Imagine two players, each with a certain amount of money, playing a series of fair games with each other. What is the probability that one player will go broke before the other? This problem has applications in gambling and finance.
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To solve this problem, we can use a technique called "absorbing Markov chains." Let's assume Player A has a total of "a" dollars, and Player B has a total of "b" dollars. We'll define the state space as the possible combinations of the players' remaining money.
Let's denote the probability of Player A going broke before Player B, starting from state (a, b), as P[a, b]. The goal is to find P[0, b] for any b≥0.
We can set up the following recurrence relations:
1. If a = 0, P[0, b] = 1, as Player A has already gone broke.
2. If b = 0, P[0, 0] = 0, as both players have already gone broke.
3. For any other state (a, b):
- P[a, b] = 0.5 * P[a-1, b] + 0.5 * P[a, b-1]
The reasoning behind this recurrence relation is that if Player A loses one dollar, the probability of Player A going broke before Player B becomes the same as the probability of going from state (a-1, b) to (0, b) as taking Player A's move. Similarly, if Player B loses one dollar, the probability of Player A going broke before Player B becomes the same as going from state (a, b-1) to (a, 0) as taking Player B's move. The factor of 0.5 in both cases assumes that the games are fair.
We can start solving this problem using dynamic programming by calculating values of P[a, b] for all a and b in a bottom-up manner. Once we know the values of P[a, b] for a=0 and all b, we can calculate P[0, b] for any b.
The final answer would be the sum of P[0, b] for all b>0, as Player A has to go broke before Player B.
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