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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.

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

joriki
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    Looks fine to me and IMV can indeed be marked as a proof of the rule of succession that exceeds the proof on e.g. wikipedia if it comes to elegancy. Well done Joriki! – drhab Jan 20 '20 at 13:23
  • This is indeed an elegant proof! However, I have a little confusion. You say that the rank of p is s+1, and for the next toss you take a number uniformly randomly in [0,1]. But what I don't understand is how do you deduce that this number has the same probability of landing in each of the n+2 slots since in general those slots cannot be of uniform length. – Ad Nazir Jan 27 '23 at 17:31
  • @AdNazir: This is by symmetry. Of course you're right that if you know the previously drawn numbers, then conditional on those numbers the probabilities of hitting the slots between them are unequal. But we don't know the numbers, and by symmetry a priori all permutations of all the numbers drawn are equiprobable. – joriki Jan 27 '23 at 17:52
  • @Joriki Yes, indeed. It did occur to me that by symmetry the result follows. But I was hoping if there's a way to amplify this (handwavy) argument into a sound mathematical argument. At any rate, it's a beautiful proof! – Ad Nazir Jan 29 '23 at 07:50
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    @AdNazir: I emphatically reject the view that symmetry is "handwavy" and not a "sound mathematical argument" :-) Some of the most beautiful mathematical arguments are by symmetry. – joriki Jan 29 '23 at 09:37