Probability of one specific die rolling the highest in a pool

Question

I have four dice: three red, one blue. I roll all four.

1) How do I find the formula to establish the probability of the blue one getting the highest result (ties included: if all four rolled 6, then the blue one still got the highest result)?

2) How do I generalize the formula above for a group of four dice, each one with a different number of faces? (ie: one red 6-sided die, two red 8-sided dice, and a black 10-sided die)?

I apologize if anything is unclear, English isn't my first language.

Answer 1

For 1), you didn't specify the numbers of faces, but I take it from $2)$ that you intended them all to have the same number of faces in $1)$. Let $n$ denote this number of faces. Then the probability for the other three dice to show at most the value shown by the blue die is

$$ \frac1n\sum_{k=1}^n\frac{k^3}{n^3}=\frac1{n^4}\left(\frac{n(n+1)}2\right)^2=\frac14\left(1+\frac1n\right)^2\;. $$

The factor $\frac1n$ is the probability for the blue die to show one of the $n$ faces, and $\frac{k^3}{n^3}$ is the probability for the other three dice not to show more. As one might expect, the probability goes to $\frac14$ for $n\to\infty$.

For $2)$, you basically do the same thing but with a bit more casework. For instance, in the example you gave, the probability for the black die to show at least as much as all the others would be

$$ \frac1{10}\left(\sum_{k=1}^6\frac{k^3}{6\cdot8\cdot8}+\sum_{k=7}^8\frac{k^2}{8^2}+\sum_{k=9}^{10}1\right)=\frac1{10}\left(\frac14\cdot\frac{6^2(6+1)^2}{6\cdot8\cdot8}+\frac{7^2}{8^2}+\frac{8^2}{8^2}+2\right)=\frac{629}{1280}\;. $$

P.S.: As requested in a comment, here's the probability for one $8$-sided blue die to show not less than one $6$-sided and two $10$-sided red dice:

$$ \frac18\left(\sum_{k=1}^6\frac{k^3}{6\cdot10\cdot10}+\sum_{k=7}^8\frac{k^2}{10^2}\right)=\frac18\left(\frac14\cdot\frac{6^2(6+1)^2}{6\cdot10\cdot10}+\frac{7^2}{10^2}+\frac{8^2}{10^2}\right)=\frac{373}{1600}\;. $$

Answer 2

For (1): Treat the dice as being distinguishable (including the three red dice).

Then count favorable outcomes conditioning on the value of the blue die:

If the blue die is $1$, there is one favorable outcome: Each of the red dice is $1$.

If the blue die is $2$ there are eight favorable outcomes. Each of the red dice is $1$ or $2$

If the blue die is $3$ there are $27$ favorable outcomes... etc.

So the total number of favorable outcomes is $1+8+27+64+125+216$ out of $1296$ possible outcomes.

Problem (2) can be handled similarly.