Paying to win with more red pairs than the dealer
Question
A shuffled standard deck is dealt two cards at a time. Two black cards go to the dealer, two red cards go to you, and a red-black pair is discarded. Once the deck is exhausted, you win $100 if you hold more cards than the dealer, and nothing otherwise. What is the fair price for this game?
Short answer
$0. Every discarded pair removes one red and one black, so the numbers of reds and blacks left for the players are always equal. You and the dealer therefore always finish with the same number of cards, and you never win.
Adapted from Zhou, A Practical Guide to Quantitative Finance Interviews, sec. 2.2 Logic Reasoning, 'Card game (pairs)', p.5; question and answer rewritten by SelfMadeQuant.com.
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Open the worked solutionLessons behind this question
- Sample Spaces, Events, and the Probability Axioms
- Conditional Probability, Total Probability, and Bayes' Rule
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