KCC's Quizzes about probability puzzle on a square

There are 4 sub-cases:  a and b on the same segments,  a and b on opposite segments,  and b on one of the two segments adjacent to the segment of a. Each of the four cases has a probability of 1/4 to occur.
In the first case, the final probability is 0.
In the second case, the probability is 1, so its contribution is 1/4 *  1 to the total (the total of the 4 cases)
The two last cases are symmetric, so their contribution to the total is 1/2 * p,  where p is the probability than, on the given figure, d is greater than c.
Total probability:  0.25 + 0.5*p.

Now, to evaluate p.

If we draw a circle of length c centered at point a, the value of b is sqrt(c^2  - a^2). Without losing in the generality, assume c = 1 and so, a, the point on the horizontal segment, can vary from 0 to 1. So, p is when b (itself of a uniform distribution from 0 to 1) is greater or equal to sqrt(1-a^2). 
At first glance, through integration from 0 to 1, p = 1 - pi/4.
So,total probability: 0.25  + 0.5*(1-3.141592/4) = 0.3573...

I will try to get a numerical confirmation before making a formal claim of the figure, but that sounds to be "in the possible" range.

Oups, it is not  p = 1- pi/4  but it is   p=pi/4. Indeed, 1- pi/4  is when the distance "d" in the sketch  is LESS than "c", not GREATER than. As confirmed by the numerical simulation (see below).

So the total probability is 0.25 + 0.5 * pi  / 4  = 0.642699...

The numerical simulation is done with Excel. 

Ask a random number between 0 and 1 in cell A1, same in B1, and in C1, have the formula:  if(A1^2  + B1^2  > 1, 1, 0).

Copy the formulas of these three cells 1000 times down.

Then, in C1001, use SUM(C1:C1000) / 1000. It should resolve around 0.783, which is approximately pi/4.  (and once again, not 1 - pi /4)

Hi vanderghast , Strangely, your first answer was correct in fact. Therefore, there is probably an error introduced in th numerical method.... I have still to investigate more deeper to see what went wrong...

You are right, my numerical simulation (the one that I used) was in error. Using the one I typed here will give 1-pi4 and so, my first answer was in agreement with the simulation. Sorry for the time spent.

You are right, my numerical simulation (the one that I used) was in error. Using the one I typed here will give 1-pi4 and so, my first answer was in agreement with the simulation. Sorry for the time spent.

no problem! and congratulation for having found the same result via 2 different methods!