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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!

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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 $$\left(\binom 54+\binom 55\right)\times \frac 1{32}=\frac 6{32}$$

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    I'm sorry, it should of course be draw instead of remis. I have a German teacher. – 2017-02-14
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    Never heard that German word...good to know! – 2017-02-14
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    But by the way, thank you for the explanation. I understood it! – 2017-02-14