Compare Two Strategies for Getting Ten Consecutive Heads

A fair coin is used in two games: - **Strategy A:** Flip the coin at most 1,000 times. You win as soon as you see 10 consecutive heads. - **Strategy B:** Play 1,000 independent rounds. In each round, flip until either the first tail appears, which loses that round, or 10 consecutive heads appear, which wins the game. Start a fresh round after every loss and stop after the first win or after 1,000 rounds. Which strategy has the larger probability of winning? Derive an exact recurrence for Strategy A, derive a closed-form expression for Strategy B, and discuss whether the comparison uses the same expected number of coin flips. ### Constraints & Assumptions - Every flip is independent with probability (1/2) of heads. - A run does not carry across round boundaries in Strategy B. - A numerical implementation may use dynamic programming; a simulation alone is not a derivation. ### Clarifying Questions to Ask - Does “1,000 rounds” mean 1,000 complete attempts rather than a total budget of 1,000 flips? - Does a Strategy B round end immediately on its first tail? - Is the goal at least one successful run of 10 heads? ```hint Track the current streak For Strategy A, a state only needs the number of flips already used and the current trailing run of heads. ``` ### What a Strong Answer Covers - Correct boundary conditions and recurrence for the finite-horizon streak probability. - The per-round and 1,000-round success probabilities for Strategy B. - A justified comparison rather than treating overlapping windows as independent. - The expected-flip-budget difference between the strategies. ### Follow-up Questions - How would the recurrence change for a biased coin with head probability (p)? - How would you compare the strategies under an equal expected-flip budget? - Can the Strategy A computation be expressed as a small Markov chain?