Guessing red from a lopsided deck

Question

A shuffled deck has 30 red and 20 black cards, dealt face-up one at a time. At any moment you may stop and bet that the next card is red; you win if it is. You always know the counts remaining. What is the best win probability any stopping rule can achieve? Prove your answer.

Accepted Answer

Answer: exactly 30/50 = 0.6. No stopping rule beats it. The lopsided deck does not change the shape of the argument, only the starting value. Martingale view. Let r and b be the reds and blacks remaining. The probability the next card is red is \frac{r}{r+b}, the current fraction of reds. Revealing one card leaves that fraction's expectation unchanged (a red card, probability \frac{r}{r+b}, lowers it; a black card raises it; the two balance exactly), so the red fraction is a martingale. It starts at \frac{30}{50} = 0.6. By the optional stopping theorem its expected value at any stopping time …

Guessing red from a lopsided deck

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