Re: Probability of two birthdays happening on the same day
PJs Ronin - That's where the OP started, though. = )
sanderj's answer really covers this nicely, but to explain the logic:
Originally Posted by RhysGM
Using your second method actually just double-counts cases where more than two people have the same birthday. You're right that for any two people, the chance would be 1/365. That's where the proper method starts, too, after all. But think about what you're testing. In the group of three, the first test is for A and B to be the same, which is a 1/365 chance, and the second test is for B and C to be the same, also a 1/365 chance, and the third for A and C, the same. If A, B, and C are all the same (a 1/365^2 chance,) that meets the conditions for both tests, so if you simply add the probabilities, it gets counted three times, once for each case - but it's still just one case where two people share a birthday (and incidentally a case where three people do, but the original question being answered doesn't care about that.) If you take 3/365 and subtract out 2/365^2, you get the proper probability.
I guess that for a group of five, you'd need to remove four cases where three but not four or five people shared a birthday, four cases where four but not five people shared a birthday, and four cases where five shared a birthday. So for a group of 28....
Last edited by Copper Bezel; May 22nd, 2013 at 07:58 AM.
Reason: Typos and stuff left in from an earlier version; I need to proofread things.
I know I shouldn't use tildes for decoration, but they always make me feel at home~