Number of Macro-States w/ 2 6-sided Dice: Formula

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Discussion Overview

The discussion revolves around the calculation of the number of macro-states when rolling two 6-sided dice, specifically focusing on the relationship between micro-states (the sums of the dice) and macro-states. Participants explore whether a formula exists for determining the number of macro-states based on the possible outcomes of the dice.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose that the only way to determine the number of macro-states is to list all possible micro-states and count the corresponding sums.
  • One participant suggests a formula for calculating the number of macro-states as n(q-1) + 1, where n is the number of dice and q is the number of states each die can take.
  • Another participant expresses uncertainty about the existence of a straightforward formula for determining combinations that yield specific sums, questioning the validity of relying solely on listing combinations.
  • There is mention of a pattern observed in the sums of three dice, indicating that if all dice rolled unique states, the combinations would differ based on the presence of duplicates.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the existence of a formula for calculating macro-states. There are competing views on whether listing micro-states is necessary and whether a more elegant solution exists.

Contextual Notes

Some assumptions about the numbering of dice and the nature of combinations are not fully explored, leaving open questions about the applicability of proposed formulas.

Alec Neeson
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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.
 
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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.
 
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.
 
HallsofIvy said:
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.
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).
 
Alec Neeson said:
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.

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.
 
Yeah that seems to work! Thank you so much.
 
DrClaude said:
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).

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?
 
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...
 
Alec Neeson said:
Yeah that seems to work! Thank you so much.

Yesssssss
 

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