How do we choose things? (at random)

(This post appeared on my previous blog, “Not Just Numbers”)

Don’t you just hate it when you are asked to pick something “at random”? No? Well let me show you what I mean.

Suppose your friend shows you three marbles, one black, one red, and one blue. You are asked to pick one, at random. Now, the must obvious thing to do is of course to pick the marble uniformly. So the probability of you picking any marble is a third (suppose you have a fair three sided coin with you to help with that).


But, you were told to pick at random, and not specifically to give each marble a fair chance. So maybe you like the color blue, you really, really like it, so you will pick blue with a chance of 90%, 99%, 99.999999%. You still chose at random, although in most cases you will pick blue. In fact, mathematically speaking even if you choose the blue marble with a 100% chance, you chose at random!

Okay, so maybe when people tell you to pick something at random, they mean that you need to pick it uniformly. So when you are told to pick a random object out of five, there is a probability of a fifth you pick each one. If you get a hundred, a probability of one over hundred, and for a general number n, you pick each one with a probability of one over n.


Now, you may be tasked to pick something at random (uniformly) from somewhere else, say pick a random point on the circumference of a circle of radius 1.

This is no simple task. A circle has infinitely many points, and in fact uncountably infinitely many points. The difference between countable and uncountable infinity is a matter for a future post, but all you need to know for now is that you can’t just assign each point on the circumference a different probability.

Luckily, there is the ever helpful field of Measure Theory to help us out. In essence, measure theory is about looking at a set (of things), and assigning its subsets a measure, which is just a non-negative number, in such way that we get nice properties. So for example, the measure of the union of two sets with no common element should be the sum of their measures.

The field of measure theory is a whole other rabbit hole, and again is a good topic for a future post. But for now, we can assign the measure of length to each arc in the circumference. Now, after dividing the measure of each arc (or, more accurately, each subset of the circumference such as a union of arcs) by the measure of the whole circumference, 2π, we get what is called a probability measure, which is just a measure where the measure of the whole set is equal to 1.

The probability of the green arc is 1/3, of the pink is 1/12, and of the blue is 1/2+1/12 (the length of the smaller blue arc is the same as the pink arc)

So we get a way to pick a point on the circumference at random and uniformly, in some sense. What we got is a way to pick a random point such that the probability of finding the point on any arc on the circumference is proportional to the length of that arc. But what this means is that the probability of picking each point individually on the circumference is zero (think of a point as an arc of length zero).

In a similar fashion, if we want to choose a point uniformly at random on a line segment, we do it by assigning each smaller line segment inside of is a probability proportional to its length. If we want to choose uniformly at random from the inside of a shape, we assign probability measures proportional to area. And if we choose uniformly at random inside a three dimensional object, we assign measures proportional to volume.

Good, now we have intuition to how to pick elements at random from an infinite set. So let us ask a bigger question: what if we want to pick a chord on the circle, uniformly at random. Piggybacking off that, what if we have an equilateral triangle embedded in the circle, and we want to find the probability that the length of the chord we chose is shorter than the length of the triangles edge.

Well, we can choose two points on the circumference uniformly at random and draw the chord between them. First we choose one point, and shift the triangle such that one of its vertices lines up with our point (this does not change the length of the triangles edge). Then we choose the second point. Note that the triangle divides the circumference into three equal length arcs, therefore the probability of landing in each one is one third. And should the second point land in one of the two arcs adjacent to the first point, we get a chord whose length is shorter than the one of the triangles edge. If the second point landed on the third arc, the chord’s length will be longer. Therefore, since the arcs only intersect in the triangles vertices (whose probability is zero), the probability of the second point landing in one of the two arcs, and so the chord length being shorter, is two thirds.


But we can choose the chord a different way. Say this time we choose one point on the circumference at random, and draw the radius from the center to it. Then we choose a random point on that radius uniformly (in such way that the probability of choosing a point on a interval on the radius is proportional to that interval’s length) and draw a chord perpendicular to the radius at the chosen point.

Because, again, the position of the embedded triangle in the circle doesn’t change it’s edge length, we can use a triangle whose edge is parallel to the chord we chose. Then the chord’s length is shorter than the edge’s if and only if the point on the radius is also outside the triangle. As it turns out, the triangles edge bisects the radius, and so the probability of choosing the second point outside the triangle, and so the chord’s length being shorter than the edge’s, is a half!


There is yet another way of (trying to) uniformly choose a random chord. That is choosing a random point inside the circle and choosing the (single) chord with that point as its middle. Then, after a bit more complicated procedure (which I will not show since this post is already getting too long), we can find out that the probability of the chord’s length being shorter than the edge’s is three fourths!

So, which way is the right way to pick a uniformly random chord? In all methods we seem to have chosen the chord uniformly since we used uniformly chosen points, and yet we got three different answers! This is called Bertrand’s paradox (not to be confused with the paradox of the same name in probability, this is not the same Bertrand), and it shows that we can’t just naively choose things “at random”

The underlying problem with choosing the chords is that, unlike choosing uniformly random points on a circumference or a square, there isn’t a geometrical intuition on what choosing a uniformly random chord means. So if we look at the set of all chords in the circle, there isn’t a geometrical tool like length or are to help us classify different sets of chords. And so, we resort to choose the chord by choosing uniform points, which, as we saw, gave us three different answers.

So, the next time someone tells you to choose something “at random”, don’t go in naïvely, and stump them by asking how you choose things randomly.

For more information on Bertrand’s paradox, and the third case I didn’t cover the proof for, check out its Wikipedia article.

2 thoughts on “How do we choose things? (at random)”

  1. Interpretation 1 contains a blunder. For a fixed orientation of the inscribed triangle, having both points on the same third of arc is sufficient but not necessary for the chord to be shorter.


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