DI. Dice Games

A young listener (or really her father, on behalf of a young listener) wrote us:

Two players each choose any 10 digits from 1 to 36.

Then you take turns
rolling 2 die with the object being trying to be the first to eliminate all
your numbers. You can eliminate any of your numbers by matching it with either
the number on one/both of the die or using the sum or product of the die.

For example:

If I chose 1, 2, 3, 4, 5, 6, 7,8, 9, 10. Then if I roll a 2 and a 4 I can
eliminate the 2 & 4, the 6 OR the 8. The other player also eliminates from his
list using your roll.

What is the best 10 numbers to choose?

Thanks for a great puzzle Anika!

Well let’s see:

The first, fairly straightforward task is to work out the probabilities that various numbers will be eliminated on a given roll. (And only 20 numbers out of the 36 can be eliminated at all!)

Some interesting variations: a given roll can eliminate multiple numbers; this is perhaps more subtle since you want to find sets of numbers that could perhaps be wiped out efficiently, and at a high probability.

An additional possibility would be to allow, say, a roll of 2 & 4 to wipe out 24. This rule would allow 30 different numbers to be eliminated.

We gave two puzzles this week; I’m going to move discussion of the second one to the next thread.


  1. strauss said,

    January 9, 2008 at 8:59 am

    To work out the probabilities of a given roll occurring, imagine that we have a red die and a blue die. Then there are 36 possible rolls altogether. We can then count, for example, how many ways that a 6 can be eliminated by looking through all 36 possibilities.

    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!

  2. MattD said,

    January 18, 2008 at 1:13 pm

    Hearing your dice problems today (great show by the way), a few thoughts came to me instantly in regards to the “prime number” puzzle that I thought I’d share with anyone else trying to solve this issue:

    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.

  3. KeithT said,

    January 22, 2008 at 10:07 am

    1 through 6 are easy enough choices. Their 1/6 probability for direct elimination eclipses the higher number’s opportunities to be sums and products.

    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.

  4. strauss said,

    January 28, 2008 at 10:05 am

    If we are just looking for the probability that a given number can be wiped out:

    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?)

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