# The expected area of a triangle formed by three points randomly chosen from the unit square

“Three points are chosen uniformly and at random from a unit square. What is the expected value of the area of the resulting triangle?”

I need to do a research about that problem and i found this suggested solution: here.

Now, I understand almost everything except anecdote (2) when he computes the expected value of $b$ and $v$. I can’t understand how he reaches those calculations.

If someone can explain to me this that would be great.

Thanks.

#### Solutions Collecting From Web of "The expected area of a triangle formed by three points randomly chosen from the unit square"

Here is an approach which is different from the one in the link, does not involve so many cases, and is self-contained.

Let $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$ be the three vertices of the random triangle $T$. It is sufficient to consider the case $a_2<b_2<c_2$, which takes ${1\over6}$ of the total “volume”. Fix $a_2$, $b_2$, $c_2$ for the moment, and write
$$b_2=(1-t)a_2+t c_2,\qquad0\leq t\leq 1\ .$$ The side $AC$ of $T$ intersects the horizontal level $y=b_2$ at the point $S=(s ,b_2)$ with
$$s=s(a_1,c_1,t)=(1-t)a_1+t c_1\ .\tag{1}$$
The area $X$ of $T$ is then given by $$X={1\over2}|b_1-s|(c_2-a_2)\ .$$

We now start integrating with respect to our six variables. The innermost integral is with respect to $b_1$ and gives
\eqalign{X_1&:=\int_0^1 X\>db_1={1\over2}(c_2-a_2)\left(\int_0^s (s-b_1)\>db_1+\int_s^1(b_1-s)\>db_1\right)\cr &={1\over4}(c_2-a_2)(1-2s+2s^2)\ .\cr}
Next we integrate over $b_2$:
$$X_2:=\int_{a_2}^{c_2} X_1\>db_2={1\over4}(c_2-a_2)^2\int_0^1(1-2s+2s^2)\>dt \ ,$$
whereby $s$ is given by $(1)$, and does not depend on $a_2$ and $c_2$. It follows that from now on the integration factorizes into
$$X_3:={1\over4}\ \int_0^1\int_{a_2}^1(c_2-a_2)^2\>dc_2\> da_2\ \times\ \int_0^1\int_0^1\int_0^1 (1-2s+2s^2)\>dt\> dc_1\>da_1\ .$$
The easy computation gives
$$X_3={1\over4}\cdot{1\over12}\cdot{11\over18}={11\over 6\cdot 144}\ .$$
From this the end result is obtained after multiplying by $6$, in order to take care of the assumption $a_2<b_2<c_2$ made at the beginning. The probability in question is therefore given by ${11\over144}$, as obtained in the source quoted by the OP.