What's the probability that 3 out of 23 people will share the same birthday?

[FlatAssembler]

The probability that 2 out of 23 people will share the same birthday is slightly higher than 50%, and that is the famous Birthday Paradox. However, what is the probability that 3 out of 23 people will? I have, like I have written in this article about something distantly related to that, estimated numerically using Monte Carlo method that it is around 1.26%, and that the probability that 4 out of 23 people will share the same birthday is around 0.018%. However, I am interested whether there is a general formula for that.

[ignoramus]

ssssssss

[FlatAssembler]

ssssssss

What does that mean?

[BrianSoddingBoru4]

It's one of those paradoxes that isn't. Here's the roughest way to look at it.

If you have a group of 23 people, the number of possible pairs is 253: 23x22/2. The odds of shared birthdays is found by dividing the possible pairs by the number of days in a year. 253/365 is just over 69% (the actual percentage is much closer to - but still above - 50%).

Boru

[ignoramus]

ssssssss

What does that mean?

[FlatAssembler]

It's one of those paradoxes that isn't. Here's the roughest way to look at it.

If you have a group of 23 people, the number of possible pairs is 253: 23x22/2. The odds of shared birthdays is found by dividing the possible pairs by the number of days in a year. 253/365 is just over 69% (the actual percentage is much closer to - but still above - 50%).

Boru

Right, it is more like the Monty Hall Problem: something which is not hard to analyze mathematically, but the result is surprising to most people.

[Jehanne]

Yes, there is a general formula, which can be reasoned as follows:

1) The probability of an event plus its complement (the event not occurring) is 1 (or, "unity").

2) While not being entirely true, we may assume, for convenience, that every day of the year is equally likely to be born on.

3) The probability of being born on any one day is 1/365 for non-leap years.

4) If a single person is born on one day, there are 364 other days for another individual to be born on, such that the two individuals do not have the same birthday. For the third individual, there would be 363 days, etc.

5) 1 - (365 * 364 * 363...) / 365 ^ n would give the probability that 2 or more individuals would have the same birthday.

P.S. Okay, it's 4 AM, and, so, I misread your question a bit. Yes, getting the probability for exactly three individuals is quite a bit more tricky.

[brewer]

I don't share my birthday with anybody, the cake is all mine damnit.

[arewethereyet]

Another fascinating thread.

[FlatAssembler]

I don't share my birthday with anybody, the cake is all mine damnit.

This is a serious question, and I do not expect joke answers.

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Another fascinating thread.

Are you being serious or ironic, I cannot tell?