**S**ome 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))<pi/2)
dchord=abs(sin(theta))*sqrt(sum(a^2))
}
}
lacorde[t]=2*sqrt(1-dchord)
if (runif(1)<.1) lines(rbind(a,b),col="wheat")
}
lecercle=cbind(sin(seq(0,2*pi,le=100)),cos(seq(0,2*pi,le=100)))
lines(lecercle,col="sienna")

**A**s 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.

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