Combined probabilty for 1 to 6 between a cube and a dodecahedron, not counting 7 to 12

Question

We have two dice, one with six sides (a cube)

and one with twelve (a dodecahedron)

The cube bears the numbers 1 to 6. The probability for each number is $\frac{1}{6}$. The dodecahedron bears the numbers 1 to 12. The probability for each number is $\frac{1}{12}$.

What is the combined probability for the numbers 1 to 6, when you roll the cube n_C times and the dodecahedron n_D times and discount all results of 7 or above?

That is, all results between 1 and 6 from both cubes are counted as valid, and all results between 7 and 12 from the dodecahedron are considered invalid and the toss is not counted (as if it had not happened).

For example, you roll 5, 3, 6 with the cube and 2, 8, 9 with the dodecahedron. As a result you have rolled n = 4 times and the results are 2, 3, 5, 6.

Is it correct to assume that the combined probability for each of the numbers 1 to 6 is $\frac{1}{6}$?

Answer 1

Looking at the dodecahedron, if we roll it and discard everything higher than six, then this "truncated dodecahedron" functions precisely as a standard six-sided die: every roll between 1 and 6 has equal probability $\frac{1}{6}$.

We thus concatenate a number of rolls from the cube (where every roll between 1 and 6 has equal probability $\frac{1}{6}$) with a different number of rolls from the truncated dodecahedron (where every roll between 1 and 6 has equal probability $\frac{1}{6}$).

Overall, we have a fancy way of simulating many rolls from a standard six-sided die: every roll between 1 and 6 has equal probability $\frac{1}{6}$.

(Some copy-pasting was involved in the crafting of this answer.)