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Summing dice problem

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    Given two unmarked, cubic dice and using the set of whole numbers, place numbers on each die such that only the following sums can be thrown: 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 and the probability of rolling each sum is the same.


    2. Relevant equations



    3. The attempt at a solution

    If I have one die with even numbers (2, 4, 6, 8, 10, and 12) and one with odd (3, 5, 7, 9, 11 and 13) it works up to number 25...but not 27. And I am lost on how the probability of each sum is the same.

    Thanks in advance for any help you can give!

    MW
     
  2. jcsd
  3. Sep 16, 2007 #2

    D H

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    There are 12 different sums and 36 ways to roll a pair of dice (counting duplicates). If the probability of rolling each sum is the same, how many ways must exist for rolling each sum?
     
  4. Sep 16, 2007 #3
    That is exactly my confusion. With the dice set up as I described, there is a 1/36 chance of rolling a five but 6/36 of rolling a 15. I don't understand how there could be the same probability for all numbers.

    MW
     
  5. Sep 16, 2007 #4

    D H

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    The dice setup as you described is incorrect since (a) it doesn't obtain all of the desired sums and (b) it does not have the same probability for each sum. You have to try something different.

    This might help: Instead of hitting 5, 7, 9, ..., 25, 27 with equal probabilities, try to find some arrangement that hits 5, 7, 9, 11, 13, and 15 (just 6 sums instead of 12) with equal probabilities. Now see how you can change this solution to meet the original problem.
     
  6. Sep 16, 2007 #5

    D H

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    If the above hint doesn't help, try to find some arrangement such that the sum is 5, and nothing else. You always roll fives.
     
  7. Sep 16, 2007 #6
    If one die has a constant and the other die has the appropriate numbers, ie,

    the constant die has 0 and the other die could have 5, 7, 9, 11, 13, 15
    or the constant die has 1 and the other has 4, 6, 8, 10, 12, 14

    Am I on the right track? But I still get confused with 2 dice and 12 sums because how could you have a constant?

    MW
     
  8. Sep 16, 2007 #7

    D H

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    Just because the number on one die cannot all be the same number to yield 12 different sumes does not mean they have to be all different numbers.
     
  9. Sep 16, 2007 #8
    Ah...so if one die had 0, 0, 0 and 12, 12, 12 and the other die had 5, 7, 9, 11, 13, and 15 it would work. Every solution has the same probability of 1/12.

    Is there a formula for this?

    MW
     
  10. Sep 16, 2007 #9

    D H

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    Not that I know of.

    BTW, there are an infinite number of solutions assuming "whole number" means "integer". Ignoring different ways to label one die, and if "whole number" means the non-negative integers, there are but six solutions.
     
    Last edited: Sep 16, 2007
  11. Sep 16, 2007 #10
    Eureka!

    One die would have the combinations: 0, 12; 1, 13; 2, 14; 3, 15; 4, 16; 5, 17 with the other die descending in the proper increments that 5 and 27 and always the beginning and ending points (ie, the last one would be 0, 2, 4, 6, 8, 10).

    So I am a little slow but finally the light bulb shined for me! LOL

    Thanks for your help,

    MW
     
  12. Sep 16, 2007 #11

    D H

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    Your'e welcome.

    A homework helper can mark this one as SOLVED.
     
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