# Probability to break deuce and win in tennis

Two players are playing tennis, and are at deuce. The first player has a 60% chance of winning each time they play a point, and the second players has a 40% chance. What is the probability that the first player wins the match?

I understand that there are two basic scenarios: Player 1 wins two consecutive points to cinch the set, OR {Player 1 Loses, Player 1 Wins (and ties it back to deuce), Player 1 Wins, Player 1 Wins}.

This sets up a sort of recursive logic, which I thought should be $$p = 0.6^2 + 0.4(0.6)p$$.

However, after attempting, I find that the solution has it as $$p = 0.6^2 + 2(0.4)(0.6)p$$, which I don't understand. Where does this 2 come from?

$$p=0.6^2+0.6\cdot 0.4\cdot p+0.4\cdot 0.6\cdot p=0.6^2+2\cdot 0.6\cdot 0.4\cdot p$$
If player 1 wins the next two points, player 1 wins the game. If player 1 wins the first point and then player 2 wins the second point, which occurs with probability $$(0.6)(0.4)$$, or player 2 wins the first point and then player 1 wins the second point, which occurs with probability $$(0.4)(0.6)$$, the game returns to deuce, so the probability that player 1 wins the game after the next deuce is again $$p$$. Hence, the probability that player wins the game if the game is tied at deuce is $$p = 0.6^2 + [(0.6)(0.4) + (0.4)(0.6)]p = 0.6^2 + 2(0.6)(0.4)p$$