= Cost of trials until the first success
{title2=$\mathbb E[S_N]=3/p-2$}
Consider a stipulated repeated-attempt model, with <independent> <Bernoulli trials> of success <probability> $p\in(0,1]$. An unsuccessful attempt costs three units of work, and a successful attempt costs one. Let $B_i$ be its success <indicator function>, $X_i=3-2B_i$ its cost, and $N=\inf\{i\geq1:B_i=1\}$ the first success. Observing $X_i$ reveals $B_i$, so $N$ is a <stopping time> for the costs' <natural filtration>. The <geometric distribution> gives $\mathbb P(N\geq i)=(1-p)^{i-1}$ and $\mathbb E[N]=1/p$, including the deterministic case $p=1$. Thus
$$
\mathbb E[X_1]=3-2p,\qquad \boxed{\mathbb E[S_N]=(3-2p)/p=3/p-2.}
$$
At $p=1/4$, the average is \b[four attempts and ten units of work]. An independent check uses the pathwise identity $S_N=3(N-1)+1$: there are $N-1$ failures and one final success. No success probabilities or costs in this example are measured data.
\Image[/wald-first-success-cost-and-survival.png]
{title=Cost and survival probabilities for stopping at the first success}
{description=Original stipulated model with success <probability> $1/4$, failure cost three, and success cost one. The top shows one illustrative stopped sequence; the lower panels show exact <probabilities> and the contributions to the <expected value> from each admitted step.}
Here $N$ is not <independent> of the observed costs. In particular, conditioning on $N=n$ forces every preceding cost to be three and the final cost to be one, so $\mathbb E[S_N\mid N=n]=3n-2$, generally different from $n\mathbb E[X_1]=(5/2)n$ at $p=1/4$. Nevertheless the unconditional identity holds. This distinction explains why conditioning on the total count as though the original cost law remained unchanged is a faulty proof.
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