Probability challenge

[mindstalk]

Yeah, I'm behind on posting news links and such. Moving on!

Given 11 people, what is the probability that 3 of the 11 share the same birthday? Assuming uniform distribution of birthdays in the population, of course.

This seems a lot harder than the classic problem of finding whether at least two have the same birthday. I debunked a couple of answers, came up with a nice one of my own for "at least 3 share, and I don't care about anyone else" which came nowhere near simulation, then someone else came up with an answer for "3 people share a birthday, and the other 8 don't share any birthdays" which matched debugged simulation, and that one's actually fairly easy in retrospect (in my defense, it wasn't the problem I'd been thinking about.) I still don't know why I'm so far off for the "at least" case.

Comments

[thomasyan.livejournal.com]

Gah, I read this just before going to bed. Damn you!

[dogofjustice.livejournal.com]

First approximation: What is the chance that at least three people share a birthday of January 1st? Let's ignore leap years.

1 - P(nobody has that birthday) - P(one person has it) - P(two have it)
= 1 - (364/365)^11 - 11 * (364/365)^10 * (1/365) - 55 * (364/365)^9 * (1/365)^2
= about 3.3378e-6

So, double-counting the cases where two birthdays are triple-shared and triple-counting the three birthday case, we get a 0.12183% chance.

For a precise answer, we'd use inclusion-exclusion. Subtract 365*364/2 times the chance a particular pair of birthdays is triple-shared, and add 365*364*363/6 times the chance a particular triplet is triple-shared. This is a bit tedious since you need to subtract six different terms from 1 to calculate the former chance, and ten different terms to get the latter.