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?
|Contact| |Front page| |Contents| |Up|
Copyright © 1996-2018 Alexander Bogomolny
a / (a + b). You can check the solution.
|Contact| |Front page| |Contents| |Up|
Copyright © 1996-2018 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(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
References
- E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995, #494
|Contact| |Front page| |Contents| |Up|
Copyright © 1996-2018 Alexander Bogomolny
64662986 |