= Accidental coincidence
{title2=$R_{\rm acc}\simeq r_1r_2w$}
Suppose two independent background streams follow <Poisson processes> with rates $r_1$ and $r_2$. Define a coincidence by $|t_2-t_1|<w/2$: \b[$w$ is the full accepted time-difference width]. Around each first-counter signal, the expected number of second-counter signals is $r_2w$. The expected pair rate is therefore $r_1r_2w$. A circuit producing at most one output per first signal instead gives
$$
R_{\rm acc}=r_1(1-e^{-r_2w})\simeq r_1r_2w\qquad(r_2w\ll1).
$$
The distinction matters when multiple detections occur in one window. If a different convention accepts $|t_2-t_1|<\tau$, its full width is $w=2\tau$ and the dilute expression is $2r_1r_2\tau$.
For a chosen numerical model, take background rates $r_1=100\,\mathrm{s^{-1}}$, $r_2=80\,\mathrm{s^{-1}}$ and $w=20\,\mu\mathrm s$. The dilute pair rate is $0.16\,\mathrm{s^{-1}}$. These are stipulated teaching inputs, not measurements of the historical chamber.
Making $w$ smaller also rejects some genuine events. Suppose true pairs arrive at rate $S_0$ before the timing cut, and their time difference has a centered <normal distribution> with <standard deviation> $\sigma_t$. Its accepted fraction, using the <error function>, is
$$
\epsilon_t(w)=\operatorname{erf}\!\left(\frac{w}{2\sqrt2\sigma_t}\right),\qquad
S(w)=S_0\epsilon_t(w).
$$
For chosen $S_0=2.0\,\mathrm{s^{-1}}$ and $\sigma_t=5\,\mu\mathrm s$, the $20\,\mu\mathrm s$ window retains about $95.45\%$ of true pairs. In the dilute background model the fraction of accepted events that are true is $S/(S+R_{\rm acc})\simeq92.27\%$. A narrow window trades <detection efficiency> against background; it does not make either quantity equal to one. Delaying one counter's signal far beyond genuine event correlations gives a proposed background check under stationarity, without assuming that the zero-delay excess itself is accidental.
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