Bertrand’s Paradox
Let’s assume you have a circle and draw a chord(line segment joining two points on a curve). Here we are going to be focusing on the distribution of the lengths of the chords. If a chord goes through the diameter it will be very long but it will be very short if it just connects to points on the edge.
Bertrand the mathematician decided to inscribe an equilateral triangle inside the circle. Where each side was labeled “s”. Now what Bertrand asks is what is the probability that the length of a random chord was greater than s. Written as P(l > s).
Let assume the circle has the radius of 1. Then you can draw a straight line down and create a thirty, sixty and ninety right triangle.
The probability discussed earlier can also be framed as what is the probability that our chord is longer than the square-root of three. Written as: P(l>√3).
Now if take the dot and follow along the blue line. If the second point chosen was behind the top point of the equilateral triangle then the chord would be shorter than the side lengths of the triangle.
If we take the dot and follow the red line it would longer than the side lengths of the triangle along with anything else in the red region.
If the line also fell on the bottom side of the equilateral triangle it would also be shorter.
Now the probability that the chord will be longer than the side of the triangle is 1/3 because the circle is evenly divided into three parts as shown above and only one of those parts(red area) creates chords longer than the side length of the triangle.
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So now lets look at this another way by see chord defined by the midpoint in the circle. So if you draw a circle and are given a midpoint. If you are given the midpoint of a chord where there is one and only one chord where that would be the midpoint. Then you draw the line that’s perpendicular the radius between than midpoint and the center of the circle. This will give you one and only one chord.
Now lets drawn an equilateral triangle inside this circle as well. Now lets inscribe a circle in this triangle. What is concluded is that when the center of the chord is inside the circle then the chord with that center is going to end up longer than the side length of the triangle. But if the midpoint of the chord is outside the circle than the chord will be shorter.
The radius of that circle is 1/2. This is because the radius is the 1/2 side of the thirty, sixty, ninety triangle that we drew earlier. This means the inner circle has an area 1/4 of the out-ter circle. So now the answer to the question Bertrand proposed is that: P(l>s) = 1/4.
Now if you do a simulation for each of the techniques you get 1/3 for the first one(around due to randomness) and 1/4 for the second one(around due to randomness).
Now there is one last way to think about this! So the last way to think about this is to think that the chord we draw it will have a random angle. The angle will be in-between 0 and pi or 0 and 2pi. So the radial line it intersects should be chosen uniformly on the circle.
Now you can again draw a triangle in the circle. Same conclusion as we had before that the length of the center of the circle down to one length of the triangle will be perpendicular and create the thirty, sixty, ninety triangle. Hence the length must be 1/2.
Now when the midpoint of the chord falls on this line then until the line and side of the triangle intersect the chord will be longer than the sides of the triangle which is 1/2 of the time.
Now if this is done in a simulation using this method you find the answer also nears 1/2! So now what is the real answer????
The paradox shows that the choice of a random process is crucial in determining the probability. Without specifying the method of choosing the chord the question cannot be answered definitively.
