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Roulette puzzle
Posted By: Timothy Chow
Date: Wednesday, 28 April 2010, at 5:15 a.m.
Here is a puzzle that is due to Sheldon Axler, with slight modifications by me. I'll let people think about it, and then when I give the solution, I'll explain what this has to do with backgammon.
I have a "3 strikes and you're out" betting strategy at roulette. That is, I bet $1 on the number 1 three times in succession. Win or lose, I then move on to betting $1 on the number 2, which again I try 3 times. This continues around the wheel, until finally I bet $1 on the number 36 three times in a row, and then I quit for good. (So in total, I place 108 bets, each $1.) What is the probability that at the end of this process, I'll be poorer than when I started? Is it
a. more than 99%;
b. 90% to 99%;
c. 75% to 90%;
d. 50% to 75%; or
e. less than 50%?The puzzle is more fun if you first try to make a shrewd guess without writing anything down, and only then resort to explicit calculation.
In case you're unfamiliar with roulette, all you need to know is that if you bet $1 on a particular number, then you'll lose $1 with probability 37/38 and you'll win $35 with probability 1/38.

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