Troubled prisoners


By CAPosts 20 November, 2020 - 05:21am 25 views

The problem of black and white balls, raised three weeks ago, sparked an intense and interesting debate that revealed, once again, how misleading and counterintuitive probabilistic estimates can be. This problem is related to another that, at first glance, seems totally different, but it is not so much: Three prisoners, whom we will call Alberto, Bernardo and Carlos, know that one of them is going to be pardoned, but they do not know which of the three. Alberto bribes the jailer to tell him the name of one of his companions in misfortune who is not going to be pardoned. The jailer accepts the bribe and tells him: "Bernardo is not going to be pardoned." Alberto feels a little better, because he thinks that now his probability of being pardoned is fifty percent, since there are only two candidates: Carlos and him. Is your relief justified? And assuming Carlos found out what the jailer has said, could he be as comparatively relieved as Alberto?

The Problem of 100 Prisoners

You cannot talk about prisoner problems without mentioning some classics, such as the famous Prisoner's Dilemma, or the riddle of the three white hats and the two black hats, or the one of the 100 prisoners and the 100 numbered drawers. For now, we will deal with the latter: In a prison there are 100 prisoners sentenced to death, each with a different number, from 1 to 100, in their prison uniform. In a prison office there is a locker with 100 drawers, also numbered from 1 to 100, and in each drawer there is a card with a number, also from 1 to 100, which does not have to coincide with the number of the drawer that contains it. contains, since the cards have been randomly distributed. The prison director proposes the following game to them: each prisoner, separately, will enter the locker's office and can open a maximum of 50 drawers to find the card with the same number as his uniform, and before leaving he will close all of them again the drawers. If all the prisoners find their number, they will all be pardoned; But it will be enough if one of them does not find their number for all of them to be executed

The probability that a prisoner finds his number by opening half of the drawers is obviously 1/2, so the probability that all of them they achieve it is 1/2 raised to the power 100, less than one in a quintillion, that is, practically nil. Prisoners cannot communicate once the game has started, but they can agree on a strategy before starting. And in fact, there is a strategy that makes the probability of success greater than 30%. What is it?

This is the original statement of the problem, as formulated in 2003 by Danish computer scientists Anna Gal and Peter Bro Miltersen; but I suggest my astute readers to start trying to solve it with fewer prisoners and drawers: 2, 4, 6, 8 ... (always an even number, due to opening half of the drawers). It is evident that reducing the number of prisoners and boxes simplifies the problem and increases the probability that the convicted will be saved; but what if we increased it? Will the more prisoners and crates decrease the probability of being saved?

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 maths' or 'The great game'. He was a screenwriter for 'La bola de cristal'.

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