PDF of an unknown probability after a successful test. Let $X$ be a random variable having the possible outcomes $\{0, 1\}$ representing failure or success, with unknown probability. We test $X$ just one time and the outcome is success. Let $X(1) = p \in [0, 1]$, ie, $p$ is the probability that another test of $X$ will be successful. Is it possible to find a probability density function for $p$? If so, how could this distribution be described?
Intuitively it would seem that $P(p = 0) = 0$, and $P(p > \frac{1}{2}) > P(p < \frac{1}{2})$, for example.
This question is very similar to the unanswered Unknown Probabilities, but that question was asked over a year ago and takes the problem in a slightly different direction.

Okay, I think I have an idea here.
Let P(1) denote the probability of the first outcome being a success. If $p$ is given then $P(1)=p$, but since $p$ is unknown and different values of $p$ are mutually exclusive events with probabilities $f(p)dp$:
$$
P(1) = \int p{f(p) dp} = \int {pd(F(p))}\\ = pF(p) - \int{F(p)dp}
$$
Let $g(p)= \int Fdp$, and $P(1)=a$, constant.Then $g'-g/p=a/p$. I multiply the homogenous equation by an integration factor $M(p)=1/p$ to get
$$
g_h'-\frac{1}{p}g_h=0 \\
\frac{1}{p}g_h'-\frac{1}{p^2}g_h=0 \\
\frac {d}{dp} \left( \frac{g_h}{p}\right) = 0 \\
g_h = cp
$$
The obvious partial solution is $g_p(p)=a$ so $g= g_h+g_p = cp+a$.
This is where I'm stuck. $F(p)=g'(p)=c$, but this doesn't make sense. I think my initial assertion that $P(1) = \int p{f(p) dp}$ is correct, but I can't spot the error in the math.
