# 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 $\frac13$. What are Player A's chances of winning the game?

Thank you!

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## marked as duplicate by joriki, Michael Greinecker♦, William, Rudy the Reindeer, 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.
@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