Here is my attempt, I would appreciate feedback:
Let $p$ be the probability that player A wins a single point.
(1) Player A can win after 40-0, with probability $p^4$
(2) Player A can win after 40-15 with probability ${4\choose 1}\times p^4(1-p)$
(3) Player A can win after 40-30 with probability ${5\choose 2}\times p^4(1-p)^2$
(4) Player A can reach deuce, with probability ${6\choose 3}\times p^3(1-p)^3$, and once deuce is reached, player A will win with a probability of $\frac{p^2}{1-2p(1-p)}$ (http://www.austinrochford.com/posts/2013-04-25-probability-and-deuces-in-tennis.html)
Therefore the probability player A will win the game is given by,
$$P(Win) = p^4 + {4\choose 1}\cdot p^4(1-p) + {5\choose2}\cdot p^4(1-p)^2 + {6\choose 3}\cdot \frac{p^5(1-p)^3}{1-2p(1-p)}$$
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:
$$P(\text{$A$ wins}) = P(7-0) + P(7-1) + P(7-2) + P(7-3) + P(7-4) + P(7-5) + P(6-6)P_D.$$
Best Answer
P(wins first point) = 3/5
P(wins second point) = 1/3
so he can win those two points with probability 1/5 and win in first two points - but if he wins the first point, then loses the second point, with probability 2/5 - he is back to square one - if he loses the first point and wins the second point (prob 2/15) - he is back to square one - if he loses two points - he lost (prob 4/15)
so after two points, he either wins, loses, or is back to square one with the same probability of winning - therefore P(winning) = M = P(wins in two points) + M x P(back to a draw after two points)
P(winning) = M = (3/5)(1/3) + (3/5)(2/3)M + (2/5)(1/3)M + 0 x(2/5)(2/3)
M = 1/5 + (2/5)M + (2/15)M
M = 3/15 + (6/15)M+ (2/15)M
M = 3/15 + (8/15)M
(7/15)M = 3/15
M = 3/7
Answer - 3/7