Give that two six sided dice are rolled once

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When rolling two six-sided dice, the odds against rolling a sum that is a multiple of 3 are calculated based on the total combinations. There are 12 combinations that yield a multiple of 3 out of 36 total combinations. This results in a probability of 24/36 for not rolling a multiple of 3, which simplifies to 2/3. However, the correct interpretation of odds is 2 to 1, reflecting that there are twice as many outcomes that do not result in a multiple of 3 compared to those that do. Thus, the final odds against rolling a sum that is a multiple of 3 are 2:1.
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1. Given that two six sided dice are rolled once, what are the odds against rolling a sum that is a multiple of 3



3. The answer I got was 2:3

Was just wondering if this is correct.
 
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The highest possible number with a single roll of two dice is 12. Multiples of three going up to 12 are 3,6,9, and 12. Looking at the grid, there are 12 out of 36 combinations that give a sum that is a multiple of three ((1,2), (1,5), (2,1), (2,4), (3,3), (3,6), (4,2), (4,5), (5,1), (5,4), (6, 3),(6,6)). This means that the probability against rolling a sum that is a multiple of three is 24/36 because 36-12=24.

24/36 can be simplified to 2/3 , thus, the odds against rolling a sum that is a multiple of 3 is 2:3.
 
Your calculation of the probability is fine. However, the odds would be 2 to 1 because there are twice as many ways to not roll a multiple of 3 as there are ways to roll a multiple of 3. The odds is the ratio of the relative probabilities of an event and its complement, so in this case you have (2/3) to (1/3) or 2 to 1.
 
Thank you!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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