Roland Bones tosses a pair of dice. What are the odds that he won't roll a $2,3,7,11,$ or $12$.

I will quietly put this out there that I have two interpretations of this problem mainly because of the phrase "pair of dice." The results $2,3,7,11$ or $12$ either mean the sum of the results rolled or simply possible outcomes on either of the two dice; in other words, a dice "could roll" a $2,3,7,11,12$.

The first case can easily be solved by listing all the cases. The probability of the attaining the undesired numbers as a sum of two numbers is $\frac{1}{3}$. So the probability of not getting one of these numbers is $\frac{2}{3}$ or an odds of $2:1$.

Two questions (or one if you bash out the first): Is the second interpretation possible? If so, what are the odds of rolling $2$ such that the a number doesn't roll a $2,3,7,11,$ or $12$?

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    $\begingroup$ The second interpretation is plausible, but would make very little sense, as the numbers "7", "11" and "12" do not appear on a standard (6-sided) die. $\endgroup$ – barak manos Aug 26 '16 at 4:18
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    $\begingroup$ Second thought, I'd replace "plausible" with "possible". It is definitely not plausible (unless whoever asked you this question was mean and evil). $\endgroup$ – barak manos Aug 26 '16 at 4:19
  • $\begingroup$ Hahaha. I'll edit that immediately. Thought what is the odds still? $\endgroup$ – Ian Limarta Aug 26 '16 at 4:40
  • $\begingroup$ Odds for that? That this is the actual question, or that you'll get either one of these numbers on either one of the dice? $\endgroup$ – barak manos Aug 26 '16 at 4:51
  • $\begingroup$ As an avid boardgamer, I am often frustrated with the lack of detail often included in "dice related" questions. There are many shapes, sizes, associated probabilities, and ways of labeling faces available for dice. They range from having three sides to having 100 sides (the most commonly used number of sides being 4,6,8,10,12,20). It should be included in the problem statement that: 1) The dice used are standard fair six-sided dice. 2) Having tossed a pair of dice we look at the sum of the faceup numbers /or/ we look at the faceup numbers individually(this would have removed confusion) $\endgroup$ – JMoravitz Aug 26 '16 at 4:56

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