There are three ways this can happen; we can roll the number on one die or the other or both; for numbers 1-6, this can occur in 11 ways out of the 36 possible rolls.

(This is even more likely than KeithT says! Keep in mind we have * two different* dice; to get a 6 outright, for example, there are eleven rolls that work– 1,6; or 2,6; or … or 6,6; or 6;5 or 6;4 or… or 6,1.)

We can roll the number as a sum:

sum 2 3 4 5 6 7 8 9 10 11 12 ways 1 2 3 4 5 6 5 4 3 2 1

Or as a product. We don’t have to worry about trivial products, where one of the dice shows ‘1’: for example, if we want to count the ways we could get a product of 6, we only need to count rolling a 2 on the first die and a 3 on the second, or vice versa— two additional ways; we’ve already counted rolling a 1,6 or a 6,1 when we counted the ways to get a six on one die or the other.

This gives:

product 4 6 8 9 10 12 15 16 18 20 24 25 30 36 ways 1 2 2 1 2 4 2 1 2 2 2 1 2 1

The grand totals then are:

num: 1 2 3 4 5 6 7 8 9 10 11 12 15 16 18 20 24 25 30 36 ways 11 12 13 15 15 18 6 7 5 5 2 5 2 1 2 2 2 1 2 1

So 6 is by far the best number to choose. KeithT is absolutely right that the probability of rolling a number outright is by far the most important consideration here.

** BUT **How do things change if we allow one roll to knock out several entries? What happens if we allow include other ways to make numbers (for example, a roll of 2 and 4 giving us 24?)

Skip 7 since it is prime and will never be on the product list. 8 would be next, then 7 since it can be summed 6 different ways. Round out the list with any two of 9, 10, and 12. Once again, 11 is skipped since it is prime.

11 and higher (excepting 12) all have 2 or fewer opportunities to be eliminated.

]]>first off, if you’re rolling two dice and calculating their sum to find a prime number, one die MUST have all even numbers and the other die MUST be all odd: (an even + an even or an odd + an odd equals an even – and not prime in all cases (except the number two))

second, and related to that, one must remember that all primes (with two exceptions) end in the numbers 1,3,7,or 9. If you were to put the first 6 odd numbers on your first die, you could have no number on the “even die” that would guarantee a prime: if the even number ended in a 0, the 0 + the 5 would = a number that ends in 5 … similarly 2+3, 4+1, 6+9, 8+7 …

the challenge is to pick numbers on the odd die that allow for numbers on the even die to work when added with all of them. I have not yet solved this, as I need to get back to work, but I will definitely post a solution if/when I find one.

My work so far suggests 1,3,7,13,15,19 on the odd die, and numbers ending in “4” on the even die. As is always the case, I’m sure there’s an easier answer than I’m making it out to be. I hope you guys mention this problem on a future episode of math factor.

]]>I’m counting that there are 18 ways to get a six, either on a die, or as a sum, or as a product:

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

(1,6) (2,6) (3,6) (4,6) (5,6) (1,5)

(2,4) (3,3) (4,2) (5,1) (2,3) (3,2)

The probability of getting a six is thus 18/36 = 1/2; half of all rolls get a six! Probably this is a good one to get on your list!

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