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Assume player A, B throw a pair of fair dices in turn. Player A is the winner if the sum of two dices is 5 at the round he throw and player B is the winner if the sum of the dices is 9 at his turn. Once any one of the player achieve their number, the game end and he will become the winner.What is the expected number of rolls such that A or B wins

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Why do you slice your question? Anyway, here is an exercise: adapt the decomposition proof on the other page, to reach an answer to the present problem which avoids the use of series and the like. – Did Nov 5 '12 at 15:34
It is asking different things, do you mean i have to ask on the same question? – Mathematics Nov 5 '12 at 15:46
up vote 1 down vote accepted

Hint: what is the chance the game ends when A tosses-call it $a$? what is the chance when B tosses-call it $b$? The expectation is then $1\cdot a + 2(1-a)b + 3(1-a)(1-b)a +4(1-a)(1-b)(1-a)b + \ldots$ where I have written it (number of rounds)(chance to not end before that round)(chance to end that round)

Added: A simpler approach is to let $n$ be the number of total rounds, with a round being a toss by A and maybe one by B. You play another round with probability $(1-a)(1-b)$ so the expected number of rounds is $\frac 1{(1-a)(1-b)}$. Given that you end on a particular round, you should be able to calculate the chance that A wins, and therefore the chance that you play only one game.

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Is the chance the game ends when A tosses = the probability A wins? – Mathematics Nov 5 '12 at 14:29
@Mathematics: it is the probability that A wins that time. that is how I read your third sentence. – Ross Millikan Nov 5 '12 at 14:34
well, i think that's true but how do you solve that summation?? That form seems complicated – Mathematics Nov 5 '12 at 14:49
It is basically a geometric series with each term multiplied by $n$. You could look at Eric Naslund's answer here for guidance. – Ross Millikan Nov 5 '12 at 15:00
Is n geometric distributed? It seems the second approach is much faster. – Mathematics Nov 5 '12 at 16:30

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