Problematic pardons

Joaquin Phoenix, in a photogram of 'Gladiator'.

Of our three prisoners from last week, Carlos is the only one who has reason to be happy, a lot.

Alberto's assumption that, after learning that Bernardo will not be pardoned, his probability has risen from 1/3 to 1/2 is wrong.

He already knew that one of the other two was not going to be pardoned, and finding out his name doesn't change anything.

However, since the probability that the pardoned was between Bernardo and Carlos was 2/3, when Bernardo is discarded, that probability is concentrated in Carlos, who thus sees his probability of pardon doubled.

The optimal strategy in the case of 100 prisoners and 100 drawers is as follows:

Each prisoner opens, first, the drawer whose number matches that of his uniform.

If the card with your number is inside, you have already passed the test;

If, as is most likely, there is another number, then he opens the drawer corresponding to that number, and so on until his number is found or his quota of 50 drawers is exhausted.

In an excellent article by Clara Grima entitled

, readers who wish to delve into this interesting problem will find a detailed development illustrated with graphs.

Since the probability of success, when the prisoners do not follow any strategy, decreases exponentially as their number increases, it could be thought that if this number went from, say, 100 to 1000, their probability of success would decrease considerably no matter how much they applied. the optimal strategy;

but surprisingly, this is not the case: no matter how much the number of prisoners and cajons increases, the probability of getting pardon always remains slightly above 30%.

The unexpected hanging

Two other classics that cannot be left out when talking about problems of prisoners and pardons are that of black and white hats and that of unexpected hanging.

Although it is likely that my sagacious readers already know them, it is worth remembering them:

A king decides to pardon one of three prisoners.

He sends for them and tells them: “In this chest there are three white hats and two black ones.

Now you will face the wall and I will have them put a hat on each of you, so that each of you will be able to see the hats of the other two, but not your own.

Whoever deduces what color his hat is, will be free ”.

They put the three white hats on them and immediately afterwards the three prisoners look at each other in silence.

One of them shakes his head in bewilderment, another shrugs his shoulders and the third announces: "My hat is white."

How did you figure it out?

The paradox (or is it a fallacy?) Of the unexpected hanging, which, incidentally, gives the title to a book by the teacher Martin Gardner, is the following:

A judge condemns a cunning criminal to death by hanging and tells him: "You will be hanged one day next week, but you will not know in advance the date of your execution."

"That means that I cannot be hanged on Sunday, because, being the last day of the week, on Saturday I would know it," argues the condemned man.

Nor can I be hanged on Saturday, because, by being ruled out on Sunday, on Friday I would know that I was going to be executed the next day.

And, continuing with this reasoning, every day they are discarded one after another, so I will not be hanged ”.

Is your reasoning correct?

If not, where is the fault?

Carlo Frabetti

You can follow