Suppose $n$ competitors in a tournament organize a sweep stake on the result of the tournament. Their names are placed in an urn, and each player pays a dollar to withdraw one name from the urn. The player holding the name that wins the tournament is awarded the pot of $n$.
(a) Show that the probability that exactly $r$ players draw their own name is $$\frac{1}{r!}\bigg\{ \frac{1}{2!} - \frac{1}{3!} + ... + \frac{(-1)^{n-r}}{(n-r)!}\bigg\}$$
This I understand and I can derive it.
(b) Given that exactly $r$ such matches occur, what is the probability that Fred draws his own name?
This I am struggling to answer. Fred is either part of the $r$ matches or he is not. This gives us an idea. I can condition on on Fred being part of the $r$: Let $F$ be the event that Fred draws his own name, $R$ be the event that exactly $r$ matches occur and $Q$ be the event that Fred is part of the group of matches.
$$P(F|R) = P(F|R,Q)P(Q) + P(F|R,Q^c)P(Q^c) = P(F|R,Q)P(Q)$$
I am a bit stumped here. As I have not achieved anything.
@user2661923 gave a simple approach. Here is an alternative approach that applies Bayes' Theorem and the result from part (a): $$\mathbb{P}(F | R) = \frac{\mathbb{P}(R | F)\mathbb{P}(F)}{\mathbb{P}(R)} = \frac{\frac{1}{(r-1)!}\left(\frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^{(n-1)-(r-1)}}{((n-1)-(r-1))!}\right) \frac{1}{n}}{\frac{1}{r!}\left(\frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^{n-r}}{(n-r)!}\right)} = \frac{r}{n}$$