Let's assume we conduct an experiment which generates two outcomes: success $A$ or failure $B$. The experiment never generates more than $N$ failures. Both $A$ and $B$ have the same probability. After $N+1$ successes the experiment stops. What is the probability to see $m\geq0$ failures before the $(N+1)$-th success?
This looks like a negative binomial distribution, i.e. $P(m)={N+m\choose m}\frac{1}{2^{N+m+1}}$. However, in a negative binomial distribution $m$ is unbounded. Further, in the experiment we ca only see the outcomes $m=0,1,\dots,N$. Hence, $$ P(m=0)+P(m=1)+\dots +P(m=N)=1. $$ However, if we consider the binomial distribution it is certainly not the case that $$ \sum\limits_{m=0}^N{N+m\choose m}\frac{1}{2^{N+m+1}}\color{red}{=}1. $$
So I am wondering if this distribution is appropriate in this setting? How could we justify the application of the negative binomial distribution (if it is possible at all)?
With Wimbledon on, it strikes me that it would be easier to model it instead as a truncated binomial distribution (thus probabilities not adding up to $1$) in which you lose a $(2N+1)$ set match
$$P(m\geq 0)=\frac{1}{2^{2N+1}}\sum_{m=o}^N\binom{2N+1}{m}$$
Added at OP's request
To mould it into a negative binomial distribution, given $N+1$ successes, we find the number of trials $T$ (variant #2 of Alternative Formulations in Wikipedia)
$$P(X=T) = \binom{T-1}{N}p^{N+1}(1-p)^{T-N-1}$$
Of course, due to the conditions laid down,
$(N+1) \leq T \leq (2N+1)$,
so again, it is a truncated NB distribution.