Problem:
I have the following problem: Richard and James are playing chess regularly. There are three possible outcomes of every match: 1) win 2) loss 3) draw. Assume that the probability of winning, losing or remis is 1/3 and that the chess games are independent.
They play 9 games. What is the probability for James to win 4 games given that Richard wins exactly 4 games?
My attempt: Because they are independent, I tried to find the total amount of combinations this can happen: $9!/(4!\cdot4!)=630\ldots$ then I just multiplied it by $(1/3)^9$. This is wrong, and I am not sure why or how to solve this problem.
What am I doing wrong, and how can I think differently to solve this?
Thanks!
Since Richard wins exactly four we can simply ignore those games. We note that James has a $\frac 12$ chance of winning any of the remaining five games, and a $\frac 12$ of tying (never heard the word "remis"). thus you are just asking about the probability that James throws at least four heads with five tosses of a fair coin. That is $$\left(\binom 54+\binom 55\right)\times \frac 1{32}=\frac 6{32}$$