Let $x$ denote player $A$'s score before a toss.
There are two strategies to consider . . .
- Player $A$ doubles only at $x=1$.$\\[4pt]$
- Player $A$ doubles at both $x=0\;$and $x=1$.
First suppose player $A$ doubles only at $x=1$.
Let $W_k$ be the probability that player $A$ wins exactly $2^k$, and let $L_k$ be the probability that player $A$ loses exactly $2^k$.
Starting at $x=0$, let $p$ be the probability of reaching $x=1$ before reaching $x=-2$.
It's easily shown that $p={\large{\frac{2}{3}}}$.
There's no way for $A$ to win exactly $1$, hence $W_0=0$.
For $A$ to lose exactly $1$, the score, starting at $x=0$, must reach $x=-2$ before reaching $x=1$, hence $L_0=1-p={\large{\frac{1}{3}}}$.
For $A$ to win exactly $2$, the score, starting at $x=0$, must reach $x=1$ before reaching $x=-2$, and once reaching $x=1$, the next toss must be $\text{H},\;$hence
$W_1=(p)\bigl({\large{\frac{1}{2}}}\bigr)=\bigl({\large{\frac{2}{3}}}\bigr)\bigl({\large{\frac{1}{2}}}\bigr)={\large{\frac{1}{3}}}$.
Considering the $2$-toss sequences $\text{TH},\text{HT},\;$we get the recursion
$$
\begin{cases}
W_k=\frac{1}{4}W_k+\frac{1}{4}W_{k-1}&\text{for}\;k\ge 2\\[4pt]
L_k=\frac{1}{4}L_k+\frac{1}{4}L_{k-1}&\text{for}\;k\ge 1\\
\end{cases}
$$
or equivalently,
$$
\begin{cases}
W_k=\frac{1}{3}W_{k-1}&\text{for}\;k\ge 2\\[4pt]
L_k=\frac{1}{3}L_{k-1}&\text{for}\;k\ge 1\\
\end{cases}
$$
It follows that
$$
\begin{cases}
W_0=0\\[4pt]
W_k=\left(\frac{1}{3}\right)^k&\text{if}\;k\ge 1\\[4pt]
L_k=\left(\frac{1}{3}\right)^{k+1}&\text{if}\;k\ge 0\\
\end{cases}
$$
Letting $e$ denote $A$'s expectation using this strategy, we get
\begin{align*}
e&=
\sum_{k=0}^{\infty}\left(W_k\right)\left(2^k\right)
-
\sum_{k=0}^{\infty}\left(L_k\right)\left(2^k\right)\\[4pt]
&=
\sum_{k=1}^{\infty}\left({\small{\frac{2}{3}}}\right)^k
-
\left({\small{\frac{1}{3}}}\right)\sum_{k=0}^{\infty}\left({\small{\frac{2}{3}}}\right)^k\\[4pt]
&=2-1\\[4pt]
&=1
\end{align*}
Next, suppose player $A$ doubles at both $x=0$ and $x=1$.
Using this strategy, let $e$ be player $A$'s expectation.
For sequences of $2n$ tosses where the game ends on toss $2n$, there is a one-to-one correspondence between sequences where $A$ loses, and sequences where $A$ wins, obtained by using the exact same toss sequence except reversing the last two tosses from $\text{TT}$ to $\text{HH}$. Call such sequences conjugates of each other. But for any pair of conjugate sequences, if the losing one loses $2^k$, the winning one wins $2^{k+1}$, hence the conditional expectation, given that the game ends after $2n$ tosses, is positive.
It follows that $e > 0$ (possibly positive infinity).
Suppose $e$ is finite.
Starting at $x=0$, and considering the $2$-toss sequences $\text{TT},\text{TH},\text{HT},\text{HH}$, we get the recursion
$$e=\left(\small{\frac{1}{4}}\right)(-2)+\left(\small{\frac{1}{4}}\right)(2e)+\left(\small{\frac{1}{4}}\right)(4e)+\left(\small{\frac{1}{4}}\right)(4)$$
which yields $e=-1$, contrary to $e > 0$.
It follows that $e=+\infty$.
Thus, if the goal is to maximize expectation, then doubling at both $x=0$ and $x=1$ is optimal.
But the conjugate of a big win is a big loss (though only half as much), so this strategy, though it has expectation positive infinity, and offers the potential for huge gains, also risks huge losses.
The strategy has no bearing on the impact of the game. You can equivalently consider this game to be a fair random walk, starting at $N$, ending at either $0$ or $2N$ and with the option to select whether the next move will take you $2$ or only $1$ step back or forward (the choice alternates between you and an opponent).
Denote $P(X)$ as the probability we hit $2N$ before hitting $0$ starting from $X$.
Now in any strategy, consider the last step where either player selects $2$ as the jump size, after which we will use $1$ as long as the game goes on. (This point must exist), and let's assume the game is at point $X$
At this point, if the player selects $2$, chances of reaching $2N$ chances are $0.5*(P(X-2)+P(X+2))$.
If player selects $1$ instead, chances of reaching $2N$ are now $0.5*(P(X-1)+P(X+1))$.
Since we have that the game afterwards is an unbiased random walk, we use (from theory) the fact that the probability of a random walk at $X$ going to $2N$ is merely $X/2N$. Hence, by substituting this into the equations above, we find that both options are equivalent.
The completion of the proof simply takes this "backwards" to show that at no point, does the selection of $2$ versus $1$ matter.
The crux of this depends on the fact that the probability of reaching $2N$ is linear in terms of $X$, ie, $f(A+B)=f(A)+f(B)$. Otherwise, the player currently winning/losing might select $2$ over $1$.
And, since we showed this game is equivalent to a random walk, starting at the midpoint, then it follows that it is a fair $1/2$ game.
Best Answer
Yes, your reasoning looks correct to me. There is a 0.5 chance of a H on the first throw. In that event A exercises the option. If the next throw is H he wins 2. If the next throw is T, then his expected win is 0.
On the other hand, there is a 0.5 chance of a T on the first throw. In that event, A does not exercise the option. If the next throw is T he loses 1. If the next throw is H, then his expected win is 0.
So adding that up, with the option his expected win is $0.25\times2-0,25\times1=0.25$, compared with an expected win of 0 without the option. So the option is worth 0.25.
Note that A the only time A benefits from exercising the option is immediately after a H on the first throw. If there is a head and he delays, then either there is a H on the next throw and it is too late - the game is over - or there is a T on the next throw and he has no advantage. Similarly, if there is a T on the first throw his position is worse, so it hurts him to exercise the option.