Gambler's ruin with one rebuy

Alice holds $\text{\char36}15$ and Bob holds $\text{\char36}10$. They repeatedly wager $\text{\char36}1$ on a fair coin, one taking heads and the other tails. If Alice runs out of money she may rejoin once with a fresh $\text{\char36}15$, while Bob keeps everything he has won; if she is wiped out a second time she is finished. What is the probability that Bob is the one who ends up broke?

$\tfrac34$. In a fair game the chance of ruin is your opponent's share of the total. Alice's first ruin has probability $\tfrac{10}{25}=\tfrac25$; after rebuying against Bob's $\text{\char36}25$, her ruin probability is $\tfrac{25}{40}=\tfrac58$. So Alice loses with probability $\tfrac25\cdot\tfrac58=\tfrac14$ and Bob with $\tfrac34$ — the same as if Alice simply started with $\text{\char36}30$.