The winner of a tennis tie break is the first to get to 7 points and lead by 2.
Let $p$ be the probability of player 1 winning when serving, and let $m$ be the probabiliity of player 1 winning when receiving the serve.
Every time the score sums to an odd number, the server changes.
What is the probability that player 1 will win the entire tie-break?
Best Answer
The ways for Player $A$ to win the tiebreaker are with scores: $7-0,\;\;$ $7-1,\;\;$ $7-2,\;\;$ $7-3,\;\;$ $7-4,\;\;$ $7-5,\;$ and then after reaching $6-6$ gaining an advantage of $2$ points.
We assume Player $A$ serves first. Then the server of each ball will be:
\begin{matrix} 1&2&3&4&5&6&7&8&9&10&11&12&\cdots \\ A&B&B&A&A&B&B&A&A&B&B&A&\cdots \\ \end{matrix}
We'll look at the probability of each scoreline separately. For example, $7-4$. This has $11$ points, the last of which Player $B$ serves and Player $A$ wins. Player $A$ wins $6$ of the other $10$ points and this can be done in the following ways:
\begin{matrix} \text{$A$ wins $5$ of $A$'s serves and $1$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{5}\binom{5}{1}p^5m^2(1-m)^4 \\ \text{$A$ wins $4$ of $A$'s serves and $2$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{4}\binom{5}{2}p^4(1-p)m^3(1-m)^3 \\ \text{$A$ wins $3$ of $A$'s serves and $3$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{3}\binom{5}{3}p^3(1-p)^2m^4(1-m)^2 \\ \text{$A$ wins $2$ of $A$'s serves and $4$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{2}\binom{5}{4}p^2(1-p)^3m^5(1-m) \\ \text{$A$ wins $1$ of $A$'s serves and $5$ of $B$'s} & \qquad\text{Prob} = &\binom{5}{1}\binom{5}{5}p(1-p)^4m^6. \end{matrix}
The probability of $A$ to win $7-4$ is the sum of those $5$ values.
A similar method is used to obtain the probability for other scorelines:
$$P(7-0) = \binom{3}{3}\binom{4}{4}p^3m^4 $$ $$\\$$ $$P(7-1) = \binom{3}{3}\binom{4}{3}p^4m^3(1-m) + \binom{3}{2}\binom{4}{4}p^3(1-p)m^4$$ $$\\$$ $$P(7-2) = \binom{4}{4}\binom{4}{2}p^5m^2(1-m)^2 + \binom{4}{3}\binom{4}{3}p^4(1-p)m^3(1-m) + \binom{4}{2}\binom{4}{4}p^3(1-p)^2m^4$$ $$\\$$ $$P(7-3) = \binom{5}{5}\binom{4}{1}p^5m^2(1-m)^3 + \binom{5}{4}\binom{4}{2}p^4(1-p)m^3(1-m)^2 + \binom{5}{3}\binom{4}{3}p^3(1-p)^2m^4(1-m) + \binom{5}{1}\binom{4}{4}p^2(1-p)^3m^5$$ $$\\$$ $$P(7-4) = \binom{5}{5}\binom{5}{1}p^5m^2(1-m)^4 + \binom{5}{4}\binom{5}{2}p^4(1-p)m^3(1-m)^3 + \binom{5}{3}\binom{5}{3}p^3(1-p)^2m^4(1-m)^2 + \binom{5}{2}\binom{5}{4}p^2(1-p)^3m^5(1-m) + \binom{5}{1}\binom{5}{5}p(1-p)^4m^6$$ $$\\$$ $$P(7-5) = \binom{5}{5}\binom{6}{1}p^5m^2(1-m)^5 + \binom{5}{4}\binom{6}{2}p^4(1-p)m^3(1-m)^4 + \binom{5}{3}\binom{6}{3}p^3(1-p)^2m^4(1-m)^3 + \binom{5}{2}\binom{6}{4}p^2(1-p)^3m^5(1-m)^2 + \binom{5}{1}\binom{6}{5}p(1-p)^4m^6(1-m) + \binom{5}{0}\binom{6}{6}(1-p)^5m^7$$ $$\\$$
To reach $6-6$, there is no restriction on who wins the $12^{th}$ point. So,
$$P(6-6) = \binom{6}{6}\binom{6}{0}p^6(1-m)^6 + \binom{6}{5}\binom{6}{1}p^5(1-p)m(1-m)^5 + \binom{6}{4}\binom{6}{2}p^4(1-p)^2m^2(1-m)^4 + \binom{6}{3}\binom{6}{3}p^3(1-p)^3m^3(1-m)^3 + \binom{6}{2}\binom{6}{4}p^2(1-p)^4m^4(1-m)^2 + \binom{6}{1}\binom{6}{5}p(1-p)^5m^5(1-m) + \binom{6}{0}\binom{6}{6}(1-p)^6m^6$$ $$\\$$
Let $P_D$ be the probability of Player $A$ winning from $6-6$. Then we have,
\begin{eqnarray*} P_D &=& pm + (p(1-m) + (1-p)m)P_D \\ && \\ \therefore P_D &=& \dfrac{pm}{1 - p(1-m) + (1-p)m} \\ && \\ &=& \dfrac{pm}{1 - p - m + 2pm} \\ \end{eqnarray*}
Therefore, the probability of $A$ winning the tiebreaker is obtained by gathering all the components above:
$$P(\text{$A$ wins}) = P(7-0) + P(7-1) + P(7-2) + P(7-3) + P(7-4) + P(7-5) + P(6-6)P_D.$$