= Posterior replacement probability after different-colour draws
{title2=$P(R\mid\overline B_1\cap B_2)=r(N-1)/(N-r)$}
An urn has $N$ balls, including $K$ of one colour and $N-K$ of the other, with $0<K<N$. After a uniform first draw, an independent decision returns that ball with <probability> $r$; the second draw is uniform from the remaining contents. Let $B_j$ mean the $j$th draw has the first colour and $R$ mean replacement. The two relevant joint <probabilities> are
$$
P(\overline B_1\cap R\cap B_2)=\frac{N-K}{N}\,r\,\frac KN,\qquad
P(\overline B_1\cap\overline R\cap B_2)=\frac{N-K}{N}\,(1-r)\,\frac K{N-1}.
$$
Applying <Bayes' theorem> and cancelling the common factors gives the displayed <conditional probability>. It is independent of $K$ and satisfies
$$
r-P(R\mid\overline B_1\cap B_2)=\frac{r(1-r)}{N-r}\ge0.
$$
Two different observed colours therefore decrease the posterior <probability> of replacement: removing the first colour makes the second more likely. The replacement decision must be independent of the first draw for this formula; its marginal <probability> alone does not specify the model.
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