Triangle area from uniformly distributed points along a line

Question

3 random numbers from independent uniform distributions between 0 and 1 are selected. How to calculate the expected triangle area if the points are arranged as in the sketch?

Answer 1

The middle line decomposes the triangle into two triangles of height 1. So it remains to find the expected length of the base in the middle line.

If the three variables are $y_1$, $y_2$, $y_3$ from left to right, then the base has length $$ \left| y_2-\frac{y_1+y_3}{2}\right| $$ and so the expected area is $$ \int_{[0,1]^3} \left| y_2-\frac{y_1+y_3}{2}\right| $$

Answer 2

I suppose the variables $x_i$s are independently and uniformly distributed over the interval $[0,1]$. The mean value of the surface area should be:

$$E(S)=\int_{[0,1]^3}S(x_1,x_2,x_3)$$

By euclidean geometry, $S(x_1,x_2,x_3)=\left|\frac{x_1+x_2}{2}+\frac{x_2+x_3}{2}-2\frac{x_1+x_3}{2}\right|$ (hint: these are areas of well-chosen trapezoids). Hence:

$$E(S)=\int_{[0,1]^3}\left| x_2-\frac{x_1+x_3}{2}\right|$$

which is a multivariable integral. Hint: consider $x_2\geq\frac{x_1+x_3}{2}$ and $x_2<\frac{x_1+x_3}{2}$.

Answer 3

Solving the integrals in the answers one gets an expected triangle area of $7/24=0.291\bar{6}$.