# (Probability) The birthday problem...P(at least 2)...DIRECT APPROACH

by Applejacks01
Tags: 2direct, birthday, probability, problempat
 P: 26 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: 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 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 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 For a triplet, there are C(5,3) ways to assign the matching day. The # of days are 365*1*1*364*363 For a quadruplet, there are C(5,4) ways to assign the matching day. the # of days are 365*1*1*1*364 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?!?!
 P: 327 Have you considered the possibility of one pair and one triplet?
P: 26
 Quote by awkward Have you considered the possibility of one pair and one triplet?
Yes! I realized last night I was missing that one! (and now my solution for the 5 person case is correct, thank you)

Okay guys, so I thought all was well, but to really test my combinatoric skills I decided to try the 8 person case. Well...once again there is a discrepancy! I will attach my chart.
Edit: Please ignore "quadrupleruple" lol...I used find/replace to change words around and that happened...oops
Edit 2: I didn't actually sum rows 21-23...oops. Answer is still off by roughly the same margin.
Attached Thumbnails

 Sci Advisor P: 1,170 (Probability) The birthday problem...P(at least 2)...DIRECT APPROACH Are the people in your group indistinguishable from each other? I mean, if you label them 1,2,3,4,5 , are you distinguishing between ,say, 1,2 having the same birthday or 2,4, etc? Also, in case you're interested in a realistic model --and not just practicing your counting--there is data that strongly suggests that birthdays are not distributed uniformly.
 P: 26 Okay for example: If 1,2 have the same birthday as 3,4 , then that is considered a quadruple, not 2 pairs. For a group of 4 people, we can have the following cases: (0 same): 1 <> 2 <> 3 <> 4 (1 pair) 1=2 and 3<>4 , 1 = 3 and 2<>4, 1=4 and 2<>3. 2=3 and 1<>4, 2=4 and 1<>3, 3=4 and 1<>2 2 pairs 1=2 and 3=4, 1=3 and 2=4, 1=4 and 2=3 (1 triplet) 1=2=3 and 4 is different, 1=2=4 and 3 is different, 1=3=4 and 2 is different, 2=3=4 and 1 is different (1 quadruplet) 1=2=3=4 And for example, suppose we have 2 birthdays: A and B _ _ _ _ = 1,2,3,4 There are only 3 ways to have 2 pairs of birthdays: AABB ABAB ABBA Note that BABA is identical to ABAB (because A can be B and B can be A, but not at the same time). Does that answer your question? And yes I am aware, but I am solving this to work on my counting skills really.

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