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Question People are arriving at a party one at a time. While waiting for more people to arrive they entertain themselves by comparing their birthdays. Let X be the number of people needed to obtain a birthday match, i.e., before person X arrives there are no two people with the same birthday, but when person X arrives there is a match. Find the PMF of X.

Attempt Method 1: If k equals the number of people at which there is a match, then choosing one out of k-1 people who share the same birthday with the kth person. There are k-1 ways to do this and each person has a probability of 1/365 to have the same birthday as the kth person, thus P(X=K) = (K-1)/365

In this method, I am considering each case to be disjoint i.e.if one of the k-1 people has the same birthday as the last person then the other k-2 people can not have the same birthday.

Method 2: In this my answer = ((k-1)/365)((364/365)^(k-2)) I have obtained the first term (k-1)/365 by choosing a person from k-1 and multiplying that by the probability of having the same birthday as the kth person And the second term (364/365)^k-2, is the probability of k-2 people not having a similar birthday to kth person.

Doubt: In method 2, I have not accounted for the fact that those k-2 people could have the same birthday, how do i do that? What is the correct approach towards this question? I would also like to know the basic fallacy in my thinking? Thanks!

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1 Answer 1

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Neither of those are correct. The best way to approach this problem is to instead work out $P(X>k)$, i.e. the probability that the first $k$ people all have different birthdays (hint: what is the probability that person $k$ has a different birthday from the first $k-1$, assuming the first $k-1$ have different birthdays?).

Then you can get what you want because $P(X=k)=P(X>k-1)-P(X>k)$.

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  • $\begingroup$ Could you please help me understand the fallacy behind both those methods @Especially Lime $\endgroup$
    – shubham kumar
    Commented Dec 6, 2017 at 9:32
  • $\begingroup$ math.stackexchange.com/a/520513/510680 This link discusses similar problem, however i don't understand if the solution is similar to what you have suggested. - @EspeciallyLime $\endgroup$
    – shubham kumar
    Commented Dec 6, 2017 at 9:49
  • $\begingroup$ @shubhamkumar the fallacy in both your methods is that you only consider the probability that the $k$th person has the same probability as exactly one previous person. You also need to make sure that there aren't two other people with the same birthday. For example, your second calculation for $k=4$ includes the possibility that the birthdays are Jan 1, Jan 2, Jan 2, Jan 1. $\endgroup$
    – Especially Lime
    Commented Dec 6, 2017 at 9:58
  • $\begingroup$ Is it correct that P(X>k-1) = (365 choose k-1)/(365^k-1) and that P(X>k) = (365 choose k)/ (365^k)? Also, is this method correct P(k) = (1 - P(k-1)) * ((k-1)/365) $\endgroup$
    – shubham kumar
    Commented Dec 6, 2017 at 10:36
  • $\begingroup$ @shubhamkumar not quite, it should be ${}^{365}\mathrm P_k$ rather than ${}^{365}\mathrm C_k$. The second formula also isn't quite right, and should be $P(k)=P(X>k-1)*((k-1)/365)$. $\endgroup$
    – Especially Lime
    Commented Dec 6, 2017 at 10:51

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