A player starts with a bankroll of `10`. Each round, a biased coin is heads with probability `0.6` and tails with probability `0.4`. On heads the bankroll increases by one; on tails it decreases by one. The player stops upon reaching `20` or `0`. What is the probability of reaching `20` before going broke? Derive the formula rather than simulating, and evaluate it numerically. ### Constraints & Assumptions - Rounds are independent and probabilities remain constant. - “Winning double a one-unit bet” means the original stake is returned plus one unit of profit, so the net step is `+1`; a loss is `-1`. - Both boundaries are absorbing. ```hint Verify the stopping states first Any general expression should return certainty at the upper boundary and impossibility at zero, and should behave continuously as the coin approaches fairness. ``` ### What a Strong Answer Covers - Consistent treatment of both stopping states and the biased transition probabilities. - A derivation that uses the stated start, target, and payoff convention. - Interpretation of the high result and sensitivity to the payoff convention. ### Follow-up Questions - What would the answer be for a fair coin? - How would the expected game duration be derived? - Why must the payoff convention be clarified before using this formula?