Help Solving for Probability in a 2-Player Multi-Round Game with Unequal Odds [duplicate]

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 1/3. 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

marked as duplicate by joriki, Michael Greinecker♦, William, Matt N., Ｊ. M.Sep 3 '12 at 23:45

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.
 +1 @André Nicolas – dato 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