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Suppose you are in a large room that random people generally enter. How many people must be in the room before the probability that some share a birthday becomes at least 50 percent?

Now, let's just think about this rationally first. Assuming there are 365 possible birthdays, there is a one in 365 chance that you share your birthday with the second person to enter the room. As long as there are just two of you, there's a 1/365 percent chance that your birthday is shared. On the opposite end of the spectrum, if there are 366 random people in the room, there is a 100 percent chance that at least one birthday is shared (this is due to the Pigeonhole Principle). So for a 50 percent chance of a shared birthday, it'll require abut halfway between 2 and 366 people. Right?

We can assume this pair have matching birthdays. |

Wrong. The answer is a mere 23 random people. And if you were looking for a 99 percent chance of any two people in a room sharing a birthday, you need only 57 people. The answer depends on pairs of people. When there are 2 people, only 1 pair is possible. With 3 people, there are 3 pairs possible. With 4 people, there are 6 possible pairs. When you get to 23 people, 253 different pairs are possible, any of which could lead to a matched birthday. For a more detailed (and designed in TeX) explanation, our math intern liked this page.

The answer is so much lower than many would expect because as humans we make assumptions that are irrelevant to the problem. Our minds automatically make the bridge to finding two particular people, or a particular birth date. We overthink automatically, when a match on

*any*date by

*any*two people is sufficient.

Yes!! I have heard about this problem too ...Interesting..

ReplyDeletewww.teachmathfree.com

I am reminded of a story of Raymond Smullyen (sp?). In a class with 19 students, he discussed this result and pointed out that the odds were considerably against two students having the same birthday in that class. Nevertheless, one student persistently insisted that he would bet that two students would have the same birthday.

ReplyDeleteSmullyan decided to show him. He began going through the list of students, asking their birthday -- then stopped in the middle humbled and embarrassed himself. Two of the students were identical twins.