Let's assume take N random numbers in the [0,1] interval. What is the expected value of the lowest number in the sample?
Running a few simulations it really seems it is
EV = 1/(N+1)
but I don't know how to prove this equation.
Farthest I went was calculating the EV for N=2 (so for two values a and b). You can easily derive that, with a known, EV is
a - (a^2)/2
and you just need an integral to come to the correct answer (1/3). Still, I don't know how to generalise to any N.
For $x \in [0,1]$, $P(\min\{X_1,...,X_N\} > x) = (1-x)^N$ (where here the $X_i$ are uniformly distributed on $[0,1]$ and independent).
Therefore $P(Y \leq x) = 1 - (1-x)^N$, where I've denoted the min by $Y$.
$Y$ therefore has density $N(1-x)^{N-1}$ concentrated on $[0,1]$. So to calculate the expected value of $Y$ you just need to compute
$$ N\int_0^1 x(1-x)^{N-1} dx, $$
which hopefully you can do.