 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 $\left(\genfrac{}{}{0ex}{}{x+r-1}{x}\right)$. Each such stopped sequence has probability $(1-p{)}^x{p}^r$ 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 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: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 $\mathrm{Var}(X)=r(1-p)/p^2$. This makes the moments transparent without a long calculation of sums. For $p=1$, every block is zero.