What do you do if given the probability of a probability? This is a weird question that just occurred to me. Assume an event occurs with a definite probability $p$. The given data is a function $f:[0,1]\rightarrow [0,1]$ whose input is a probability for the given event and whose output is the probability that that probability is $p$. My question is whether or not you can use this data to find $p$, and if so, how? My first thought is that you can and that this would be accomplished by integrating the function over the interval, but I'm not sure how to justify this.
 A: You need a detailed sense of what "the probability that that probability is the true probability for the event occurring" means.
To flesh this out, imagine that the "true probability distribution" (which I'll call a model) is chosen randomly. Imagine we have some discrete set $\mathcal{M}$ of possible models, with a probability distribution $P(M=m)$ for $m\in \mathcal{M},$ where $M$ is a random variable representing what the true model is. Then if your event is $A$ you would have conditional probabilities $P(A|M=m)$ for each possible model $m$. $P(A|M=m)$ is the probability the event $A$ occurs given that the true model is $m$. Then the probability that the event occurs $P(A)$ is given by $$P(A) = \sum_{m\in\mathcal{M}} P(A|M=m) P(M=m).$$
To make contact with your original thoughts, say we have $x = P(A|M=m)$ for some model $m$. We'd like to know 'the probability $x$ is the true probability'. This could be complicated since more than one model might produce this same probability for event $A$. However, if $m$ were the only model that produced a probability of $x$,  then we would have $f(x) = P(M=m).$
We can deal with the complication as follows: Since I've been discrete here, assume that $x$ is a discrete variable. Then you'd have $$f(x) =\sum_{m\in\mathcal{M_x}}P(M=m)$$ where $\mathcal{M_x} = \{m\in \mathcal{M} : P(A|M=m) = x.\}$ This just says that 'the probability $x$ is the true probability for $A$' is the probability that the true model has $P(A|M)=x.$ 
Then we can see that $$\sum_{x}f(x) = \sum_{m\in\mathcal{M}}P(M=m) = 1,$$ since we will just wind up adding up all the $P(M=m).$  So your original idea of just integrating $f(x)$ is off. 
But look at $$ \sum_x xf(x).$$ That will weight each $P(M=m)$ by $P(A|M=m)$ and will come out to $P(A).$
