Probability of Reaching a Chip Target Under Bold Betting

You start with $A$ chips and want to reach $B$ chips, where $A$ and $B$ are integers with $B > A > 0$. Each round you place an even-money bet that you win with probability $p$: a win adds the amount you staked, and a loss removes it. You play boldly: holding $x$ chips, you stake whichever is smaller, all $x$ chips or the $B - x$ chips you still need. Play stops when you reach $B$ chips or run out. What is the probability that you reach $B$ chips before going broke? The specific values of $A$, $B$ and $p$ are generated randomly, and each question states the required output format. ```hint Follow the chip count Write the success probability from each chip count in terms of the counts one round can lead to, and check whether that chain of counts always ends. ``` ### Constraints and Clarifications - $A$ and $B$ are integers with $B > A > 0$, and $p$ is a probability; write $q = 1 - p$. - Bets pay even money, and rounds are independent. - With $x$ chips the stake is $\min(x, B - x)$, so every stake is a whole number of chips. ### Clarifying Questions - Does a winning bet pay exactly the stake, or some other multiple of it? - When you hold exactly half the target, is the stake all of your chips, which is also exactly what you need? - Is the answer wanted as an expression in $p$, or as a number for the given values? ### What a Strong Answer Covers - The transition from each chip count under bold play, written as one recurrence with both boundary values - Recognizing that the chain of chip counts can revisit a state, and solving the resulting linear equation instead of recursing forever - A sanity check in the fair case - A comparison with betting one chip per round, and when each strategy is better ### Follow-up Questions - If you bet one chip per round instead, what is the probability of success, and for which $p$ is bold play better? - What is the expected number of rounds under bold play? - How does the analysis change if a winning bet pays twice the stake?