If you don't like maths, don't read the stuff below. It's only here to assist anyone thinking about where (and if) I made an error.
In Three New Wrong Answers for Bertrand, I talk about arriving at four numbers, 1/2, 3/4, something close to 4/9 and something close to 2/3. What I don’t get into is the mathematics by which I arrived at this numbers. I’ll do that here.
My analysis was done in Excel, using the rand() function. I assumed a circle of radius 1 (y2 + x2 = 1), so I looked for a random point within a box defined by x=-1 and x=1 and y=0 and y=1 and a random point within a box defined by x=-1 and x=1 and y=0 and y=-1.
So x1=2*rand()-1, y1=rand(), x2=2*rand()-1 and y2=rand()*-1.
Then I determined the line defined by these two points, y=mx+c where m= (y2-y1/ (x2-x1) and c=y1-mx1= y2-mx2.
Then I looked for the points at where this line intersected the circle, which I did by inserting y=mx+c into y2 + x2 = 1 and solving for two values of x, x3 and x4 and using those to generate y3 and y4, from which I could determine the length of the chord and compare it to √3.
Just for completeness, after inserting y=mx+c into y2 + x2 = 1, I arrived at:
(m2+1)x2 + 2mcx + (c2-1) = 0 = αx2 + βx + γ
Which I solved using (-β ± √(β2 - 4αγ))/2α.
When I used two random pairs without limiting them to different halves, I checked against (β2 - 4αγ)>0 to ensure that the line intersected the circle. If the pairs didn’t meet this criterion, I rejected them as not resulting in a chord.
For that effort, I used x1=2*rand()-1, y1=2*rand()-1, x2=2*rand()-1 and y2=2*rand()-1 to generate the two random pairs.