On a roll of 9 dice (d6), which are the probabilities of getting:

- three 3 of a kind (xxxyyyzzz configuration)

- at least 4 of a kind?

Many thanks to whoever could help

To clarify, you're asking about a single roll of each die? (Possibly all at once)

That's correct, a single roll of the 9 dice at once.

The effects of turkey overdose and liquid holiday cheer are doing a tapdance on my mind.

For 3 of a kind in xxxyyyzzz format there should be a 20 * 84 * 20 / 6^9 chanceQuote:MorgSort of an odd (and quite specific) question.

On a roll of 9 dice (d6), which are the probabilities of getting:

- three 3 of a kind (xxxyyyzzz configuration)

- at least 4 of a kind?

Many thanks to whoever could help

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Quote:Ace2For 3 of a kind in xxxyyyzzz format there should be a 20 * 84 * 20 / 6^9 chanceQuote:MorgSort of an odd (and quite specific) question.

On a roll of 9 dice (d6), which are the probabilities of getting:

- three 3 of a kind (xxxyyyzzz configuration)

- at least 4 of a kind?

Many thanks to whoever could help

link to original post

link to original post

Many thanks Ace2!

Could you guide me through the resolution?

Quote:DieterI'm going to have to recuse myself from math until at least Saturday, I'm afraid.

The effects of turkey overdose and liquid holiday cheer are doing a tapdance on my mind.

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No worries at all of course, many thanks for looking into this!

I hope you had a wonderful Thanksgiving! :)

Assuming this is correct. But this is the method of counting valid permutations

Verified. This is correct. The answer for xxxyyyzzz is 0.003334.Quote:Ace220 ways to chose 3 numbers from 6 times 84 ways to place first triple among 9 places, times 20 ways to place second triple among 6 places. Divide by 6^9 total permutations

Assuming this is correct. But this is the method of counting valid permutations

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The answer for xxxxabcde is 0.284003