Question

If independent trials, each resulting in a success with probability p, are performed, what is theprobability of r successes occurring before m failures?

Answer 1

The events represented by the terms of your summation are not pairwise disjoint. Consider, for example, the outcome in which the first n trials are successes; that situation is counted both in the k=0 term and in the k=1 term, since the latter includes the sequence of n successes followed by 1 failure.

We can also use the fact that

∑k≥0(n+kn)xk=1(1−x)n+1

to observe that if p=12, then

∑k≥0(n+kn)pn(1−p)k=12n⋅1(1/2)n+1=2,

so your summation must be greater than 1 for sufficiently large m; this doesn’t explain why it’s wrong, but it does show that it can’t be right.