Probability that the difference of the max and min of three random numbers between 0 and 2 is less than 1/4?

Basically along the same lines as provided answer.

First note that the answer won't change if we assume $X_1,X_2,X_3$ are IID $\mbox{U}[0,1]$, and we're asking what is the probability that the difference between the maximum and the minimum is at most $\frac 18$.

More generally, suppose that $X_1,\dots,X_n$ are IID $\mbox{U}[0,1]$, and we're asked for the probability that the difference between the maximum and minimum is at most $\alpha$. The answer then is

$$ n \times \int_0^1 P( \bigcap_{j=2}^n \{t<X_j < t+\alpha\}|X_1=t) dt = n \times \int_0^1 \min (\alpha,1-t)^{n-1} dt.$$

The expression on the right is

$$ n \times \left ( \alpha^{n-1}\times (1-\alpha) + \int_0^\alpha u^{n-1} du\right)=n \alpha^{n-1} (1-\alpha) + \alpha^n = n \alpha^{n-1} -(n-1) \alpha^n. $$

In the case $n=3$ and $\alpha=\frac 18$, this formula gives $\frac{11}{256}$.