Suppose that $A$ plays strategy $2$ with probability $p_H$ when the coin comes up heads and probability $p_T$ when it comes up tails, and suppose that $B$ plays $Y$ with probability $q$. The expected payoff to $A$ when the coin comes up heads is then
$-(1-p_H)+p_H(1-q)+2p_Hq=2p_H+p_Hq-1\;,$
and the expected payoff to $A$ when it comes up tails is
$-(1-p_T)+p_T(1-q)-2p_Tq=2p_T-3p_Tq-1\;,$
so the expected payoff to $A$ is
$p_H+p_T+\frac{q}2(p_H-3p_T)-1\;.$
For a given $q$ and $p_H$ this is clearly maximized when $p_H=1$, when it is $f(p_T,q)=\left(1-\frac32q\right)p_T+\frac{q}2\;.$
Now $f_{p_T}(p_T,q)=1-\frac32q\quad\text{ and }\quad f_q(p_T,q)=-\frac32p_T+\frac12\;,$
so the unique critical point of $f$ is at $p_T=\frac13$ and $q=\frac23$. Let $p'=p_T-\frac13$ and $q'=q-\frac23$. Then
$\begin{align*} f(p_T,q)&=\left(1-\frac32\left(q'+\frac23\right)\right)\left(p'+\frac13\right)+\frac12\left(q'+\frac23\right)\\ &=p'-\frac32p'q'-p'+\frac13-\frac12q'-\frac13+\frac12q'+\frac13\\ &=\frac13-\frac32p'q'\;. \end{align*}$
At the critical point, therefore, the expected payoff to $A$ is $\frac13$. However, this is a saddle point: if $p_T$ and $q$ are either both increased or both decreased, the expected payoff to $A$ decreases, but if they are changed in opposite directions, it increases. Thus, if the opponent’s probabilities are known, either player can punish the other for deviating from those at the critical point, and in that sense the value of the game is $\frac13$ to $A$.