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. $(365*364*1*364)/(365*364^3) = 1/364$
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. $(365*364*1*364)/(365*364^3) = 1/364$
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 $=\frac{364}{364+364*363}=\frac{364}{364^2}=\frac{1}{364}$
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.
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.