What the Deuce? Djokovic V/s Federer

Another approach which doesn't require summing an infinite series is to look at it as a Markov chain.

let $p$ be the probability of Djokovic winning when a Deuce exists. Well there are 3 possibilities.

1) He gets the next two points and wins 2) he gets one of the next two points, and returns to deuce 3) he doesn't get either point and thus loses

In other words, the probability of his winning in each of these 3 cases are $1,p,$ $and$ $0$ respectively. Thus we can set up the following equation

$p=0.75^2*1+2*0.75*0.25*p$ which can be easily solved for $p=\frac{9}{10}$

As a side note I would like to mention that you did a very nice job of explaining this problem, I have neither played or watched a single game of tennis in my entire life and yet I was able to easily understand all that was needed to solve the problem. I will admit that when I first started reading it I was concerned that I would have to abandon it due to my lack of knowledge and was pleasantly surprised to find that was not the case.

To win from deuce, one of the players has to win 2 points in a row, otherwise we return to deuce.

The chance of Djokovic winning 2 points a row from deuce is 9/16.

The chance of Federer winning two points in a row from deuce is 1/16.

If we return to deuce, we're back where we started and we can reset the scenario. So this possibility can be ignored.

Therefore, the ratio of Djokovic's chances to Federer's is 9:1, giving Djokovic 9 chances out of 10 -- or 0.9 -- probability of winning.

(P.S. I'm a huge tennis fan and those are my two favorite active players.)

Let the probability of a player winning a single point be $p$ . Two points after a deuce, the game must either end or return to deuce. Thus such a game will end an even number of points after the first deuce.

The probability that the player wins two points after the first deuce is $p^2$

The probability that game returns to deuce after two points is $p\times (1-p)$ + $(1-p)\times p$ = $2p\times (1-p)$

The probability of winning the game in four points after the first deuce is therefore $2p\times (1-p)\times p^2$ .

The probability of winning a game after six points after the first deuce (meaning that the game has returned to deuce twice) is therefore $(2p\times(1-p))^2\times p^2$ .

The general pattern is probability for winning after $2n$ points =

$(2p\times(1-p))^(n-1) \times p^2$

So the probability of winning after deuce is the summation of the probabilities of winning after $2n$ points. That works out to

$\dfrac {p^2}{1- 2p\times (1-p)}$