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Macro-states and dice

  1. Mar 2, 2015 #1
    If there are 2 6-sided dice. What is the formula used to calculate the number of macro-states where a micro state is the sum of the two dice? I know there are 11 macro-states but I was curious if there was a formula for calculating this.
     
  2. jcsd
  3. Mar 2, 2015 #2

    DrClaude

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    I don't think there is any way to find this out apart from listing all microstates and counting the corresponding macrostates.
     
  4. Mar 3, 2015 #3

    HallsofIvy

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    Combine 1, 2, 3, 4, 5, 6 with 1, 2, 3, 4, 5, 6. There are 6 x 6= 36 such combinations:
    1+ 1= 2, 1+ 2= 3, 1+ 3= 4, 1+ 4= 5, 1+ 5= 5, 1+ 6= 7.
    2+ 1= 3, 2+ 2= 4, 2+ 3= 5, 2+ 4= 6, 2+ 5= 7, 2+ 6= 8. etc.
     
  5. Mar 3, 2015 #4

    DrClaude

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    Can you confirm that I am right in thinking that there can be no way of knowing how many combinations will add to a certain number without actually listing them all (or some variant thereof).
     
  6. Mar 3, 2015 #5
    microstates = q^n, where q = number of states the dice can be in, and n = number of dice

    Totally not sure about this, but if we're assuming that dice are always numbered using integers starting at 1 and moving upward sequentially, and label each macrostate with its dice-sum, it seems like it will always be true that the smallest (as it were) macrostate would be n, and the largest macrostate would be q*n. It would also seem true that every integer between n and q*n would also be a legitimately possible macrostate given the integer assumption. If so, the number of possible macrostates in a dice setup could be stated as n(q-1) +1. (edit for the +1)

    Disclaimer: Just getting back into thinking and have grown pretty stupid in the last several years.
     
  7. Mar 3, 2015 #6
    Yeah that seems to work! Thank you so much.
     
  8. Mar 4, 2015 #7
    What makes you think that would be true?

    I don't see an easy formula off the top of my head, but I'd be awfully surprised if there wasn't a somewhat elegant way to phrase it.

    But maybe you're seeing something that makes you think it's not possible?
     
  9. Mar 4, 2015 #8
    Don't have the notebook in front of me, but I did a 5^3 in the passenger seat and I think it went:

    1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1

    Somewhat interesting to note, if all 3 sites(n) rolled unique states(q), there are 6 combos; if one duplicate, 3; if triplets, 1. Didn't do a n=4, but if it scales formulaically...
     
  10. Mar 4, 2015 #9
    Yesssssss
     
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