# Probability that Alice Wins n Games Before Bob Wins m Games ## Problem 746

Alice and Bob play some game against each other. The probability that Alice wins one game is $p$. Assume that each game is independent.

If Alice wins $n$ games before Bob wins $m$ games, then Alice becomes the champion of the game. What is the probability that Alice becomes the champion. Add to solve later

## Solution.

If Alice won at least $n$ games in the first $n+m-1$ games, then Bob won at most $m-1$. On the other hand, if Alice won at most $n-1$ games in the first $n+m-1$, then Bob won at least $m$ games. Hence, it is necessary and sufficient for Alice to be the champion that Alice wins at least $n$ games in the first $n+m-1$ games.

(Remark that even if Alice won $n$ games before finishing $n+m-1$ games, we may assume that they continue playing until $n+m-1$ games.)

The probability that Alice wins exactly $k$ games in the first $n+m-1$ games is given by
${n+m-1 \choose k} p^k(1-p)^{n+m-k-1}$ since each game is independent.
Therefore, the probability that Alice becomes the champion is obtained by summing this probability over $k=n, n+1, \dots, n+m-1$.

Hence, we get
$P(\text{Alice becomes the champion}) = \sum_{k=n}^{n+m-1}{n+m-1 \choose k} p^k(1-p)^{n+m-k-1}.$ Add to solve later

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