Tennis Match of Probability Part 2

There are 4 ways Peter can win the game:

Jacob wins 0 rounds

Jacob wins 1 round

Jacob wins 2 rounds

The score becomes a deuce, and Peter wins afterward

The probability of Peter winning 4 rounds with Jacob winning 0 rounds is ${(0.7)}^4$ .

If Peter wins the game while Jacob wins 1 round, then 5 rounds are played in total. The last round has to be won by Peter, so there are 4 possibilities for the round Jacob won. Therefore, the probability of Peter winning the game with Jacob winning 1 round is $4 \times 0.3 \times {(0.7)}^4$ .

If Peter wins the game while Jacob wins 2 rounds, then 6 rounds are played in total. The last round has to be won by Peter, so there are ${5 \choose 2}=10$ possibilities for the 2 rounds Jacob won. Therefore, the probability of Peter winning the game with Jacob winning 2 rounds is $10 \times {(0.3)}^2 \times {(0.7)}^4$ .

We can calculate the probability of the fourth case happening by multiplying the probability of the score becoming a deuce at some point by the probability Peter wins afterward. For the score to become a deuce, 3 rounds must be won by each player. There are ${6 \choose 3}=20$ ways to choose the 3 rounds Peter wins, so the probability of the score becoming a deuce is $20 \times {(0.3)}^3 \times {(0.7)}^3$ . We can call the probability that Peter wins after a deuce $p$ , and then find an expression equal to $p$ to solve for it. Let's consider the first 2 rounds that are played after the deuce. The probability of Peter winning both of them is ${(0.7)}^2=0.49$ , and if Peter does win both of them, the probability of him winning the game is 1 (because he has already won). Now, let's determine the probability that one round is won by each player. There are 2 possibilities for the round Peter won, so the probability of this happening is $2 \times 0.3 \times 0.7=0.42$ . If each player wins one round, the score goes back to a deuce, so the probability of Peter winning is $p$ once again. So, $p=0.49+0.42p$ . Solving for $p$ gives that $p$ is approximately 0.8448276. Then, we multiply 0.8448276 by the probability the score will become a deuce at some point, so the probability of the fourth case happening is $0.8448276 \times 20 \times {(0.3)}^3 \times {(0.7)}^3$ .

To find the overall probability Peter wins the game, we add up the probabilities we determined in each of the 4 cases, to get ${(0.7)}^4+4 \times 0.3 \times {(0.7)}^4+10 \times {(0.3)}^2 \times {(0.7)}^4+0.8448276 \times 20 \times {(0.3)}^3 \times {(0.7)}^3$ , which simplifies to be about 0.9007889, which is $\boxed {90.079}$ as a percentage rounded to the nearest thousandth.