Paradox of the three coins

The Paradoxe of the three coins rests on a reasoning subtly fallacious, but clearly and incontestably identifiable. In that, it is by no means a Paradoxe, but a good exercise of probabilistic reasoning .

Statement

Three coins are launched. Which is the probability that all three fall down on the same side, that it either pile or face? One on four. That is to say.

However, if I launch three parts, there are inevitably two of them which will be already same side; the third will be there with a chance on two. I thus have a chance on two that all three fall on the same side.

Interpretations

The good answer is of course: 1/4. Let us detail the involved reasoning:

The énumératif reasoning

There are 8 possibilities: PPP, PF, PFP, PF, FP, FPF, FP, FF; and 2 successful outcomes: PPP, FF. The probability is thus 2/8, that is to say one on four.

The fallacious reasoning

Let us illustrate that through a dialog between Candide and professor Cosinus:

- Ingenuous: If I launch three parts, there are inevitably two of them which will be already same side.

- Cosine: Indeed.

- Ingenuous: There is thus well a chance on two that the third is same side.

- Cosine: At all!

- Ingenuous: But yes. There are 4 possibilities: 2xP+P, 2xP+F, 2xF+P, 2xF+F; and 2 successful outcomes: 2xP+P, 2xF+F.

- Cosine: Yes but, contrary to the énumératif reasoning, the possibilities are not equiprobable.

- Ingenuous: I.e.?

- Cosine: For example: 2xP+P only corresponds to PPP; but 2xP+F corresponds to PF, PFP or FP. The case 2xP+F is thus three times more probable than 2xP+P.

- Ingenuous: But finally, the first two parts give pile or face with a probability 1/2; the third gives also pile or face with a probability 1/2; thus the probability that the result of the first two parts is identical to the result of the third is 1/2.

- Cosine: Your premises are right; but the conclusion is false in fact. That does not have anything to see. What allows you such a conclusion?

- Ingenuous: Euh… Good… But let us suppose that one numbers the parts without looking at them, then one discovers that the parts n°1 and n°2 give pile. Which is the probability so that the part n°3 gives pile?

- Cosine: 1/2.

- Ingenuous: Ah! I am right thus.

- Cosine: At all. The situation is different.

- Ingenuous: In what?

- Cosine: When the parts are numbered in the ignorance of (or before) pulling, the part n°3 is not correlated to both others. The third part, on the other hand, is selected or given a posteriori . It is correlated to both others. It is, coarsely the part from which the result can be different from the other .

- Ingenuous: But then, how to precisely continue my reasoning?

- Cosine: I should have added that, “the probability so that the part n°3 gives pile when the parts n°1 and n°2 give pile” means “the probability so that the part n°3 gives pile knowing that the parts n°1 and n°2 give pile. ” It is thus necessary to make use of the conditional probabilities.

The reasoning using the conditional probabilities

Let us note $P1$, $P2$, $P3$ the result (pile/face) the parts numbered according to the experiment of Ingenuous. Probability so that the three parts " are semblables" knowing that the parts n°1 and n°2 " are semblables" is: $P\; (P3=P2|P2=P1)\; =\; P\; (P3=P1|P2=P1)\; =\; P\; (P3=P2=P1|P2=P1)\; =\; \backslash ,$
$\{P\; (P3=P2=P1)\; \backslash \; over\; P\; (P2=P1)\}\; =\; \{P\; (P3=P2)\; P\; (P3=P1)\; \backslash \; over\; P\; (P2=P1)\}\; =\; \{(1/2)\; (1/2)\; \backslash \; over\; \{1/2\}\}\; =\; \{1\; \backslash \; over\; 2\}$. But this problem does not correspond to the initial statement. The good reasoning, built by analogy, leads to an artificially complicated calculation: Let us note $P$, $D$, $T$ the result (pile/face) of the first, second and third parts defined according to the process of the statement. Probability so that the third parts " that is to say semblable" with the two first is: $P\; (T=D=P|D=P)\; =\; \{P\; (T=D=P)\; \backslash \; over\; P\; (D=P)\}\; =\; \{P\; (T=D)\; P\; (T=P)\; \backslash \; over\; P\; (D=P)\}\; =\; \{(1/2)\; (1/2)\; \backslash \; over\; 1\}\; =\; \{1\; \backslash \; over\; 4\}$. Here $P\; (P=D)=1$ because the two first " are semblables" by definition.

To finish, it is important to notice that if $P\; (T=D\; \backslash \; hbox\; \{and\}\; T=P)\; =\; P\; (T=D)\; P\; (T=P)$; on the other hand $P\; (T=D\; \backslash \; hbox\; \{and\}\; D=P)\; =\; P\; (T=D)\; P\; (D=P)$ is false. In a general way, there is not $P\; (has\; \backslash \; hbox\; \{and\}\; B)=P\; (A).P\; (B)$. This is why, it is advised to reason in terms of intersections of whole rather than in terms of conjunctions of proposals:

$P\; \backslash \; left\; (\backslash \; left\; \backslash \; \backslash \; right)\; =\; P\; (\backslash \; \{X\; \backslash \; in\; \backslash \; Omega|With\; (X)\; \backslash \}\; \backslash ;\; \backslash \; course\; \backslash ;\; \backslash \; \{X\; \backslash \; in\; \backslash \; Omega|B\; (X)\; \backslash \})\; =$

See also: Axioms of the probabilities

See too

Article related

• List of paradoxes
• probabilistic Paradox
• Paradox of the prisoners
• Paradox of the two children
• Problem of Monty Hall
• Problem of Beautiful with wood sleeping

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