expected area of a triangle determined by randomly placed points Three points are placed at independently and at random in a unit square. What is the expected value of the area of the triangle formed by the three points?
 A: This isn't a full solution, but it goes most of the way there.
The area of a triangle $(x_1, y_1),(x_2,y_2),(x_3,y_3)$ is given by the formula
$A(x_1,x_2,x_3,y_1,y_2,y_3)= | \frac{x_1(y_2-y_3) + x_2 (y_3-y_1) + x_3 (y_1-y_2)}{2}|$
If $x_1, x_2, x_3, y_1, y_2, y_3$ are all independently identically uniformly distributed over $[0,1]$, then the average area is just given by:
$\int_0^1 \int_0^1 \int_0^1 \int_0^1 \int_0^1 \int_0^1 A(x_1,x_2,x_3,y_1,y_2,y_3) d x_1 d x_2 d x_3 d y_1 d y_2 d y_3$ 
At this point, it's a fairly simple, though tedious, calculation. I recommend using Mathematica or some other computational software if you have access to it. There are also ways to simplify the computation based on inherent symmetries in the problem. 
I can't post the final answer because I want to avoid giving a wrong answer, which is totally possible (I don't claim to be able to do the above integral by hand without errors). I can check the answer in Mathematica if you don't have access to it, though it will have to wait until at least Monday.
A: See here.
A: User 'Shai Covo' (who seems no longer active) gave a solution as the link to an old webpage maintained by Dr. Les Reid of the math department at Missouri State University. The solution seems to be valid after being fixed by the comments @joriki, and the rigorous integration approach @achille hui supports it (and so much more).
I think the solution is nice and worthy of displaying with a good format so here it is.

The argument below is due to Philippe Fondanaiche (Paris, France), directly lifted from here.

Let $A$ with coordinates $(a,u)$, $B(b,v)$ and $C(c,w)$ be the three points chosen uniformly and at random from a unit square.
As the probability to have 2 or 3 coincident points is nil, we will
consider hereafter the strict inequalities between the abscissas/coordinates of the 3 points.
There are 6 positions of the 3 abscissas having the same probability:
$$\begin{align}
a&<b<c & a&<c<b & b&<a<c \\
b&<c<a & c&<a<b & c&<b<a
\end{align}$$
For each of them, there are 6 equally likely positions of the cordinates:
$$\begin{align}
u&<v<w & u&<w<v & v&<u<w \\
v&<w<u & w&<u<v & w&<v<u
\end{align}$$
So there are globally 36 possible configurations having the same
probability 1/36.
These 36 configurations can be split into 2 subsets $S_1$ and $S_2$
representing respectively 12 and 24 possibilities:


*

*$S_1$: one of the 3 points ($B$ for example) is within the square one of the diagonals of which joins the 2 other points ($A$ and $C$ for example). In other words, $\min(a,c)< b < \max(a,c)$ and $\min(u,w)< v < \max(u,w)$.

*$S_2$: any other of the 24 remaining configurations.
It is well known that:


*

*If $a,b,c$ on one hand and $u,v,w$ on the other hand are independent to the ones from the others, then 
$$\begin{align}
\min(a,b,c) &= \min(u,v,w) = \frac14 \\
\max(a,b,c) &= \max(u,v,w) = \frac34 \\
\text{med}(a,b,c) &= \text{med}(u,v,w) = \frac12
\end{align}$$

*If $\min(a,c) < b < \max(a,c)$, $\min(u,w) < v < \max(u,w)$, and $b < v$, then the expected value of $b$ is $\frac12 + \frac13 \frac14 = \frac7{12}$ and the expected value of $v$ is $\frac12- \frac13 \frac14 = \frac5{12}$.


In these conditions, it is easy to check that the "average" position of
the $\triangle ABC$ with the subset $S_1$ is the $\triangle T_1$ whose vertices are $(\frac14,\frac14),\,  (\frac7{12},\frac5{12}),\,  (\frac34,\frac34)$ or one of the three triangles obtained from $\triangle T_1$ by successive rotations of angle $90^\circ$. The 4 triangles are the same area equal to $\frac1{24}.
As far $S_2$ is concerned, the average position of the $\triangle ABC$ is the $\triangle T_2$ whose vertices are $(\frac14, \frac12),\, (\frac12, \frac34),\,  (\frac14, \frac34)$ or one of the three triangles obtained from $\triangle T_2$ by successive rotations of angle $90^\circ$. The 4 triangles are the same area equal to $\frac3{32}$.
Therefore the expected area of the resulting triangle is equal to
$$ \frac{12}{36} \frac1{24} + \frac{24}{36} \frac3{32} = \frac{11}{144}$$

For completeness, below is the transcription of jorki's comment correcting the error of the derivation above.
Small error in the case distinction: The answer is based on the fact that the for fixed orientation area of the triangle is linear in each of the coordinates of the points, since it's the absolute value of a polynomial that's linear in each of the coordinates and changes sign when the orientation changes. Thus the case distinction has to be such that each case only covers triangles with the same orientation. For the case $S_1$, the author seems to have had in mind a mapping from the rectangle formed by $A$ and $C$ to the unit square.
That may be why in "within the square one of the diagonals of which joins the 2 other points" it says "square" where it should say "rectangle", and, crucially, further down in item 2) the condition is stated as $b<v$ where in fact it should say that $B$ is above the diagonal $AC$. With that condition, it's OK to use linearity of expectation, since this is the condition that separates the two orientations of the triangle.
A: Here's a perl script that confirms the answer Shai linked to via a Monte Carlo approach.
#!/usr/bin/perl -w

$numTrials = 1000000 ;

sub distance {
   my $point1 = $_[0] ;
   my $point2 = $_[1] ;
   return sqrt(($x[$point1]-$x[$point2])**2 + ($y[$point1]-$y[$point2])**2) ;
}

sub heron {
   my $a = $legLength[$_[0]] ;
   my $b = $legLength[$_[1]] ;
   my $c = $legLength[$_[2]] ;
   my $s = ( $a + $b + $c ) / 2 ;
   return sqrt( $s * ( $s - $a ) * ( $s - $b ) * ( $s - $c ) ) ;
}

sub doAtriangle() {
   for ( my $j = 0; $j <= 2 ; $j++ ) {
      $x[$j] = rand(1) ;
      $y[$j] = rand(1) ;
   }   
   $legLength[0] = distance(0,1) ;
   $legLength[1] = distance(1,2) ;
   $legLength[2] = distance(2,0) ;
   return heron(0,1,2) ;   
}

for ( $i = 0 ; $i < $numTrials ; $i++ ) {
   $sum += doAtriangle() ;
}

print $sum/$numTrials . "\n" ;

