My intention is to add a twist to the dice game called "10,000 dice" by using three colored pairs of casino dice. My thought is that more fun probability-scoring could be incorporated. My problem is that I am not a mathematician.
What is the probability of simultaneously rolling three sets of colored 6-sided fair dice (a red pair, a green pair, and a blue pair) and the red pair rolling the lowest pair-set, the green pair rolling a middle-numbered pair-set, and the blue pair rolling the highest numbered pair set?
For example, the red pair rolls a pair of twos; the green pair rolls a pair of fours; and the blue pair rolls a pair of sixes?
I'm thinking that if you were to roll such a combination you should earn quite a bonus. I'm also interested in hearing your thoughts about what the bonus should be if you are familiar with the game and its scoring.
Metta,
Crunchy Guero
First, let us figure how many ways we can pick three distinct numbers $r,g,b$ from $1$ to $6$ with $r < g < b$. That would be: $${6 \choose 3} = 20$$
where we choose $3$ of the $6$ die faces without repetition in the usual way, and then let $r$ be the lowest of those numbers, $g$ the next lowest, and $b$ the highest.
Now, for each of those choices, the odds of throwing a pair of $r$'s on the red dice and a pair of $g$'s on the green dice and a pair of $b$'s on the blue dice is
$$\frac{1}{36}\frac{1}{36}\frac{1}{36} = \frac{1}{6^6}$$
So the total probability is
$$\frac{20}{6^6} = \frac{5}{11664} \approx 0.00042867$$
So - pretty unlikely! About once every $2332$ rolls...
Note: the odds of just throwing a pair on each of the red, green and blue dice of any kind is $(\frac{1}{6})^3 = \frac{1}{216}$.