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I have a probability of Player1 (1) winning a point with a probability of $\frac{3}{4}$, P2 (2) then then has a probability of winning of $\frac{1}{4}$. I need to find the probability that a deuce will be reached in the game (meaning the score will be 40-40, it goes 0, 15, 30, 40). One example of such scenario would be $\frac{3}{4}\times \frac{3}{4}\times \frac{3}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$ which equals to $\frac{27}{4096}$. It can be observed if I am seeing correctly that any game resulting in deuce at some point will be a product of $(\frac{3}{4})^{3}$ and $(\frac{1}{4})^{3}$ in various orders. My question is, how do I compute the overall probability, not just for one specific outcome, that the game will be in a deuce stage? I want to say it's 2*2*2*1*1*1 times $\frac{27}{4096}$, but I do not think that is correct way of accounting for the different orders that can be.

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You have a deuce after 6 points if each player wins exactly 3 of them. This happens with probability: $\binom{6}{3} (1/4)^3 (3/4)^3 = \frac{135}{1024} \sim 13.18\%$

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Deuce will occur after 6 points have been played in the game and each player winning 3 each. The number of ways of each player winning $3$ points each is given by $\binom{6}{3}$. Hence, the desired probability is $$\binom{6}{3} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^3 = \frac{135}{1024}$$

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What does $\binom{6}{3}$ stand for, and how is it calculated? Could you give me a link to an article etc.? – user9466 Apr 12 '11 at 6:06
@Davania: $\binom{6}{3}$ is the number of ways of choosing $3$ out of $6$. $\binom{n}{r} = \frac{n!}{r! (n-r)!}$ where $k! = 1 \times 2 \times \cdots k$. You can look up the wiki article on combination for more details. – user17762 Apr 12 '11 at 6:08

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