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# KCC's Quizzes about probability puzzle on a square

1. Quote of the week: "A good rule to remember for life is that when it comes to plastic surgery and sushi, never be attracted by a bargain" - Graham Norton

2. New quiz AQQ263 about probability puzzle on a square (this one is a complex one...)

Good luck!

P.S. Please spread those quizzes among colleagues, friends or family members; we want more participants!

Suppose you randomly select point a first. Let's consider an xy axis with the side where point a seats on the x (and the left end of the side coincident with the origin of the axis).

Now you have 3 different cases for point b:

1) point b is randomly selected to be on the same side of a. In this case you have zero probability that d>c.

2) point b is randomly selected to be on the opposite side. In this case the probability to have d>c is 1. The weight of this probability is 1/4 (just one side over four).

3) point b is randomly selected to be on one of the two vertical sides. Let's consider just the vertical side on the y axis, there is a symmetry argument here for the other one. Referring to the cartesian plane, we can write

a= k*c

b=h*c

with h,k belonging to the real interval [0,1]

At this point, the condition d>c can be translated (Pitagora) into sqrt(h2 + k2) > 1, or (h2 + k2) > 1

Now, this is the area external to an arc of circle of radius 1 and contained in the square with the following vertices: (0,0), (1,0), (0,1), (1,1).  This area is equal to the square area, which is equal to 1, minus 1/4 of the circle, that's:

Probability (d>c, case 3)= 1-PI/4 = 0.215

Now you have to take into account also the other vertical side, so you multiply the above probability by 2.

In summary, the requested probability is:

P(d>c) = 1/4 + 2 * 0.215 = 0.68.

• 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.

• Actually total probability is about 36%.

I made a mistake in the final calculation of probability. Summing up the probabilities for each side of the square you have:

p= 1*0.25 + 0*0.25 + 2* 0.25* 0.215= 0.3575

• 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)

• Probability was never my thing but here’s my try at the problem.

Each point (A or B) are independent events so they have a 0.25 chance of landing on one of the four sides.  This creates 16 states, each with the same chance of (0.25)x(0.25)=0.0625.  If they land on the same side (4 states), the distance between them cannot be >C so they don’t contribute to the result.  If they land on opposite sides (4 states) the distance between them can be >= C so they add ~0.25 to the result.  The last group of 8 land on adjacent sides so the distance between them is contingent on where they locate and this is the hard part of the problem.  Refer to the problem figure. The points form a right triangle as shown.  Fix length d at C so anything outside of it forms a line >C.  The probability of b outside the triangle is dependent on x, the length of the triangle base (0 to a).

• Bingo Gaetano! You get it, big applause!

• Hi  , 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...

• Thanks  for your prompt answer. However, it seems 0.6427 is quite too optimistic. I have still to analyze more deeper your equations 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.

• I reviewed my analysis and found an error in my evaluation of the integral.  It should be 0.2146, not 0.7854.  That said, my answer is 0.3573, which agrees with the above.