There are four people in a room, namely P, Q, R and S.
Q's birthday is different from everyone else. What is the probability that P and R share the same birthday?
I'm getting $1/364$ as answer. $(365*364*1*364)/(365*364^3) = 1/364$
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3$\begingroup$ What does the presence of $S$ change to this problem?b $\endgroup$– Marc van LeeuwenCommented Jul 12, 2012 at 16:18
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$\begingroup$ We may assume that $Q$ was born on December $31$, leaving the other $364$ days for the others. Whatever $R$'s birthday is, the probability $P$'s matches it is $1/364$. $\endgroup$– André NicolasCommented Jul 12, 2012 at 18:08
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4 Answers
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If Q's birthday is different from everyone else's, then there are 364 choices for P and R. Thus, the probability that they are the same is indeed 1/364; the calculation need not be more complicated than that.
answered Jul 12, 2012 at 16:27
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The way I see it is to fix P's birthday, and consider R. Since we know that neither of them share a birthday with Q, there are 364 different possibilities for R, each with equal probability. One of those possibilities is P's birthday, and thus the probability is 1/364.
Alternatively, you could just note that since neither share a birthday with Q, there are 364^2 ways to choose birthdays for P and R, 364 of which result in the two of them having the same birthday. Thus, the probability is 364/(364^2) = 1/364.
answered Jul 12, 2012 at 17:08
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Two cases: If $P$ and $R$ share same birthday, the number of choices $=364$, otherwise, the number of choices $=2{364\choose 2}=364*363$. Thus, the probability that $P$ and $R$ same birthday $=\frac{364}{364+364*363}=\frac{364}{364^2}=\frac{1}{364}$
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1$\begingroup$ While this is true, it is needlessly complicated. $\endgroup$ Commented Jul 12, 2012 at 17:02
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$\begingroup$ i agree with you, but it's not so complicated and above that it's better to have solutions approached with different thoughts. $\endgroup$– AangCommented Jul 12, 2012 at 17:06
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$\begingroup$ Sure, different approaches are good. I'd say that if you're going to involve ${364\choose 2}$, though, then it would be clearer to keep it in factorized form as $\frac{(364)(363)}{2}$ rather than multiplying it out. At a glance, it isn't obvious—at least not to me—that $132496 = 364^2$. $\endgroup$ Commented Jul 12, 2012 at 18:18
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$\begingroup$ I too got $1/364$ but my teacher says its wrong. I guess he's wrong this time. $\endgroup$– BazingaCommented Jul 13, 2012 at 2:53
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The solutions proposed need to be corrected for leap year, resulting in a slightly lower probability, $p=\frac{1}{364.2}$ instead of $p=\frac{1}{364}$. My calculation follows.
In four consecutive years there are 3 years with 365 days and 1 year with 366 days, for a total $365*3+366=1461$ days. We have two possibilities for Q's birthday, Feb-29 with $p=\frac{1}{4061}$, and any other day of the year with $p=\frac{1460}{1461}$.
If Q's is Feb-29, then P and R have the other 365 days for a probability $p=\frac{1}{365}$, and the joint probability is $p=\frac{1}{1461}\frac{1}{365}$
If Q's is not on Feb-29, then P and R have each a probability of having birthdays on Feb-29 $p=\frac{1}{1460}$, and a probability of not having birthdays on Feb-29 $p=\frac{1}{364}$.
The joint probability is $p=\frac{1460}{1461}\left( \frac{1}{1460} \frac{1}{1460}+ \frac{1}{364}\right)$
Adding the two joint probabilities obtained, the result is $p=\frac{1}{1461}\frac{1}{365}+\frac{1460}{1461}\left( \frac{1}{1460} \frac{1}{1460}+ \frac{1}{364}\right) = \frac{1}{364.2}$
