I saw in http://people.missouristate.edu/lesreid/AdvSol41.html?][1]
It is well known that:
1) If $a$, $b$, $c$ on one hand and $u$, $v$, $w$ on the other hand are independent ones from the others, then $$\mathrm{min}(a,b,c)=\mathrm{min}(u,v,w)=\frac14,$$ $$\mathrm{max}(a,b,c)=\mathrm{max}(u,v,w)=\frac34\text{, and}$$ $$\mathrm{med}(a,b,c)=\mathrm{med}(u,v,w)=\frac12.$$
What does this mean?
Best Answer
The person who wrote what you have cited has clearly never heard of the concept of WOLOG.
We have three points $A = (a,u)$ and $B = (b,v)$ and $C=(c,w)$. As the points are picked randomly the write figures the probability of any two of $a,b,c,u,v,w$ be equal is $0$.
So there are six ways ways to order $a,b,c$ from smallest to lowest and six ways to order $u,v, w$ to lowest and thirty six ways order $a,b,c$ and $u,v,w$ each equally likely.
Now the author is dividing this into cases, and expressing it poorly:
S1) The point, $A,B$ or $C$, associated with the middle $x$ coordinate is the same as the point associate with the middle $y$ coordinate. Or in other words if you traced the points from top to bottom, this would occur in the same order as tracing the points from one side to the other.
The writer INCORRECTLY associates this with one point being within the square formed by taking the other two points to be opposite corners of a square. If I'm not mistaken that will always occur.
There's twelve ways this could occur.
S2) That isn't the case. There are 24 ways this could occur.
....
Instead I would say let $A = (A_x,A_y)$ and $B=(B_x,B_y)$ and $C= (C_x, C_y)$.
Wolog (without loss of generality) assume $A_x < B_x< C_x$ (otherwise just rename the points).
Now there are $6$ ways to order $A_y, B_y, C_y$ in two of them; $A_y < B_y < C_y$ and $C_y < B_y < A_y$ we have $B_y$ between $A_y$ and $C_y$, and four where it isn't.
If we assume reading top to bottom is the same as bottom to top (we can just flip to picture over) then we really only have three possibilities.
S1: $A_x < B_x < C_x$ and $A_y < B_y < C_y$ and
S2: $A_x < B_x < C_x$ an either $B_y< A_y < C_y$ or $A_y< C_y < B_y$.
(Thus with wolog and equating up with down we can reduce $36$ cases, $12$ or one type and $24$ of another, done to $3$ cases $1$ of one type and $2$ of another.
....
Anyway back to the authors comments.
..... Okay... I think the author is just wrong. S/he seems to be claiming in case S1 that expected value of the leftmost/topmost value is $\frac 12$, the expected alue of the middlemost x/y value is $\frac 12$ and of the rightmost/bottommost value is $\frac 34$.
This may (or may not be true) but I don't think combining expect values of variables linearly transposes to give expectied values of the area of a geometric shape.