Problem 145 Solution

The expected gain by A on the first roll is (1/36) * $1 =~ 2.78 cents. On all future rolls they have an equal chance of winning so there is no gain on the next 999,999. Thus A's expected gain over any number of rolls is 2.78 cents.

The probability A wins the first bet is 7/13.

The probability A wins the second bet is (7/13)*(7/13)+(6/13)*(6/13) = 85/169, using logic from earlier problems.

The probability A wins the second bet is (85/169)*(7/13)+(84/169)*(6/13) = 1099/2197.

Continuing this pattern, the probability A wins the n^th bet is (((13ⁿ+1)/2)/13ⁿ)*(7/13)+(((13ⁿ-1)/2)/13ⁿ)*(6/13) =

((13ⁿ+1)/2)/13ⁿ⁺¹.

The expected gain by A on the first bet is (7/13)*(+1) + (6/13)*(-1) = 1/13.

The expected gain by A on the second bet is (85/169)*(+1) + (84/169)*(-1) = 1/169.

The expected gain by A on the n^th bet is 1/13ⁿ.

Taking the sum from 1 to one million the total gain by A is:

$∑$ (for i=1 to 1,000,000) 1/13ⁱ =~ (1/13)/(12/13) = 1/12 =~ 8.33 cents.

Thanks to Extra Stuff: Gambling Rambling by Peter Griffin for this problem. See chapter 6.

Michael Shackleford, ASA, August 19 1999

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