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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

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  • $\begingroup$ A question can be best answered if complete. An answer cannot be given without knowing the rules of 10,000 dice. Perhaps you could give a synopsis of the rules. $\endgroup$ – robjohn Dec 10 '16 at 1:02
  • $\begingroup$ Are you saying that you roll a pair (2) of each colour all at the same time? $\endgroup$ – IntegrateThis Dec 10 '16 at 1:09
  • $\begingroup$ James Dickens: yes, all at once. $\endgroup$ – Crunchy Guero Dec 10 '16 at 1:44
  • $\begingroup$ Robjon: here is a link for the rules: en.m.wikipedia.org/wiki/Dice_10000 $\endgroup$ – Crunchy Guero Dec 10 '16 at 1:57
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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 $(\frac{1}{6})^3 = \frac{1}{216}$.

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  • $\begingroup$ Chas Brown, you are a genius. :-) Thank you very much. $\endgroup$ – Crunchy Guero Dec 10 '16 at 1:52

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