Apparently same probability questions with different answers. I was reading A First Course in Probability by Sheldon Ross when I came up with this question.
This is how he introduces the famous problem of points:

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? ....1

He gives two approaches to solve this. One, by Pascal, by conditioning the outcome of the first trial and obtaining a relatively easily solvable functional equation. And the second, by Fermat; which argues that this is possible if and only if at least $n$ outcomes occur before $n+m-1$ trials.
Without going much detail to something I believe as classic, I state without explanation the answer (as per text, as per internet etc)
$$P_{n,m}=\sum\limits_{k=n}^{m+n-1}\binom{m+n-1}{k}p^k(1-p)^{m+n-1-k}$$
Where $P_{n,m}$ be the probability that $n$ successes occur before $m$ failures.
The same book has something I found similar to this after about 50 pages.
The context is negative binomial random variable. (Sentences in brackets are mine)

If independent trials, each resulting in a success with probability $p$, are performed, what is the probability of $r$ successes occurring before $m$ failures?
Solution The solution will be arrived at by noting that $r$ successes will occur before $m$ failures if and only if the $r$th success occurs no latter than the $r+m-1$ trial (same principle as the other one). This follows because if the $r$th success occurs before the $m$th failure, and conversely. Hence from equation 8.2 (or applying principles from negative random variable). The desired probability is
$$\sum\limits_{n=r}^{r+m-1}\binom{n-1}{r-1}p^r(1-p)^{n-r}$$

I found no visible difference in the two problems, expect the notation and different answers. What am I missing? What is the difference between the two questions?

1: Rest of the parts are irrelevant to this question and hence skipped. Yes, the 'original' problem of points is skipped. 
 A: The questions are identical and both answers are correct (i.e they are equal) but they use different approaches.
Solution 1:
The idea is to consider a series of $m+n-1$ trials and calculate the probability of getting exactly $k$ successes. The Binomial Theorem tells us that this is
$$\binom{m+n-1}{k}p^k(1-p)^{m+n-1-k}.$$
To satisfy the required criterion of at least $n$ successes we allow $k$ to range from $n$ to $m+n-1$. The resulting probabilities can be summed to give the result:
$$\sum_{k=n}^{m+n-1}{\binom{m+n-1}{k}p^k(1-p)^{m+n-1-k}}.$$
Solution 2:
I'll keep the same variables as in Solution 1 to help in any comparison between them.
The idea here is to consider a series of trials ending at the $n^{th}$ success and to calculate the probability of getting exactly $n$ successes and $k$ failures in this series. There are therefore $n+k$ trials and we will denote this number by $t$. The Binomial Theorem tells us that this probability is
$$\binom{t-1}{n-1}p^n(1-p)^{t-n}.$$
Note that we have replaced the normal coefficient, $\binom{t}{n}$, with $\binom{t-1}{n-1}$ because we know the $t^{th}$ trial must be a success so we only need to arrange $n-1$ successes in $t-1$ trials.
To satisfy the required criterion of fewer than $m$ failures we allow $k$ to range from $0$ to $m-1$, meaning that $t$ ranges from $n$ to $n+m-1$. The resulting probabilities can be summed to give the result:
$$\sum_{t=n}^{n+m-1}{\binom{t-1}{n-1}p^n(1-p)^{t-n}}.$$
Finally, we can map the variables to those used in the text:
\begin{eqnarray*}
n &\mapsto& r \\
m &\mapsto& m \\
t &\mapsto& n
\end{eqnarray*}
Giving: $$\sum_{n=r}^{r+m-1}{\binom{n-1}{r-1}p^r(1-p)^{n-r}}.$$
