Paradox of the prisoners

The Paradoxe of (three) captive the proposed by J. Pearl is a simple calculation of Probabilité S. It should not be confused with the Dilemme of the prisoner invented by Merrill Mr. Flood and Melvin Dresher in 1950 and which concerns the Game theory.

Statement

Three prisoners are in a cell. They know that two will be condemned to dead and pardoned, but they do not know which. One of them will see the guard and asks him: “I know well that you can nothing say to me, but you can at least show me one of my companions who will be carried out”. The guard reflects, thinks that in all manners at least one of the two other prisoners will be condemned, and is carried out. The prisoner answers him then: “Thank you, front, I had a chance on three to be pardoned, and now, I have a chance on two. ”

The answer still appears more clearly paradoxical, if it is known that the name of pardoned was registered in the minutes of the lawsuit. The chances of survival of the prisoner cannot thus vary any more.

Interpretations

Equiprobable the chances of the prisoners will be supposed. One also excludes the lie or a form preferably in the answer from the guard. Let us appoint by R the prisoner who answers (the arguer), D the appointed prisoner and T the third, and note G the prisoner who is pardoned.

Value 1/2 corresponds then (or seems to correspond) to the probability:

$P\; (G=r\; |\; G\; \backslash \; not=d)\; =\; 1/2$. This probability takes well into account the answer of the guard G ≠ D . But, actually the arguer occults important information here: its own request. The reasoning would be valid if its request had been: “Can you indicate one of us three who will be condemned? ” But such is not the case.

Taking into account the whole of information which one has with the end the dialog, the chances of survival of the arguer are, not P ( G=r | G≠d ), but P ( G=r | I=d ) where I is the response of the guard at the request of the arguer. A calculation of conditional probabilities gives

$P\; (G=r\; |\; I=d)\; =\; \{P\; (G=r)\; P\; (I=d\; |\; G=r)\; \backslash \; over\; P\; (I=d)\}$; where in addition $P\; (I=d)\; =\; P\; (G=r)\; P\; (I=d\; |\; G=r)\; +\; P\; (G=t)\; \backslash ,$. According to the assumptions $P\; (I=d\; |\; G=r)\; =\; 1/2\; \backslash \; quad\; \backslash \; hbox\; \{and\}\; \backslash \; quad\; P\; (G=r)\; =\; P\; (G=d)\; =\; P\; (G=t)\; =\; 1/3$; thus $P\; (I=d)\; =\; 1/2\; \backslash \; quad\; \backslash \; hbox\; \{and\}\; \backslash \; quad\; P\; (G=r\; |\; I=d)\; =\; 1/3$.

P ( I=d | G=r ) = 1/2 translated the absence preferably in the answer of the guard. This a priori consists in supposing that the guard is neutral in his choice. This assumption is not of nature different from that of the equiprobability. However, without this assumption, the answer of the arguer can be justified by his conviction (unfounded) that the guard indicates D as soon as it can it (i.e., P ( I=d | G=r ) = 1).

On the other hand the chances of survival of the other prisoners evolved/moved: P ( G=d | I=d ) = 0 expresses that the guard does not lie, and

$P\; (G=t\; |\; I=d)\; =\; \{P\; (G=t\; \backslash \; hbox\; \{and\}\; I=d)\; \backslash \; over\; P\; (I=d)\}\; =\; \{P\; (G=t)\; \backslash \; over\; P\; (I=d)\}\; =\; \{1/3\; \backslash \; over\; 1/2\}\; =\; 2/3$ because G=t ⇒ I=d .

Conclusions

Therefore, the prisoner has always only one chance on 3d' to be pardoned, on the other hand, information profits with the prisoner not appointed, who sees his chance to be pardoned to go up to 2/3. If this problem resembles the Paradoxe of the two children (even values of probability), it remotely by nature. It is of a fallacious reasoning and not about a true paradox. Although the semantic blur is obvious: two values of probability are advanced by the arguer without clearly specifying the associated random variables; he does not justify of anything the value 1/2, which reveals an internal contradiction in the remarks of the arguer.

J. Pearl introduced the paradox of the three prisoners with an aim of showing that the analyzes bayésienne provides a powerful tool for formalization of the reasoning in the dubious one. This example illustrates especially at which point this tool is delicate to employ.

Prolongation

Let us suppose now that the prisoners are in three numbered individual cells. One of the numbers was drawn with the fate and the prisoner occupying the cell associated with this number will be pardoned. Finally the guard makes it possible to the arguer to exchange his place with one of his congeneric. What has to make the arguer?

And well, while permuting with the third prisoner, it adapts the chances of survival of this last; its chances of survival thus pass from 1/3 to 2/3. To be convinced some, it should be considered that the arguer finds himself in the situation of a player confronted with the Problème of Monty Hall.

See too

Article related

• List of paradoxes
• probabilistic Paradox
• Paradox of the three coins
• Paradox of the two children
• Problem of Monty Hall

External bonds

• Patrick Fabiani. the paradox of the three prisoners , 1996. Expose various reasoning.

References

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