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.
finding a probability of a deuce in a tennis game
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probability