Multiple dice probabilities
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
link to original post
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
Quote: teliotWhere are you making sure that x, y & z are different?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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link to original post
"20 ways to choose 3 numbers from 6" refers to the three different numbers.
There are C(9,6) = 84 ways to place the lowest of the three numbers among the nine dice, and C(6,3) = 20 ways to place the second of the three numbers among the remaining six dice.
The calculation for “at least 4 of a kind” requires many more steps since you must consider the following possible combinations.Quote: 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
AAAAAAAAA
AAAAAAAAB
AAAAAAABB
AAAAAAABC
AAAAAABBB
AAAAAABBC
AAAAAABCD
AAAAABBBB
AAAAABBBC
AAAAABBCC
AAAAABBCD
AAAAABCDE
AAAABBBBC
AAAABBBCC
AAAABBBCD
AAAABBCCD
AAAABBCDE
AAAABCDEF
Calculating them all would be good practice for someone trying to improve their combinatorial skills
Yep -- I was just thinking about it differently. But concur, hence edited my OP.Quote: ThatDonGuyQuote: teliotWhere are you making sure that x, y & z are different?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
link to original post
link to original post
"20 ways to choose 3 numbers from 6" refers to the three different numbers.
There are C(9,6) = 84 ways to place the lowest of the three numbers among the nine dice, and C(6,3) = 20 ways to place the second of the three numbers among the remaining six dice.
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On second thought, there’s a more efficient way to calculate this using inclusion-exclusion, much easier than calculating all the possible permutations listed aboveQuote: Ace2The calculation for “at least 4 of a kind” requires many more steps since you must consider the following possible combinations.Quote: 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
AAAAAAAAA
AAAAAAAAB
AAAAAAABB
AAAAAAABC
AAAAAABBB
AAAAAABBC
AAAAAABCD
AAAAABBBB
AAAAABBBC
AAAAABBCC
AAAAABBCD
AAAAABCDE
AAAABBBBC
AAAABBBCC
AAAABBBCD
AAAABBCCD
AAAABBCDE
AAAABCDEF
Calculating them all would be good practice for someone trying to improve their combinatorial skills
link to original post
For any individual number, there’s a 483,946 / 6^9 =~ 4.8021% chance it will appear four or more times in nine rolls. This is easily calculated by subtracting the ways of getting 0-3 appearances of the individual number from 6^9 possibilities
6 * 4.8021% would be a 28.8129% chance of getting four or more of any of the six possible numbers. But that double counts instances where two numbers have hit four or more times .
It’s easy to calculate there are 2,772 ways in 6^9 =~ 0.02751% chance of rolling two numbers four or more times. That times c(6.2) ways to choose the two numbers is 0.4126%. Be careful with this step…there 2,772 ways to roll two numbers including the one that’s already been selected (not any two)
28.8129% - 0.4126% = 28.4003% chance of rolling at least one 4-of-a-kind with nine dice
