# Probability of winning a match with different probability of independently winning a game in the match?

I am playing in a tennis tournament up against a player. There are three scenarios: we are equally talented, and each of us is equally likely to win each game; I am slightly better, and therefore I win each game independently with probability 0.6; or he is slightly better, and thus he wins each game independently with a probability 0.6. In our match we play until one player wins three games. In each scenario, compute the probability that he wins the match. I get the answer, $\frac{1}{2}, \frac{1}{3},\frac{2}{3}$. Could someone let me know, if the answer is correct?

The first scenario can be answered on base of symmetry: $\frac{1}{2}$ is okay.

Let $p$ denote the probability of winning a game and $q=1-p$ the probability of loosing it.

Probability of winning the match in $3$ games: $p^{3}$

Probability of winning the match in $4$ games: $\binom{3}{2}p^{3}q=3p^{3}q$

Probability of winning the match in $5$ games: $\binom{4}{2}p^{3}q^{2}=6p^{3}q^{2}$

So the total probability of winning is: $$p^{3}\left[1+3q+6q^{2}\right]$$ Check it for $p=0.6$ (hence $q=0.4$).

• Thanks, it comes out neat, I had initially wanted to ask for the winner to win three games continuously to win the match, but the answer that you have given for the said problem is right. Thanks – Satish Ramanathan Nov 24 '14 at 14:15