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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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no subject
Date: 2010-02-18 08:03 (UTC)From:no subject
Date: 2010-02-18 08:08 (UTC)From: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.
no subject
Date: 2010-02-18 08:20 (UTC)From: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
no subject
Date: 2010-02-18 17:51 (UTC)From:Choose(11,3)*(1/365)^2*Prod(364,356,-1)/365^8
no subject
Date: 2010-02-18 09:32 (UTC)From:Though a couple of incorrect logics give approximately correct answers. Like "there are 165 triplets, and the chance of a random triplet having the same birthday is 1/365^2, so 165/365^2" Or the slightly better "1-(1-1/365^2)^165"
no subject
Date: 2010-02-18 17:34 (UTC)From:(would you post the code?)
no subject
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.