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

[dogofjustice.livejournal.com]

Oh, this deals with the "at least 3" case. "Exactly 3" is actually a bit cleaner, though:

A = P(exactly three people share a given birthday) = 11*10*9/6 * (364/365)^8 * (1/365)^3
B = P(exactly three people share a given birthday, and exactly three people share another given birthday) = 11!/(5!3!3!) * (363/365)^5 * (1/365)^6
C = P(three different given birthdays exactly triple-shared) = 11!/(2!3!3!3!) * (362/365)^2 * (1/365)^9

Exact answer = 365 * A - 365*364/2 * B + 365*364*363/6 * C

[mindstalk.livejournal.com]

Even cleaner is "exactly 3 people share a birthday, and no one else shares any"

Choose(11,3)*(1/365)^2*Prod(364,356,-1)/365^8