You are playing a game where both you and your opponent have each put $10 in the pot. Your opponent now bets another $10. What is the minimum probability of winning required for you to call this $10 bet?

Free worked solution · Quant Blueprint

25% to break even; more than 25% for a strictly profitable call, ignoring ties, fees and further betting.

1. Your original $10 is already in the pot, so it is sunk when choosing whether to call. After the opponent’s bet, the pot is $30.
2. A call costs a new $10. Winning returns the $40 final pot, a $30 gain relative to folding now; losing costs the extra $10.
3. The incremental expected value is 30p − 10(1−p) = 40p − 10. Set it to zero to obtain p = 1/4.

Common mistake

Charging the original $10 again when evaluating the current decision double-counts a sunk cost. Compare calling with folding at this point.

Follow-up

For a pot P before your call and a call cost C, the break-even probability is C/(P+C). Future betting or unequal tie payouts require a fuller model.

Original educational explanation for this practice question. Reviewed 2026-10-13. Firm-style practice; the label describes the collection’s preparation context.