I'm trying to do the following problem in Sheldon Ross' A First Course In Probability (page 86-87).
EXAMPLE $4k$. Suppose that initially there are $r$ players, with player $i$ having $n_i$ units, $n_i > 0, i = 1,\dots,r$. At each stage, two of the players are chosen to play a game, with the winner of the game receiving $1$ unit from the loser. Any player whose fortune drops to $0$ is eliminated, and this continues until a single player has all $n = \sum_{i=1}^r n_i$ units, with that player designated as the victor. Assuming that the results of successive games are independent and that each game is equally likely to be won by either of its two players, find $P_i$, the probability that player $i$ is the victor.
I've also just solved another problem from Ross' book, the solution of which I thought could help with the one above.
Independent trials resulting in a success with probability $p$ and a failure with probability $1 − p$ are performed. What is the probability that $n$ successes occur before $m$ failures?
The answer to this problem is $$P_{n,m} = \sum_{k=n}^{m+n-1} {{m+n-1} \choose k} p^k(1-p)^{m+n-1-k}$$
Now back to example $4k$. I was considering some player $i$. This player can win anything between $0$ and $n-n_i$ units, and lose anything between $0$ and $n_i$ units. Furthermore, the player would win the game if she can get $n-n_i$ wins before losing $n_i$ times. Applying the formula above, with the corresponding $p = 1 - p = \frac{1}{2}$, we would have: $$P_i = \sum_{k=n-n_i}^{n-1} {{n-1} \choose k} \left({\frac{1}{2}}\right)^k \left({\frac{1}{2}}\right)^{n-k-1} = \frac{1}{2^{n-1}}\sum_{k=n-n_i}^{n-1} {{n-1} \choose k}$$
However, the correct answer turns out to be $P_i = \frac{n_i}{n}$, which is different from what I got. I'd like to know where I went wrong with the above argument.