= Red-ball prediction with one certain red ball
A box contains one certainly red ball and $n\geq1$ independently coloured extra balls, each red with <probability> $1/2$. Conditional on the first draw being red, a second draw without replacement is red with <probability> $(n+3)/(2(n+2))$. To see the effect of using a <conditional probability>, let $R=1+B$, where $B$ has a <binomial distribution> with parameters $n,1/2$. The <conditional probability> is $\mathbb E[R(R-1)]/(n\mathbb E[R])$: histories with more red balls are more likely to produce the first red draw and therefore receive greater weight.
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