= Negative binomial stopping argument
For the <negative binomial distribution> with positive <integer> target $r$ and success <probability> $0<p<1$, the $r$th success occurs after exactly $x$ failures only if the final trial is a success, while the preceding $x+r-1$ trials contain $x$ failures and $r-1$ successes. Choosing the failure locations gives the <binomial coefficient> $\binom{x+r-1}{x}$. Each such stopped sequence has <probability> $(1-p)^xp^r$ by <independence>, and the sequences describe mutually exclusive <events>, so their <probabilities> add. This proves the <probability mass function>. Counting arrangements among all $x+r$ positions would wrongly include sequences that already reached the target earlier. The same argument is used for three heads in https://ocw.mit.edu/courses/18-05-introduction-to-probability-and-statistics-spring-2022/mit18_05_s22_pset09_sol.pdf[MIT 18.05, spring 2022, problem set 9 solutions, problem 2(b)].
A short route to normalization and moments is to split the failures count into successive waiting blocks:
$$
X=F_1+\cdots+F_r,
$$
where $F_i$ counts failures after success $i-1$ and before success $i$, with $F_1$ starting at the first trial. Fresh <independent> <Bernoulli trials> make the blocks <independent>, each with the failures-count <geometric distribution>. Each block has total mass $p\sum_{j=0}^{\infty}(1-p)^j=1$ by the <geometric series>, so their sum also defines a normalized <probability distribution>. The <linearity of expectation> and <variance additivity for independent random variables> give $\mathbb E X=r(1-p)/p$ and $\operatorname{Var}(X)=r(1-p)/p^2$. This makes the moments transparent without a long calculation of sums. For $p=1$, every block is zero.
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