What is the expected value of $\max(a, b)$ where $a, b$ are uniformly distributed from $[0,1]$?
My intuition for this question was to say $E[a] = 0.5$. This means there is probability $0.5$ that $b > a$. If $b > a$ we can say $E[b] = 0.75$ where $b$ would now be the max. Given this, my calculation for $\max(a, b)$ is $0.5(0.5) + 0.5(0.75) = 0.625$, but this does not appear to be the correct answer. Why is this approach wrong?
Yes...ish. It is by symmetry that we may say: $\mathsf P(a>b)=\mathsf P(b>a)$ when the variables are independent and identically distributed.
Well, now, by the Law of Total Expectation we do have:
$\qquad\begin{align}\mathsf E(\max(a,b)) &=\mathsf E(a\mid a>b)~\mathsf P(a>b)+\mathsf E(b\mid a\leqslant b)~\mathsf P(a\leqslant b)\\&=\tfrac 12\big(\mathsf E(a\mid a>b)+\mathsf E(b\mid b \geqslant a)\big)\end{align}$
However, we would not expect the conditional value of $a$ to be $1/2$ when given $a>b$. Indeed, knowing that $a>b$ should raise our expectations for $a$.
$\qquad\mathsf E(a\mid a>b)\gneq \mathsf E(a)$
Likewise we cannot say $\mathsf E(b\mid b>a)=0.75$. Rather it is that $\mathsf E(b\mid b>0.5)=0.75$, which is not the same thing at all.
Indeed, by the symmetry of the random variables:
$\qquad\mathsf E(a\mid a>b)=\mathsf E(b\mid b>a)$
How do we figure how much to raise our expectation under the condition of largeness? Integration is one way.
$\displaystyle\qquad\mathsf E(a\mid{a>b}) = \dfrac{\int_0^1\int_0^a a\, \mathrm d b\,\mathrm d a}{\mathsf P(a>b)}$
Intuitively
Consider the expected placement for the maximum and minimum of $a,b$ on the $[0,1]$ interval. By symmetry and unbiasedness of the uniform distributions, these points should be expected to partition the interval into three equal segments.
So...