# Birthday probability problem

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. $(365*364*1*364)/(365*364^3) = 1/364$

• What does the presence of $S$ change to this problem?b – Marc van Leeuwen Jul 12 '12 at 16:18
• 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 $1/364$. – André Nicolas Jul 12 '12 at 18:08

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 $=\frac{364}{364+364*363}=\frac{364}{364^2}=\frac{1}{364}$
• 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 $132496 = 364^2$. – Théophile Jul 12 '12 at 18:18
• I too got $1/364$ but my teacher says its wrong. I guess he's wrong this time. – Bazinga Jul 13 '12 at 2:53