finding a probability of a deuce in a tennis game

Question

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.

Answer 1

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

Answer 2

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 $$\binom{6}{3} \left(\frac{3}{4}\right)^3 \left(\frac{1}{4}\right)^3 = \frac{135}{1024}$$