Isn't hammer throw was a field sport?

You and your opponent are competing in a golf match. On any given hole you play, each of you has a $50$ percent chance of winning the hole (and a zero percent chance of tying). That said, scorekeeping in this match is a little different from the norm.

Each hole is worth $1$ point. Before starting each hole, either you or your opponent can “throw the hammer.” When the hammer has been thrown, whoever did not throw the hammer must either accept the hammer or reject it. If they accept the hammer, the hole is worth $2$ points. If they reject the hammer, they concede the hole and its $1$ point to their opponent. Both players can throw the hammer on as many holes as they wish. (Should both players decide to throw the hammer at the exact same time—something that can’t be planned in advance—the hole is worth $2$ points.)

The first player to reach $3$ points wins the match. Suppose all players always make rational decisions to maximize their own chances of winning.

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

Let's define your win probability as the function $w : \mathbb{N}^2 \to [0,1],$ so that $w(s_1, s_2)$ is the probability of winning the match given that the current score is $s_1$ points for you and $s_2$ points for your opponent, assuming that you are both playing perfectly rationally, maximizing you own chances of winning. Let $u$ be the indicator of whether or not you throw the hammer when the score is $(s_1, s_2).$ Let $v$ be the indicator of whether or not your opponent throws the hammer when the score is $(s_1,s_2).$ Then we can define the recursion formula $$\begin{align*} w(s_1,s_2) &= \max_{u \in \{0,1\}} \min_{v \in \{0,1\}} \left[uv \frac{ w(s_1+2, s_2) + w(s_1, s_2 + 2) }{2} \right. \\ &\quad\quad\quad + (1-u)(1-v) \frac{ w(s_1 + 1,s_2) + w(s_1, s_2 + 1)}{2} \\ &\quad\quad\quad\quad + u(1-v) \min \left\{ \frac{w(s_1+2, s_2) + w(s_1, s_2 + 2)}{2}, w(s_1+1,s_2) \right\} \\ &\quad\quad\quad\quad\quad \left.+(1-u)v \max \left\{ \frac{w(s_1+2, s_2) + w(s_1, s_2+2)}{2}, w(s_1, s_2+1) \right\} \right],\end{align*}$$ where the coefficients of $u(1-v)$ and $(1-u)v,$ represent the optimal choice of whether or not to accept or reject the hammer, by your opponent and you, respectively. As boundary conditions, we have $$w(s_1,s_2) = \begin{cases} 1, &\text{if $s_1 \geq 3 \gt s_2$;}\\ 0, &\text{if $s_1 \lt 3 \leq s_2.$}\end{cases}$$

In particular, we see that for instance $$\begin{align*}w(2,2) &= \max_u \min_v \left[ uv \frac{ 1 + 0}{2} + (1-u)(1-v) \frac{1 + 0}{2} + u(1-v) \min \left\{ \frac{1+0}{2}, 1 \right\} + (1-u)v \max \left\{ \frac{1+0}{2}, 0 \right\}\right] \\ &= \max_u \min_v \frac{1}{2} = \frac{1}{2}.\end{align*}$$ We also find that $$\begin{align*}w(1,2) &= \max_u \min_v \left[ uv \frac{1+0}{2} + (1-u)(1-v) \frac{\frac{1}{2} + 0}{2} + u(1-v) \min \left\{ \frac{1+0}{2}, \frac{1}{2} \right\} + (1-u)v \max \left\{ \frac{1+0}{2}, 0 \right\} \right] \\ &= \max_u \min_v \left[\frac{1}{4} + \frac{1}{4}u + \frac{1}{4}v - \frac{1}{4}uv\right] = \frac{1}{2},\end{align*}$$ with optimizers $u^*=1$ and $v^*=0.$ Similarly, $w(2,1) = \frac{1}{2},$ as well.

Working our way backward we see that $w(2,0) = \frac{3}{4}$, $w(1,1) = \frac{1}{2}$ and $w(0,2) = \frac{1}{4}.$ Therefore, we can finally arrive at the probability of winning given that you and your opponent play optimally and that the current score is $1-0$ is $$\begin{align*}w(1,0) &= \max_u \min_v \left[ uv \frac{1 + \frac{1}{2}}{2} + (1-u)(1-v) \frac{\frac{3}{4} + \frac{1}{2}}{2} + u(1-v) \min \left\{ \frac{1+\frac{1}{2}}{2}, \frac{3}{4} \right\} + (1-u) v \max \left\{ \frac{1 + \frac{1}{2}}{2}, \frac{1}{2} \right\} \right] \\ &= \max_u \min_v \left[ \frac{5}{8} + \frac{1}{8} u + \frac{1}{8} v - \frac{1}{8} uv \right] = \frac{3}{4},\end{align*}$$ with optimizers $u^*=1$ and $v^*=0.$