Is the result of comparing samples from 2 log normal distributions, uniform? I have sample values coming from two truly random log normal distributions, A and B.   They are continuous (analogue voltages actually).  They are totally independent of one another, but theoretically identical in all aspects of size and shape.   I then repetitively compare sample value A with sample value B, as in
IF valueA > valueB THEN RESULT = TRUE

IF valueA < valueB THEN RESULT = FALSE

The comparison is done with analogue circuitry (not digital) so A = B cannot happen.  What is the distribution of RESULT?  I think that it has to be uniform.  I think that it's effectively a Von Neumann extractor that will remove all bias /asymmetry from the log normal input distributions.   Does anyone concur?
 A: You have two independent random variables $X, Y$ with identical distributions (we also assume the distribution is continuous, so exact equality has probability zero).  You ask for 
$$  \DeclareMathOperator{\P}{\mathbb{P}}
   p= \P (X < Y)
$$
But the assumption of identical distributions tell you that
$$
  1-p = \P (Y < X)
$$
must be equal!  So, $p=1-p$ which has the solution $p=\frac12$. 
Without that insight, it could be done by "brute force", by calculus.   We will use integration by parts:  https://en.wikipedia.org/wiki/Integration_by_parts
$$
\P(X < Y) = \int \P(X < Y \mid Y=y) f(y)\; dy = \\
\int \P(X < y) f(y) \; dy = \int F(y) f(y) \; dy = \\
F(y) F(y) \bigg\rvert_{-\infty}^\infty - \int f(y) F(y)\; dy =\\
1- \int F(y) f(y)\; dy
$$
so we got the same equation, with the same solution.  By the way, we didn't use the lognormal assumption at all.  Moreover, our first argument neither did use independence, it only needs exchangeability, so the result is also valid with a symmetric kind of dependence.  The second (calculus) argument did use independence, though, showing that the simple argument is really superior. 
