Numbers $x_1,x_2,\ldots,x_{2n}$ are drawn independently and uniformly from the interval $[0,1]$. A game proceeds in $n$ steps. In the $i$th step, we choose the highest number left and remove it, and another number is removed uniformly at random from the remaining numbers. What is the expectation of the sum of the $n$ numbers that we choose?
For $n=1$ we choose the maximum of the two numbers, giving an expectation of $2/3$. For $n=2$ the first number we choose has expectation $4/5$, but after that how can we continue the calculation?
The $2n$ numbers have expected value $\frac j{2n+1}$, for $j=1..2n$. So the expected sum is $1/(2n+1)$ times the sum you would get if the $j$th number were simply the integer $j$.
For $n=2$, let the numbers be $a<b<c<d$. The first number you pick will be $d$, which has expected value $4/5$. If you remove $c$, the second number you pick will be $b$, with expected value $2/5$. If you remove $a$ or $b$, the second number you pick will be $c$, with expected value $3/5$. So the expected sum is $$\frac45+\frac35\frac23+\frac25\frac13=\frac{20}{15}$$