Probability of binomial n success before m failures? problem of n success before m failures where binomial probability of success is p has a standard textbook solution as follows $$P = \sum_{k=n}^{m+n-1} \binom{m+n-1}k p^k (1-p)^{m+n-1-k}$$
I am however unable to come up with this solution and i am not sure where i deviate and how?
I start with combinatorial logic that for any t trials
(a) last trial has to be a success => probability of p
(b) in t-1 trials before last trial, n-1 have to be success and (t-1)-(n-1) failures => probability of $\binom{t-1}{n-1} p^{n-1} (1-p)^{t-n}$
Combining (a) and (b) gives me probability that t trials have t-n failures before n success
$P_{t} = \binom{t-1}{n-1} p^{n} (1-p)^{t-n}$
I understand that this is negative binomial pmf as well.
From here on, i say that total probability is sum of all the probabilities with various t i.e.
$P = \sum_{n}^{m+n-1} P_{t}$ that can be written as 
$$P  = \sum_{t=n}^{m+n-1} \binom{t-1}{n-1} p^{n-1} (1-p)^{t-n} $$
I believe something is off with my last step.. may be these events of different $t$ trials are not mutually exclusive but i am not able to see it.
For example: n = 3 success and m = 2 failures; p = binomial probability
(a) I can have a 3 trial solution SSS with probability $p^3$ 
(b) 4 trial solution {SSFS,SFSS,FSSS} with probability $ \binom{3}{2} p^{3} (1-p)^1$
(c) 5 trial solution is not possible since n=2 would have happened
I would therefore add probabilities from (a) and (b) as solution but that would give 
 $$ P = p^3 + \binom{3}{2} p^{3} (1-p)^1 $$
and of course, this is not right.
Can someone help and point out why this is not correct and what can i fix here?
 A: The book answer (or your transcription of it) appears to be incorrect  
By the way, it is simpler to count failures, and  look at the results in reverse.
With $0 \le k < m$, the last trial must be a success, and the $k$ failures can be distributed any which way in the remaining $(k+n-1)$ trials, thus
$$P = \sum_{k=0}^{m} \binom{k+n-1}k p^n (1-p)^k$$
If you want to count successes (as the book has done), you should now be able to correct the formula you have transcribed.
A: I actually came across this similar problem in Bertsekas' "Introduction to Probability" 2nd edition (Ch 6 Exercise 4c), where it was making the same mistake. May I ask which textbook are you referring?
Original formula:
$$P = \sum_{k=n}^{m+n-1} \binom{m+n-1}k p^k (1-p)^{m+n-1-k}$$
The problem of this original formula is that 
1) as you vary the iterating variable k, the total number of trials should also be changing, but the term $m+n-1$ is kept constant
2) the failure exponent term $m+n-1-k$ will be counting the right number of failure.  
With (1) and (2) above, this original formula is actually accumulating the probability of getting k success and the rest failure while running m+n-1 trials for the range of $k>=n$. This is totally not what the question is asking.
Instead,
I would say the following: minor correction to @{true blue anil}'s formula:
$$P = \sum_{k=0}^{m-1} [\binom{k+n-1}k p^{n-1} (1-p)^{k} ]p = \sum_{k=0}^{m-1} \binom{k+n-1}k p^n (1-p)^{k} $$
This is saying that we want to accumulate within the range $0<=k<=m-1$, the probability that there are n successes (last trial is forced to be success, the first n-1 success happen within in the k+n-1 trials, the rest of the k trials are failure).
For you example,
n = 3 success, m = 2 failures, p = prob of success
If we do it by hand,
3 trials: XXP => $p^3$
4 trials: XXXP => $\binom{3}{2}p^2 (1-p) p = \binom{3}{2}p^3 (1-p) $
$P = p^3 + \binom{3}{2}p^3 (1-p) = p^3 +3p^3(1-p)$
And if you apply the formula I have above: $\sum_{k=0}^{m-1} \binom{k+n-1}k p^n (1-p)^{k} $
You would be iterating k = 0 to 1. 
So $P = \binom{0+3-1}0 p^3 (1-p)^{0} +  \binom{1+3-1}1 p^3 (1-p)^{1} = p^3 + 3p^3(1-p)$, which should agree with the solution by hand.
