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Some may have had reservations about the “randomness” of the straws I plotted to illustrate Bertrand’s paradox. As they were all going North-West/South-East. I had actually made an inversion between cbind and rbind in the R code, which explained for this non-random orientation. Above is the corrected version, which sounds “more random” indeed. (And using wheat as the proper, if weak, colour!) The outcome of a probability of 1/2 has not changed, of course. Here is the R code as well:

lacorde=rep(0,10^3)
plot(0,0,type="n",xlim=c(-2,2),ylim=c(-2,2))
for (t in 1:10^3){
#distance from O to chord
dchord=10
while (dchord>1){
#Generate "random" straw in large box till it crosses unit circle
a=runif(2,-10,10)
b=runif(2,-10,10)
#endpoints outside the circle
if ((sum(a^2)>1)&&(sum(b^2)>1)){
theta=abs(acos(t(b-a)%*%a/sqrt(sum((b-a)^2)*sum(a^2))))
theta=theta%%pi
thetb=abs(acos(t(a-b)%*%b/sqrt(sum((b-a)^2)*sum(b^2))))
thetb=thetb%%pi
#chord inside
if (max(abs(theta),abs(thetb))

As a more relevant final remark, I came to the conclusion (this morning while running) that the probability of this event can be anything between 0 and 1, rather than the three traditional 1/4, 1/3 and 1/2. Indeed, for any distribution of the “random” straws, hence for any distribution on the chord length L, a random draw can be expressed as L=F⁻¹(U), where U is uniform. Therefore, this draw is also an acceptable transform of a uniform draw, just like Bertrand’s three solutions.