A Fair Game of Chance

Alice and Bob play a fair game repeatedly for one nickel each game. If originally Alice has a nickels and Bob has b nickels, what is Alice's chances of winning all of Bob's money, assuming the play goes on until one person has lost all her or his money?

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Solution

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Copyright © 1996-2017 Alexander Bogomolny

a / (a + b). You can check the solution.

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Copyright © 1996-2017 Alexander Bogomolny

Let p(n) be Alice's chances of winning the total amount of a + b, provided she has n nickels in her possession. Obviously p(0) = 0. If she is left with a non-zero capital, Alice may, at every trial, win or lose one nickel, both with the probability of 1/2,

  p(n) = p(n + 1)/2 + p(n - 1)/2, n > 0.

In other words, 2p(n) = p(n + 1) + p(n - 1), or p(n + 1) - p(n) = p(n) - p(n - 1). From here, recursively,

 p(n + 1) - p(n)= p(n) - p(n - 1)
 = p(n - 1) - p(n - 2)
 = p(n - 2) - p(n - 3)
 ...
 = p(2) - p(1)
 = p(1) - p(0)
 = p(1).

It follows that p(n) = n p(1) and, since, p(a + b) = 1, p(1) = 1 / (a + b). It follows that p(a) = a / (a + b).

References

  1. E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995, #494

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