## Task 24/70

What is the probability of rolling a sum of 12 with three dice?
For the calculation of the probability, one can use the classical definition, as only a finite number of possibilities exist. It is thus the quotient $\frac{g}{m}$ of the number $g$ of favorable cases and the number $m$ of possible cases.

The number $m$ of possible cases is $m=6^{3}=216$. This is because it involves all combinations of the 6 numbers in groups of three, taking into account the order, with repetition (thus, variations of 6 elements of the 3rd class with repetition).

The number $g$ of favorable cases results from the following consideration: Favorable are all cases where the addition of three natural numbers between 1 and 6 (inclusive) results in the sum of 12, with the order being taken into account. Without considering the order, the following possibilities arise first:

1) $6+5+1=12$
2) $6+4+2=12$
3) $6+3+3=12$
4) $5+5+2=12$
5) $5+4+3=12$
6) $4+4+4=12$

When the order is taken into account, it becomes clear: In cases 1., 2., and 5., there are exactly $3!=6$ different possibilities each, in cases 3. and 4. exactly $\frac{3!}{2!}=3$ each, and in case 6. only 1. Thus, $g=3 \cdot 6+2 \cdot 3+1=25$ and it is $\frac{g}{m}=\frac{25}{216} \approx 0.116$.