A Game
George plays a game with two 6-sided fair dice:
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Rolling a total of 2 on any turn results in a loss (and the game ends).
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On the first turn, rolling a total of 12 results in a win (and the game ends).
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On the first turn, if a total other than 2 and 12 is rolled, then this becomes the target total.
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In the following turns, the target total must be rolled again for a win (and the game to end). However, if a total of 2 is rolled first, then this still results in a loss (and the game ends).
The probability that George wins can be expressed as a fraction , where and are coprime positive integers.
What is the value of ?
The answer is 4483.
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1 solution
(1) Let n = the total of the roll of the 2 dices. P(n) = 3 6 1 , 3 6 2 , 3 6 3 , 3 6 4 , 3 6 5 , 3 6 6 , 3 6 5 , 3 6 4 , 3 6 3 , 3 6 2 and 3 6 1 for n = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 respectively.
(2) P(George wins) = P(n=12 in the 1st turn) + P(n=target total in a subsequent turn | 3<=n<=11 in the 1st turn) , and P(n=12 in the 1st turn) = 3 6 1
(3) To determine P(George wins | 3<=n<=11 in the 1st turn), note the probability that n is NOT equal to the target total in any subsequent turn = 1-P(n)- 3 6 1 , since we need to remove the probability that n = 2. The equation for calculating P(George wins) for each n(3<=n<=11) actually turns out to be a geometric series :
(4) P(George wins | n=3 in the 1st turn) = 3 6 2 * 3 6 2 + 3 6 2 * 3 6 3 6 − 2 − 1 * 3 6 2 + 3 6 2 * 3 6 3 6 − 2 − 1 * 3 6 3 6 − 2 − 1 * 3 6 2 ........., = 3 2 4 1 * 12 = 2 7 1
(5) Similarly, P(George wins | n=4 in the 1st turn) = 3 6 3 * 3 6 3 + 3 6 3 * 3 6 3 6 − 3 − 1 * 3 6 3 + 3 6 3 * 3 6 3 6 − 3 − 1 * 3 6 3 6 − 3 − 1 * 3 6 3 ........., = 1 6 1
(6) P(George wins | n=5 in the 1st turn) = 3 6 4 * 3 6 4 + 3 6 4 * 3 6 3 6 − 4 − 1 * 3 6 4 + 3 6 4 * 3 6 3 6 − 4 − 1 * 3 6 3 6 − 4 − 1 * 3 6 4 ........., = 4 5 4
(7) P(George wins | n=6 in the 1st turn) = 3 6 5 * 3 6 5 + 3 6 5 * 3 6 3 6 − 5 − 1 * 3 6 5 + 3 6 5 * 3 6 3 6 − 5 − 1 * 3 6 3 6 − 5 − 1 * 3 6 5 ........., = 2 1 6 2 5
(8) P(George wins | n=7 in the 1st turn) = 3 6 6 * 3 6 6 + 3 6 6 * 3 6 3 6 − 6 − 1 * 3 6 6 + 3 6 6 * 3 6 3 6 − 6 − 1 * 3 6 3 6 − 6 − 1 * 3 6 6 ........., = 7 1
(9) P(George wins | n=8 in the 1st turn) = 3 6 5 * 3 6 5 + 3 6 5 * 3 6 3 6 − 5 − 1 * 3 6 5 + 3 6 5 * 3 6 3 6 − 5 − 1 * 3 6 3 6 − 5 − 1 * 3 6 5 ........., = 2 1 6 2 5
(10) P(George wins | n=9 in the 1st turn) = 3 6 4 * 3 6 4 + 3 6 4 * 3 6 3 6 − 4 − 1 * 3 6 4 + 3 6 4 * 3 6 3 6 − 4 − 1 * 3 6 3 6 − 4 − 1 * 3 6 4 ........., = 4 5 4
(11) P(George wins | n=10 in the 1st turn) = 3 6 3 * 3 6 3 + 3 6 3 * 3 6 3 6 − 3 − 1 * 3 6 3 + 3 6 3 * 3 6 3 6 − 3 − 1 * 3 6 3 6 − 3 − 1 * 3 6 3 ........., = 1 6 1
(12) P(George wins | n=11 in the 1st turn) = 3 6 2 * 3 6 2 + 3 6 2 * 3 6 3 6 − 2 − 1 * 3 6 2 + 3 6 2 * 3 6 3 6 − 2 − 1 * 3 6 3 6 − 2 − 1 * 3 6 2 ........., = 2 7 1
(13) Now, add (4) through (12) and 3 6 1 , we get 2 7 2 + 1 6 2 + 4 5 8 + 2 1 6 5 0 + 7 1 + 3 6 1 = 2 5 2 0 1 9 6 3
(14), Since the two numbers 1963 and 2520 are co-primes, the final answer is 1963 + 2520 = 4 4 8 3