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What is the mean of max(U(0,1),U(0,1))? Judging by computer simulations, it must be at or around 2/3, but I have no idea how to compute the precise value.

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Let $T$ be the quantity from OP, and let $x$ be another $U(0,1)$ variable.

The probability that $x > T$ is 1/3rd by symmetry. It is also $1 - E[T]$, by observation*. so we must have $E[T] = \frac{2}{3}$.

*$P(x \leq T) = \int T P(T) dT = E[T]$

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