# The paradox of Bertrand | Science

On the occasion ofto delivery number 200 of *The game of science*, we asked ourselves last week if changing a figure of this number so round and "excessive" we can turn it into a cousin. And the answer is no. Obviously, the figure to change must be that of the units, because a number ending in 0 is divisible by 10, and that last number can not be even, and neither can it be 1, 5 or 7, because if the sum of the Figures of a number is divisible by 3, it is the number itself. The only candidates are 203 and 209; but 203 is divisible by 7, and 209 is divisible by 11; therefore, you can not convert 200 into a prime by changing one of its numbers. And, by the way, 200 is the least number with this property.

### Sponsored Ads

Advertise HereAs for the sequence 2, 10, 12, 16, 17, 18, 19, 200 ..., it is that of the numbers whose name begins with the letter d.

### Infinity and chance

We all have - or believe we have - an intuitive idea of what chance is, and we also know that there are infinite numbers and that there are infinite points on a line. But when we try to operate with these elusive concepts and reach clear conclusions, we often find ourselves with puzzling paradoxes.

One of the most famous is Bertrand's paradox, named after the French mathematician Joseph Bertrand, who exhibited it at the end of the 19th century in its now classic *Calcul des probabilités*.

To illustrate his paradox, Bertrand used an example that readers brought up last week and that led to a lively and interesting debate (see comments on "The Number 200"). The example is the following:

What is the probability that a string drawn at random in a circle is greater than the side of the equilateral triangle inscribed in it?

The cords of all possible lengths can start from the same point, since all the points of the circumference are equivalent; we will consider, then, that one of the ends of the string is one of the vertices of the inscribed equilateral triangle; if the rope falls inside the triangle, its length will be greater than the side of the triangle, and if it falls outside it will be smaller, and as the angle of the equilateral triangle is 60º and all possible strings cover an angle of 180º (since the limit is in the tangent to the circumference in the vertex that we have taken as origin), the requested probability will be 60/180 = 1/3. Impeccable reasoning; but…

Now consider a radius perpendicular to one side of the inscribed triangle. All the strings perpendicular to this radius that lie between the center of the circumference and the side of the triangle are larger than it, and all the strings between the side and the other end of the radius are lower. And since the side of the inscribed equilateral triangle divides the radius into two equal parts, the requested probability will be 1/2.

And there is still another criterion that gives us a probability of 1/4, and all the approaches seem valid.

I invite my astute readers to reflect on this disconcerting paradox and to propose other examples of paradoxical or surprising probabilities. Like this:

If we draw a random circle on a piece of graph paper, what is the probability that it will pass through any of the points of intersection of the grid?

**Carlo Frabetti** He is a writer and mathematician, a member of the New York Academy of Sciences. He has published more than 50 scientific dissemination works for adults, children and young people, among them *Damn physics*, *Damn mathematics* or *The big game*. He was a screenwriter *The Cristal ball.*