(Probability) The birthday problem...P(at least 2)...DIRECT APPROACHby Applejacks01 Tags: 2direct, birthday, probability, problempat 

#1
Feb2412, 10:40 PM

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 1P(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:




#2
Feb2512, 08:53 AM

P: 325

Have you considered the possibility of one pair and one triplet?




#3
Feb2512, 05:21 PM

P: 26

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 2123...oops. Answer is still off by roughly the same margin. 



#4
Feb2512, 06:51 PM

Sci Advisor
P: 1,168

(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 countingthere is data that strongly suggests that birthdays are not distributed uniformly. 



#5
Feb2512, 06:56 PM

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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