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.
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: $$\sum_{n=r}^{r+m-1}{\binom{n-1}{r-1}p^r(1-p)^{n-r}}.$$