Coin Toss: Who Wins With Extra Heads

This interview question sets up a simple head-to-head coin-tossing game where two players flip different numbers of fair coins and compare how many heads they obtain. The candidate is asked to determine the chance that the player with more coin flips wins the game, framing it as a probability comparison between two related random totals. It looks elementary on the surface, but it probes whether the candidate can see beyond brute-force enumeration and recognize that the only structural difference between the players is a single extra coin toss.

To tackle it cleanly, the problem leans on discrete probability, binomial distributions, and symmetry arguments. An elegant solution reframes the situation in terms of identical random components plus a tiebreaker, avoiding heavy combinatorial sums. Interviewers watch for recognition of symmetry between the two players' shared coins, correct identification of when outcomes are "tied," and the ability to reason abstractly about distributions rather than computing every case. They also look for clear conditioning or counting logic, sanity checks on probabilities, and justification that no hidden bias favors either side apart from the extra trial.

What it tests

When comparing two random processes that are nearly symmetric—such as two players each tossing coins—the key insight is that the probability distribution of outcomes often exhibits a deep symmetry, especially when the underlying random variables are identically distributed and independent. If the only asymmetry is a single extra trial (like one player tossing an extra coin), the problem can often be reframed: the extra trial serves as a tiebreaker when the symmetric part of the process results in a draw. The symmetry of the binomial distribution with $p=1/2$ means that, for any number of coins, the chance of one player getting more heads than the other is the same as the reverse. The only remaining task is to account for ties, which, if resolved fairly (like by a final coin toss), preserves the overall symmetry and leads to equal probabilities of winning for both players. This principle holds because the binomial distribution is mirror-symmetric about its mean when $p=1/2$, and independent trials ensure that no player has an inherent advantage except for the tiebreaker.