How to prove the rule of succession without calculus?

Solution 1:

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 $\frac{s+1}{n+2}$.