# Chevalier de mere paradox with game with three dice

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?

• Yes, 27 versus 25 looks correct to me. – almagest May 28 '16 at 15:15
• The calculation is right. I would say that the calculation that yields equiprobable uses the wrong probability model. – André Nicolas May 28 '16 at 15:45