According to the rule of succession, if we have a uniform prior over $[0,1]$ for the probability $p$ of a coin to show heads and it has shown heads in $s$ out of $n$ trials, then the probability for the next trial to yield heads is $\frac{s+1}{n+2}$. This is typically derived by integration (e.g. in the Wikipedia article linked to above), but it seems like it should have a more elegant proof not involving calculus, as in the case of Why are all subset sizes equiprobable if elements are independently included with probability uniform over $[0,1]$?
By the way, this would also yield a calculus-free proof that the Pólya urn models a coin with probability $p$ uniformly randomly chosen from $[0,1]$, since the balls drawn from the Pólya urn follow the rule of succession by construction.
The same approach as in my answer to Why are all subset sizes equiprobable if elements are independently included with probability uniform over $[0,1]$? can be used. We choose the probability $p$ for the coin uniformly randomly from $[0,1]$, and then we simulate tossing this coin by choosing a number $r$ uniformly randomly from $[0,1]$, with the result heads if $r\lt p$. After $n$ trials, we’ve chosen $n+1$ numbers (including $p$) independently uniformly from $[0,1]$, and if $s$ of the trials yielded heads, the rank of $p$ among these $n+1$ numbers is $s+1$. For the next trial, we will again choose a number uniformly from $[0,1]$, and it has the same probability to be inserted into the order of the other $n+1$ numbers at any of the $n+2$ possible places. Of these, $s+1$ are below $p$, so the probability for it to be less than $p$ is $\frac{s+1}{n+2}$.