Red-ball prediction with one certain red ball

A box contains one certainly red ball and $n\ge 1$ 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.