1. The problem statement, all variables and given/known data What is the probability that given a group of 5 people, at least 2 will share the same birthday? 2. Relevant equations I know that 1-P(0 matches) = answer, but there is a reason I am going about this the head on approach. I am trying to develop my combinatoric skills. 3. The attempt at a solution EDIT: I think I am missing some cases. What about the case where we have exactly 1 triplet and 1 pair? Hmm.. OKAY so IF I am correct, then here is what happened in my original work: When I did the case of 1 triplet, I REALLY did the case of 1 triplet and 2 different birthdays. Which means I was missing the case of 1 triplet and 2 same birthdays! So there are C(5,2) ways to choose the people to be part of the pair = 10 ways. And similar logic yields the final piece of the puzzle. My solutions are equivalent now. Anybody want to verify? I am calculating the probabilities of the following to solve this: 0 matches 1 pair 2 pairs 1 triplet 1 quadruplet 1 quintuplet Here is my table of work: The method is as follows: Code (Text): No matches has 1 way to assign the lack of matches. The # of days for first person is 365, 2nd is 364,etc.. Multiply everything together and divide by (365^5) to get the probability Code (Text): 1 pair has C(5,2) ways to assign the pairs. The # of days for first person is 365. 2nd is 1 way. 3rd is 364, 4th 363, 5th 362. Multiply all together and divide by 365^5 Code (Text): 2 pairs has 3*5 =15 ways of combinations. this is derived by observing: A = pair 1 B = pair 2 C = standalone AABBC ABABC ABBAC There are 5 ways of assigning the standalone(no match) for each group, hence 15 ways. The way of choosing days is as follows: 365*1*364*1*363 Multiply that with 15, divide by 365^5 Code (Text): For a triplet, there are C(5,3) ways to assign the matching day. The # of days are 365*1*1*364*363 Code (Text): For a quadruplet, there are C(5,4) ways to assign the matching day. the # of days are 365*1*1*1*364 Code (Text): For a quintuplet, there is only 1 way to assign the matching day, and there is 365 ways to choose the day. Please see my chart, and note that the probabilities differ by an extremely small #. I can't figure out where the error is?!?!