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Chevalier de Mere asked Blaise Pascal why in a game with three dice the sum $11$ is more favorable than $12$, when both sums have exactly the same possible combinations:

For $11$ we have $(5,5,1), (5,4,2), (5,3,3), (4, 4, 3), (6,4,1), (6,3,2)$ and for $12$ we have $(6,5,1), (6,4,2), (6,3,3), (5,4,3), (4,4,4), (5,5,2)$, so both sums should be equiprobable.

My attempt: I think Chevalier de Mere made the mistake of thinking all the dice are indistinguishable. I tried to compute the exact probabilities.

Let $$\Omega = \left\{(x,y,z) \mid 1 \leq x,y,z \leq 6, \quad 3 \leq x + y + z \leq 18\right\} $$ be the sample space. We are interested in the events $$A = \left\{(x,y,z) \mid x + y + z = 11 \right\} $$ and $$ B = \left\{(x,y,z) \mid x + y + z = 12 \right\}. $$ For the sum $11$, we have $27$ possible permutations of all the triples. For the sum $12$, there are two less, that is $25$. So $\#A = 27$ and $\#B = 25$. Since $\#\Omega = 6 \cdot 6 \cdot 6 = 216$, we have $$ P(A) = \frac{27}{216} = 0.125, \qquad P(B) = \frac{25}{216} = 0.1157. $$ Is this reasoning correct?

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  • $\begingroup$ Yes, 27 versus 25 looks correct to me. $\endgroup$ – almagest May 28 '16 at 15:15
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    $\begingroup$ The calculation is right. I would say that the calculation that yields equiprobable uses the wrong probability model. $\endgroup$ – André Nicolas May 28 '16 at 15:45
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You're making the very common mistake of confusing issues of distinguishability with issues of equiprobability. If you roll three indistinguishable dice, the probabilities of the various sums are exactly the same as if you roll three distinguishable dice. The dynamics of the dice are not influenced by your ability to distinguish among them.

Chevalier de Méré's error was not to think of the dice as indistinguishable, but to regard the wrong sorts of events as equiprobable. Absent prior information, events should be regarded as equiprobable when there is symmetry among them. In the present case the six possible results of each individual die should be regarded as equiprobable, since there is symmetry among the six sides. It then follows that each combination of these independent equiprobable events, that is, each ordered triple of dice results should also be considered equiprobable, as you did in your calculation. Instead, Chevalier de Méré regarded each unordered triple as equiprobable. This is wrong since unordered triples with different numbers of coinciding dice correspond to different numbers of ordered triples, and thus to different numbers of equiprobable elementary events.

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