← Jane Street Interview Insights
My first instinct was that they're basically the same because the expected number of successes is 2/3 in both cases.
Calculate the probability of winning each game by finding the complement of losing (no successful outcome in any trial) and subtracting from 1. Compare the resulting probabilities to determine which game is better, and explain the reasoning clearly.
Pro tip: Mention that this is a classic problem often called the 'Chevalier de Méré' problem, which historically motivated the development of probability theory. This shows depth and connects to the role's quantitative nature.
For each game, define the success event per trial and the overall win condition. Game 1: success per roll is rolling a 6; win if at least one success in 4 rolls. Game 2: success per roll is rolling double sixes; win if at least one success in 24 rolls.
Game 1: P(no 6) = 5/6. Game 2: P(no double sixes) = 35/36.
Game 1: (5/6)^4. Game 2: (35/36)^24.
Game 1: 1 - (5/6)^4 ≈ 0.5177. Game 2: 1 - (35/36)^24 ≈ 0.4914.
Game 1 has a higher probability of winning (≈51.8% vs ≈49.1%). Therefore, Game 1 is better.
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