Two friends each stake $$10$ on opposite teams. Partway through the game, the first friend raises the stake to $$20$; if the second friend refuses to match, he forfeits the $$10$ already staked. Let $p$ be the probability, at the time of the raise, that the second friend's team wins. What is the smallest $p$ at which he should accept?

$1/4$. Folding loses $$10$ for sure. Accepting wins or loses $$20$, worth $20p-20(1-p)=40p-20$. Accept when $40p-20\ge-10$, i.e. $p\ge\tfrac14$.

Interview prep / Probability

When to call a doubled sports wager

Question

Two friends each stake $\text{\char36}10$ on opposite teams. Partway through the game, the first friend raises the stake to $\text{\char36}20$ ; if the second friend refuses to match, he forfeits the $\text{\char36}10$ already staked. Let $p$ be the probability, at the time of the raise, that the second friend's team wins. What is the smallest $p$ at which he should accept?

Short answer

$1/4$ . Folding loses $\text{\char36}10$ for sure. Accepting wins or loses $\text{\char36}20$ , worth $20p-20(1-p)=40p-20$ . Accept when $40p-20\ge-10$ , i.e. $p\ge\tfrac14$ .

Adapted from QuantQuestions.io; question and answer rewritten by SelfMadeQuant.com.

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