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If a player is 50% as good as I am at a game, how many games will it be before she finally wins one game?

Can anyone help me solve the following problem:

Player A and Player B are playing a game with multiple rounds. The game stops once one of them wins $10$ rounds and is declared the winner. Player A's chances of winning in each round are $\frac13$. What are Player A's chances of winning the game?

Thank you!

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To be precise, the question isn't an exact duplicate, but some of the answers are. –  joriki Aug 9 '12 at 6:41
    
Thanks for pointing that out! –  malicerejoins Aug 9 '12 at 9:33
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marked as duplicate by joriki, Michael Greinecker, William, Matt N., J. M. Sep 3 '12 at 23:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

up vote 3 down vote accepted

We assume independence, which for certain sports may not be realistic. Modify the game by stipulating that whether or not somebody wins $10$ rounds earlier, the game goes on to $19$ rounds. Then A wins the original game if and only if she wins $10$ or more rounds in the modified game. Finding the probability of this is a straightforward "binomial" problem. The answer is $$\sum_{k=10}^{19}\binom{19}{k}\left(\frac{1}{3}\right)^k\left(\frac{2}{3}\right)^{19-k}.$$ Getting a numerical answer out of this by hand is a little unpleasant. Some calculators and many programs can handle it easily.

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+1 @André Nicolas –  dato datuashvili Aug 9 '12 at 7:21
    
The edit cleared up a question I was about to ask. Thanks! I'm afraid I'm not familiar with solving equations like this, whether by hand, by calculator, or by program. I tried running the equation through an online calculator and got 0.064766. How does this translate to a percentage of Player A's odds? Is it 6.48% or do I need to do something more to calculate for the percentage? Sorry, I'm not too good at math.... –  malicerejoins Aug 9 '12 at 7:43
    
@malicerejoins: Your result is correct, and yes, it means player A has a $6.48\%$ chance to win. This is also calculated in some of the answers to the question I linked to above. –  joriki Aug 9 '12 at 8:12
    
Thanks! I just got a bit confused because in the question you linked above, the computed value was 0.9352 but the chances of the girl winning the game was 7%. If I understand right, her odds were 7% based on the statement "she should win about one game out of... 15 games." In hind sight, I realized that 93% was the guy's odds. Haha –  malicerejoins Aug 9 '12 at 9:08
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