Give that two six sided dice are rolled once

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The odds against rolling a sum that is a multiple of 3 with two six-sided dice is definitively 2:1. When rolling two dice, the possible sums range from 2 to 12, with the multiples of 3 being 3, 6, 9, and 12. There are 12 combinations that yield these sums out of a total of 36 possible combinations. Therefore, the probability of rolling a sum that is a multiple of 3 is 1/3, leading to odds of 2 to 1 against rolling such a sum.

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SeththeBaller
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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!
 

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