reply | Report Abuse | Link to thisNo, you are still treating each sets of odds individually. The odds are cumulative. The addition of all possibilitiuers.
The suggestion of using MLB players is not a good one. MLB players birthdays are weighted to earlier in the year and are not a random sample.
Reply | Report Abuse | Link to thisSee http://www.socialproblemindex.ualberta.ca/RelAgeMLB.pdf