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 August 27th, 2017, 05:39 PM #1 Newbie   Joined: Aug 2017 From: Hong Kong Posts: 3 Thanks: 0 Math Focus: Calculus 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 05:42 PM.
August 27th, 2017, 06:20 PM   #2
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Quote:
 Originally Posted by Toucanf16 Example: First roll is 2-2-2-3-3-3.
Is that SAME as 2-3-3-2-2-3 ?

 August 27th, 2017, 07:32 PM #3 Newbie   Joined: Aug 2017 From: Hong Kong Posts: 3 Thanks: 0 Math Focus: Calculus Yes, The roll can be any combination as long as it's three of one number and three of another number.
August 27th, 2017, 07:34 PM   #4
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Quote:
 Originally Posted by Toucanf16 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 $p = \dfrac{6^4}{6^{12}} = 6^{-8} \approx 5.59\cdot 10^{-7}$

 August 27th, 2017, 07:50 PM #5 Newbie   Joined: Aug 2017 From: Hong Kong Posts: 3 Thanks: 0 Math Focus: Calculus Romsek, thanks! Matt
 August 29th, 2017, 11:04 AM #6 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 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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