One extra coin: chance of out-heading the opponent

Question

Player A flips $n+1$ fair coins and player B flips $n$ fair coins. What is the probability that A ends up with strictly more heads than B?

Short answer

$1/2$ . A has either more heads than B or more tails than B, never both and never neither, since A has one extra coin. Swapping heads and tails maps one event to the other, so each has probability $\tfrac12$ .

Adapted from Zhou, A Practical Guide to Quantitative Finance Interviews, sec. 4.1 Basic Probability Definitions and Set Operations, 'Coin toss game (n+1 vs n)', p.60; question and answer rewritten by SelfMadeQuant.com.

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Lessons behind this question

Sample Spaces, Events, and the Probability Axioms
Conditional Probability, Total Probability, and Bayes' Rule

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