The probability of something happening is given by the relationship between the favorable cases and the possible cases, and this is expressed colloquially when speaking, for example, of “a probability in a thousand”. In mathematics the same thing is said, but in the form of a fraction: thus, there is a one in six chance of getting a 5 when throwing a die, and that is why we say that said probability is 1/6 (that is, one sixth of the possibilities). in Game).
But it is not always easy to evaluate the favorable cases and/or the possible cases. d’Alembert’s mistake mentioned last week was that he failed to realize that the three possible cases: heads-heads, heads-tails, tails-tails, are not equally likely (equally likely, in mathematical parlance), since two heads or two tails can only come out one way – if both coins land on the same side – while the combination of one head and one tail can come out two ways: heads on the first coin and tails on the second or vice versa. Therefore, there are actually four possible cases: heads-heads, heads-tails, tails-heads, and tails-tails. In three of these cases there is at least one head, so the requested probability is 3/4, and not 2/3 as d’Alembert estimated.
As for Bertrand’s paradox, last week’s figure suggests that the probability of randomly drawing a chord larger than the side of the inscribed equilateral triangle is 1/3, since the blue chords occupy two of the three equal arcs in that the triangle divides the circumference, while the green ones only have one. But we could also approach it like this:
Consider a radius perpendicular to one side of the inscribed triangle. All the chords perpendicular to said radius that remain between the center of the circumference and the side of the triangle are greater than it, and all those that remain between the side and the other end of the radius are inferior. And since the side of the inscribed equilateral triangle divides the radius into two equal parts, now the green strings in the figure cover the same space as the blue ones, and therefore the requested probability will be 1/2. And there are also other possible approaches with different results (I invite my astute readers to look for an approach in which the requested probability turns out -or seems- to be 1/4).
As far as the haggadah of the sweaty walkers is concerned, the point (as the rabbi explains to his disciple) is to question the statement: since they have been walking on a dusty path in the sun for the same amount of time, what is it like? Is it possible that one of them has a dirty face and the other keeps it clean?
The currency problem
From d’Alembert’s currency we can move on to that of another illustrious mathematician, Ferdinand Georg Frobenius (1849-1917), who made important contributions to group theory and is known above all for the so-called “currency problem” (or , in his honor, “Frobenius problem”), which consists of finding the greatest amount of money that cannot be obtained using only coins (and/or bills) of a certain value; that maximum amount is the Frobenius number for that set of coins. For example, with 2-euro coins and 5-euro bills, the Frobenius number is 3, since any even quantity can be obtained with 2-euro coins and any odd quantity greater than 3 can be expressed in the form 5+2n, for what with a bill of 5 and coins of 2 we can pay any whole amount of euros (except 1 and 3) without having to give us change.
The cases with real coins are very simple, so I invite my astute readers to find the Frobenius number for hypothetical coins of 3 and 5 euros, 7 and 9 or other atypical pairs.
Carlo Frabetti is a writer and mathematician, member of the New York Academy of Sciences. He has published more than 50 popular science works for adults, children and young people, including ‘Damn Physics’, ‘Damn Mathematics’ or ‘The Great Game’. He was a screenwriter for ‘The Crystal Ball’.