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Suppose you roll three 4-sided dice to arrive at one of 64 combinations.

Of these 64 combinations, 40 are "good," 15 are "bad," and 9 are "mixed."

You have to repeat the process three times, which makes for ten possible outcomes, e.g., 3 good; 2 good and 1 mixed; 2 good and 1 bad, etc. down to 3 bad.

What are the probabilities of each of the ten possible outcomes, given that one is more likely to receive a good combination (.625) versus a mixed (.14) or bad (.23) combination? What is the formula?

(In case you're wondering, this problem arises from a form of dice divination practiced in India and elsewhere with long dice called pashakas.)

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There are ten combinations. $\sf\{GGG,GGM,GGB,GMM,GMB,GBB,MMM,MMB,MBB,BBB\}$

The counts of good, mixed, and bad results among three trials will follow what is known as a multinomial distribution.

So the joint probability mass function is:$$\mathsf P(G{=}g, M{=}m, B{=}b) = \dfrac{3!}{g!~m!~b!}\frac{{40}^{\raise{0.5ex}g}{9}^{\raise{0.5ex}m}{15}^{\raise{0.5ex}b}}{64^3}~\mathbf 1_{g\geq 0, m\geq 0, b\geq 0,g+m+b=3}$$

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For each outcome (consisting of three fortunes) multiply the individual fortunes' probabilities together, and then multiply by the number of ways of permuting the fortunes. For example, the probability of getting one of each fortune (good, bad, mixed) is $$6×\frac{40}{64}×\frac{15}{64}×\frac9{64}=\frac{2025}{16384}=0.1236$$ The probability of getting one bad and two good fortunes is $$3×\frac{40}{64}×\frac{40}{64}×\frac{15}{64}=\frac{1125}{4096}=0.2747$$ The probability of getting three mixed fortunes is $$1×\frac9{64}×\frac9{64}×\frac9{64}=\frac{729}{262144}=0.002781$$ The number of ways of permuting the fortunes is six in the good + bad + mixed case, one if all fortunes are the same and three otherwise.

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