= Solution
Let $I$ be the number of heads in the initial allocation tosses. Then $I$ has the <binomial distribution> with parameters $n$ and $1/2$. Given $I=i$, the winning <probability> is $i/n$, including the already finished games $i=0,n$. The <law of total probability> and <linearity of expectation> give
$$
\begin{aligned}
\mathbb P(W)&=\sum_{i=0}^n\mathbb P(I=i)\frac in
=\frac{\mathbb E I}{n}
=\frac{n/2}{n}.
\end{aligned}
$$
Thus, for $n\ge1$,
$$
\boxed{\mathbb P(H\text{ finishes with all coins})=\frac12.}
$$
Back to article page
