For background (if you care), see this post (which links to the first one as well).

I have proven that balanced, non-transitive *n*-sided dice exist for every *n*>2 (and the proof is quite slick, I think). The trick to this is a particular kind of labeling followed by concatenation. I have also proven several nice things (if a set is balanced, all 3 dice add to the same total!) on top of my original thoughts. My task now is attempting to find an upper bound for the victorious probability. My conjecture is 2/3. I also conjecture this is unattainable, that is, it is asymptotic.

I’m going to be writing a lengthy paper on this topic, so it’ll be easier to link to that once it’s done (in April or so).

That’s enough on dice. Back to your regularly scheduled blogramming.

Thanks for reading.

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Can it be shown that the set of unique balanced, non-transitive dice triplets for n-sided dice can be pulled from the space [1,3n] (allowing duplicates)?

That is to say that for example on 3 sided dice, each die is equivalent (when used as one of 3 dice in a balanced non-transitive triplet in terms of probabilities that A beats B beats C beats A) to one of the 729 possible 3 sided dice using the numbers [1,9].

If that were true, it would be pretty easy for a computer to enumerate through all possible combinations and find the dice with a maximum winning probability for any given n.

As for a proof that such dice exist for every n>2, wouldn’t it be sufficient to show:

1. that they exist for n=3, 4 and 5(by example)

2. that any dice beyond that can be found by adding the 3-dice sets to the others (6=3+3, 7=4+3, 8=5+3, 9=6+3, …)

3. that the dice created by combining sets are also balanced and non-transitive for the example dice (it wouldn’t matter if there could be sets generated which no longer matched this condition, simply that the example sets do match)

I don’t know why you would allow duplicates, but even at that, the equivalence structure is more complicated. Not for 3, 4, or 5, but starting with 6, you need to define precisely what an “indecomposable” solution is, which is a little tough.

As for the “solutions for every n” bit, that’s exactly how I proved it, after a careful definition of concatenation.

Well done though.