Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
    
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.