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August 27th, 2017, 06:39 PM   #1
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Dice question, roll a double triple twice in a row

New question...
6 dice.
What are the odds of rolling a douple triple twice in a row?
Example:
First roll is 2-2-2-3-3-3.
Next roll is also a double triple; say 1-1-1-6-6-6.
Thanks,
Matt

Last edited by Toucanf16; August 27th, 2017 at 06:42 PM.
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August 27th, 2017, 07:20 PM   #2
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Quote:
Originally Posted by Toucanf16 View Post
Example:
First roll is 2-2-2-3-3-3.
Is that SAME as 2-3-3-2-2-3 ?
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August 27th, 2017, 08:32 PM   #3
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Yes, The roll can be any combination as long as it's three of one number and three of another number.
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August 27th, 2017, 08:34 PM   #4
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Quote:
Originally Posted by Toucanf16 View Post
New question...
6 dice.
What are the odds of rolling a douple triple twice in a row?
Example:
First roll is 2-2-2-3-3-3.
Next roll is also a double triple; say 1-1-1-6-6-6.
Thanks,
Matt
so essentially you are rolling 12 dice and looking for a pattern

$aaabbbcccddd$ where it's possible that triplets repeat.

there are $6^{12}$ total roll combinations

of those $6^4$ will have the above pattern.

so $p = \dfrac{6^4}{6^{12}} = 6^{-8} \approx 5.59\cdot 10^{-7}$
Thanks from Denis and Toucanf16
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August 27th, 2017, 08:50 PM   #5
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Romsek, thanks!

Matt
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August 29th, 2017, 12:04 PM   #6
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I get 101^2 / 6^10 = 0.000168705... = 1.68705... * 10^-4.

Hope we did the same interpretation of the question!

The change of rolling a double triple is 101/6^5;
say the double triple is aaabbb
Then, if a = b, we have aaaaaa which happens in $\displaystyle \frac{6}{6^6} = \frac{1}{6^5}$ of the time
OR if a isn't b, we have aaabbb which can be arranged in $\displaystyle {6 \choose 3} = 20$ ways. This happens in $\displaystyle \frac{6 \cdot 5 \cdot {6 \choose 3}}{6^6} = \frac{100}{6^5}$

This adds to a total of $\displaystyle \frac{101}{6^5}$

Rolling a double triple is independent from earlier rolls so we can use the productrule for probabilities which says the answer is $\displaystyle \left( \frac{101}{6^5} \right)^2 = 1.68705... \cdot 10^{-4}$
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