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Conditional on the pooled number $s$ of successes in $m+n$ exchangeable random variables that are binary, all placements of those successes are equally likely under a common success-chance mixture. The count in a chosen batch of size $n$ therefore has a hypergeometric distribution, with mass $\left(\genfrac{}{}{0ex}{}{s}{y}\right)\left(\genfrac{}{}{0ex}{}{m+n-s}{n-y}\right)/\left(\genfrac{}{}{0ex}{}{m+n}{n}\right)$. Under the uniform-count predictive property, $s$ is uniform on $0,\dots ,m+n$. Dividing the joint pooled-and-batch probability by the first batch's uniform marginal gives the posterior predictive distribution, with factor $(m+1)/(m+n+1)$. Substituting $s=x+y$ in the hypergeometric expression does not give a single normalized hypergeometric mass as $y$ varies, since its population success total then varies too.

With a uniform beta distribution prior, the posterior predictive distribution for another batch of the same size isBut $\mathrm{B}(x+1,m-x+1)=1/[(m+1)(\genfrac{}{}{0ex}{}{m}{x})]$. HenceThe displayed expression is symmetric in $x,y$. This is the uniform-count predictive property: the joint batch counts are exchangeable and both marginal counts are uniform. Conditional reversal is not a general property of arbitrary priors; it holds here because those marginal probabilities coincide.

Before collecting any observations, a uniform success-chance prior distribution givesThus the uniform-count predictive property assigns equal probability to every possible number of successes in a fixed-size batch under the prior predictive distribution. It is uniform over counts, not over all ordered binary sequences. After observing $x$ successes in $m$ trials, its familiar one-step posterior predictive probability is $(x+1)/(m+2)$; the count-uniform property refers to prediction before those data are observed.

For the uniform prior, write $s=x+y$ for the pooled success count in $m+n$ trials. Conditional on $s$, exchangeability makes all placements of the successes equally likely. The successes in the second batch therefore have a hypergeometric distribution, and the factor in question isThis is hypergeometric allocation of exchangeable Bernoulli counts: sample $n$ positions without replacement from a population of $m+n$ positions containing $s$ successes. The uniform-count predictive property gives $\mathrm{Pr}(X+Y=s)=1/(m+n+1)$ and $\mathrm{Pr}(X=x)=1/(m+1)$, soAs $y$ varies with $x$ fixed, $s=x+y$ also varies. Therefore $B$ alone is not a normalized hypergeometric distribution over $y$; its normalization factor is essential.