What is the probability that the square $U$ is inside the square $S$ Given a square $S$ with size $1 \times 1$. Two randomly selected points $A$ and $B$ are inside the square. Let $U$ be a square with diagonal $AB$. How to find out the probability $P$($U$ is inside $S$).

 A: Assume that the square $S$ is $(0,1)\times(0,1)$ and that the selected points $A$ and $B$ are $(X,Y)$ and $(Z,T)$. Then, $X$, $Y$, $Z$ and $T$ are i.i.d. and uniform on $(0,1)$ and
$$
[U\subset S]=C\cap D,
$$
where
$$
C=[2W\lt X+Y+Z+T],\qquad D=[X+Y+Z+T\lt2+2V].
$$
and
$$
V=\min(X,Y,Z,T),\qquad W=\max(X,Y,Z,T).
$$
It remains to compute $P[C\cap D]$.
The symmetry 
$$
(X,Y,Z,T,V,W)\to(1-X,1-Y,1-Z,1-T,1-W,1-V)
$$ 
exchanges $C$ and $D$ hence it leaves the event $[U\subset S]$ invariant and one has $P[C]=P[D]$. Since $[2W\gt2+2V]=\varnothing$, $C^c\cap D^c=\varnothing$, that is, $C\cup D=\Omega$ hence 
$$
P[C\cap D]=2P[C]-1.
$$
It remains to compute $P[C]$.
Assume without loss of generality that $W=T$, then, conditionally on $[T=t]$, $X$, $Y$ and $Z$ are i.i.d. and uniformly distributed on $(0,t)$. Thus, $P[C\mid W=T=t]=1-|\Delta_t|/t^3$, where 
$$
\Delta_t=\{(x,y,z)\in(0,t)^3\mid x+y+z\lt t\}.
$$
By scaling, $P[C\mid W=T=t]=1-|\Delta_1|$. This does not depend on $t$ hence $$
P[C]=1-|\Delta_1|.
$$
It remains to compute $|\Delta_1|$.
If $X$, $Y$ and $Z$ are once again i.i.d. and uniform on $(0,1)$, the distribution of $X+Y+Z$ has CDF $P[X+Y+Z\lt u]=\frac16u^3$ for every $u$ in $(0,1)$ (for higher values of $u$, the formula changes). In particular, $P[X+Y+Z\lt1]=|\Delta_1|=\tfrac16$. 
Finally, $P[C]=\tfrac56$ hence
$$
P[U\subset S]=\tfrac23.
$$
