= Solution
Let $W$ be the <event> that $H$ wins, and $F$ the <event> that his first toss is a tail. For a nontrivial starting state $1\le i\le n-1$, a first tail has <probability> $1/2$ and changes his holding to $i-1$. By the <chain rule for probabilities>,
$$
\mathbb P(F\cap W)=\frac12p_{i-1}=\frac{i-1}{2n}.
$$
Since $\mathbb P(W)=i/n>0$, the definition of <conditional probability> gives
$$
\boxed{\mathbb P(F\mid W)=\frac{i-1}{2i}.}
$$
In particular, when $i=1$, losing the first coin would end the game in defeat, so the conditional answer is zero. At $i=0$ conditioning on victory is undefined, while at $i=n$ the game has already ended and no first toss occurs.
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