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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.
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.
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Date: 2010-02-18 17:34 (UTC)From:(would you post the code?)
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Date: 2010-02-18 17:46 (UTC)From:It's nothing much:
from random import * triplets=0 for j in xrange(100000): bdays={} for i in range(11): b=randint(1,365) if b in bdays: bdays[b]+=1 else: bdays[b]=1 # if max(bdays.values()) == 2 and len(bdays)==8: if max(bdays.values()) >= 3 triplets+=1 print tripletsCommented line is for finding the probability of exactly 3 pairs, vs. at least one triplet-or-more.
First version was less elegant and ultimately less flexible:
from random import * triplets=0 for j in xrange(100000): bdays=[] for i in range(11): bdays.append(randint(1,365)) if len(bdays)==len(set(bdays)): continue else: for i in range(11): match=0 for k in range(i+1,11): if bdays[i]==bdays[k]: match+=1 if match>=2: triplets+=1 print tripletswhich actually inflates 'match' but for these purposes it doesn't matter because we only care about the threshold.
Python code, if that's not obvious.