Extra credit hammer throwing

Instead of playing to $3$ points, now the first player to $5$ points wins the match.

Good news (again)! You have won the first hole, and now lead $1-0.$ What is your probability of winning the match?

This game will have the same recurrence relationship, but the initial conditions will be $$\tilde{w}(s_1, s_2) = \begin{cases} 1, &\text{if $s_1 \geq 5 \gt s_2;$}\\ 0, &\text{if $s_1 \lt 5 \leq s_2.$}\end{cases}$$ Since $\tilde{w}(s_1,s_2) = w(s_1-2,s_2-2),$ we automatically have $\tilde{w}(4,4) = \tilde{w}(4,3) = \tilde{w}(3,4) = \tilde{w}(3,3) = \frac{1}{2}$ along with $\tilde{w}(4,2) = \tilde{w}(3,2) = \frac{3}{4}$ and $\tilde{w}(2,3) = \tilde{w}(2,4) = \frac{1}{4}.$ We also have $\tilde{w}(4,1) = \tilde{w}(3,1) = \frac{3}{4},$ $\tilde{w}(1,4) = \tilde{w}(1,3) = \frac{1}{4},$ $\tilde{w}(4,0) = \tilde{w}(3,0) = \frac{7}{8},$ $\tilde{w}(1,1) = \tilde{w}(2,2) = \tilde{w}(1,2) = \tilde{w}(2,1) = \frac{1}{2},$ and $\tilde{w}(2,0) = \frac{11}{16}.$

Finally, we arrive at the probability of winning the match to $5$ points if you each are playing optimally as $$\begin{align*} \tilde{w}(1,0) &= \max_u \min_v \left[ uv \frac{\frac{7}{8} + \frac{1}{2}}{2} + (1-u)(1-v) \frac{\frac{11}{16} + \frac{1}{2}}{2} \right. \\ &\quad\quad\quad\quad\quad\left. + u(1-v) \min \left\{ \frac{\frac{7}{8} + \frac{1}{2}}{2}, \frac{11}{16} \right\} + (1-u)v \max \left\{\frac{\frac{11}{16} + \frac{1}{2}}{2}, \frac{11}{16} \right\} \right] \\ &= \max_u \min_v \left[ \frac{11}{16} uv + \frac{19}{32} (1-u)(1-v) + \frac{11}{16} u(1-v) + \frac{11}{16} (1-u)v \right] \\ &= \max_u \min_v \left[ \frac{19}{32} + \frac{3}{32} u + \frac{3}{32} v - \frac{3}{32} uv \right] = \frac{11}{16},\end{align*}$$ with optimizers $u^* = 1,$ $v^* = 0.$