So sadly, no one above has a truly correct answer due to missing factors in the problem. The first answer didn't give an answer. The second one assumed only one scenario (winning two points in a row), I can only assume that the third answer is attempting to factor in a chance of loss (most likely through the concept of dynamic programming), but the logic appears to be very flawed. In order to correctly answer this question, it must be understood that a tennis match can go on forever until one player has two points over the other player no matter how unlikely it is to occur. We're also assuming that the two players can continue to play without any change to their point win rate. If player 1 is to win, then player one needs two points, with the chance at .6^2 But, when player 2 scores a point, player 1 would need to score a point to return to deuce and then the two points would need to be scored. this would create a series that looks something like: 0.6^2 + 0.4*0.6^3 + 0.4^2*0.6^4 ... which can be written as the sum of 0.6^(x+2)*0.4^x for x from 0 to infinity. As it is, it would be incredibly difficult to solve by hand, but we can do a few transformations to the formula: 0.6^(x+2)*0.4^x=0.6^x*0.4^x*0.6^2=(0.6*0.4)^x*0.6^2=0.36*(0.24)^x Now we can factor out the 0.36 from the summation so it would be 0.36 * the sum of 0.24^x from 0 to infinity. We now have a geometric series where the r is 0.24 and the k would be x. Luckily the sum of a geometric series from 0 to infinity is 1/(1-r), so we sub in 0.24 as r and get 1/(1-0.24)=1.31578947368 The last thing we do is apply the factor of 0.36 to the series and we'll get 0.47368421052 as the chance. BUT WAIT, why is the answer less than 0.5 you ask? If you look at player 2's chances of winning (0.21052631578), you'd get a total of 0.6842105263, which is far less than the 100% that is expected. Once you factor in the chance that it's a perpetual deuce with no one winning (0.31578947368), you get the 100% that you were looking for. This is an interesting problem, but it is far more math related than it is programming related. tl;dr the correct answer is 0.47368421052.

For a shorter answer, we can describe that the probability of player one winning from a deuce with the following recursive-like definition. p = (0.6)^2 + (0.6)*(0.4)*p => p = 9/19 In other words, the probability of winning the game is equal to the probability of winning two consecutive matches plus the probability of both players winning once and returning to the double deuce state times p, the probability of winning from a deuce state.

All answers posted thus far are wrong, although the last one is the closest. The correct probability is p = (0.6)^2 + 2*(0.6)*(0.4)*p. This reduces to Player 1 having 69.2308% of winning the game. Anyone with any degree of common sense will realise that an infinitely long games is ludicrous and would require a highly improbabley sequence of points.

All answers posted thus far are wrong, although the last one is the closest. The correct probability is p = (0.6)^2 + 2*(0.6)*(0.4)*p. This reduces to Player 1 having 69.2308% of winning the game. Anyone with any degree of common sense will realise that an infinitely long games is ludicrous and would require a highly improbabley sequence of points.

I arrive at the same answer as EliteTrader, although his last sentence is somewhat odd since the recursive formula he uses does allow for the possibility of infinitely long games. The reasoning behind the recursive formula is as follows. Let's call the probability of player 1 winning the game starting at deuce "p". Starting at deuce, player 1 can win by winning two points in a row (with probability 0.60^2). Alternatively, he can win one of the next two points to arrive back at deuce again. The probability of winning one of the next two points is 0.60*0.40 + 0.40*0.60 = 2*0.60*0.40 (that is, for win-lose and lose-win). This leads to the recursive formula p = 0.60^2 + 2*0.60*0.40*p with solution p = 0.60^2/(1-2*0.60*0.40) = 0.6923076923.