Expected Maximum of Three Numbers that Sum to 1 I've been working on a programming project on classification with 3 classes, and I'm interested in comparing my results to what I'd expect from pure noise. So I have the following question:
Let $X,Y,Z$ be uniformly distributed over $\{(x,y,z)\in [0,1]^3:x+y+z=1\}$. What is $E \max (X,Y,Z)$?
Thanks
Sorry for editing after the fact. I'm also interested in $E(-X \log ( X )-Y\log ( Y )-Z \log (Z))$
 A: If $(X,Y,Z)$ is uniformly distributed in $T=\{(x,y,z)\in[0,1]^3 : x+y+z=1\}$, then $(X,Y)$ is uniformly distributed in $T'=\{(x,y)\in[0,1]^2 : x+y\leqslant 1\}$, and $Z=1-X-Y$ a.s., thus
$$\mathbb{E}f(X,Y,Z)=2\iint_{T'}f(x,y,1-x-y)\,dx\,dy.$$
A simplification by symmetry, $\mathbb{E}\max\{X,Y,Z\}=6\,\mathbb{E} Z[X\leqslant Y\leqslant Z]$, gives
$$\mathbb{E}\max\{X,Y,Z\}=12\iint_{T''}(1-x-y)\,dx\,dy,$$
where
$$T''=\{(x,y) : 0\leqslant x\leqslant y\leqslant 1-x-y\}=\{(x,y) : 0\leqslant x\leqslant y\leqslant(1-x)/2\}.$$
This is easily computed, giving $\color{blue}{\mathbb{E}\max\{X,Y,Z\}=11/18}$.
For $\mathbb{E}(-X\log X-Y\log Y-Z\log Z)=3\,\mathbb{E}(-X\log X)$ (as noted in comments),
$$\iint_{T'}(-x\log x)\,dx\,dy=\int_0^1x(1-x)(-\log x)\,dx=\Gamma(2)(2^{-2}-3^{-2})=\frac{5}{36}$$
which gives $\color{blue}{\mathbb{E}(-X\log X-Y\log Y-Z\log Z)=5/6}$.
A: I'm interpreting "$X$, $Y$, $Z$ uniformly distributed over $S:=\bigl\{(x,y,z)\in[0,1]^3\bigm| x+y+z=1\bigr\}$" as uniformly distributed with respect to surface measure on the equilateral triangle $S$. This measure is proportional to the euclidean measure of the projection $S'$ of $S$ to the $(x,y)$-plane. 
The three medians $x=y$, $y=z$, $z=x$ partition $S$ into six congruent parts corresponding to the order $\leq$ of $x$, $y$, $z$. It is sufficient to compute the expected value $E$ in the part $T\subset S$ corresponding to $y\leq x\leq z=1-x-y$. This part $T$ projects to the triangle
$$T'=\bigl\{(x,y)\bigm| 0\leq y\leq x, \ x\leq 1-x-y\bigr\}\subset S'\ .$$
The vertices of $T'$ are $(0,0)$, $({1\over2},0)$, $({1\over3},{1\over3})$.
On $T$ we have $$\max\{x,y,z\}=z=1-x-y\ .$$ We therefore obtain
$$E={\int_{T'}(1-x-y)\>{\rm d}(x,y)\over{\rm area}(T')}=\ldots={11\over18}\ .$$
