Simultaneous roll of Non Transitive Dice

[FallenApple]

So say that we roll three of them and assume that the fates of the roll are all decided simultaneously. I've seen the baysean network analysis for games where one player goes first and another goes second. But what I'm talking about is a roll of three dice such that they have to land simultaneously. Say we have dice A, B, and C, say with properties P(A > B) = P(B > C) = P(C > A) =5/9. where A, B , C are the sum of the numbers on a die's faces. So does the probability not work anymore if all three have to land at the same time?

The probability of die A being greater than die B is 5/9 and the probability of die B being greater than die C is also 5/9. So A is more likely than B and B is more likely than C, so it's common to assume that A is more likely than C. But it's not since C is more likely than A. So it's non transitive and we can't tell which one is the most likely to land with the largest sum out of the three.

It seems one of them have to land first before the others in order for the other two to have meaningful probabilities.

 

[FactChecker]

All the probabilities you give are correct [CORRECTION: I actually neglected to check these probabilities. I assumed that they were standard die probabilities and that they were correct. But those probabilities would be 5/12.], but they can not all happen at the same time. The probabilities described is not a preorder relationship. (see https://en.wikipedia.org/wiki/Preorder )

 

[FallenApple]

All the probabilities you give are correct, but they can not all happen at the same time. The probabilities described is not a preorder relationship. (see https://en.wikipedia.org/wiki/Preorder )

Ah ok. So then a related question is that when all three land simultaneously, there is no notion of probability? So then it has undefined measure?

 

[FactChecker]

Ah ok. So then a related question is that when all three land simultaneously, there is no notion of probability? So then it has undefined measure?

The probability that A > B > C, and all other orderings of the three are well defined. The order of the rolls, or doing them simultaneously, makes no difference.

 

[FallenApple]

The probability that A > B > C, and all other orderings of the three are well defined. The order of the rolls, or doing them simultaneously, makes no difference.

I mean if I want to define the random variable M:=max(A,B,C), would it be defined for this variable?

 

[Dale]

Say we have dice A, B, and C, say with properties P(A > B) = P(B > C) = P(C > A) =5/9

Is this possible?

 

[FallenApple]

Is this possible?

Oh, I forgot to mention. They are not regular.

• Die A has sides 2, 2, 4, 4, 9, 9.
• Die B has sides 1, 1, 6, 6, 8, 8.
• Die C has sides 3, 3, 5, 5, 7, 7.

 

[Dale]

Oh, I forgot to mention. They are not regular.

• Die A has sides 2, 2, 4, 4, 9, 9.
• Die B has sides 1, 1, 6, 6, 8, 8.
• Die C has sides 3, 3, 5, 5, 7, 7.

Oh, it is possible. That is interesting!

So there is nothing undefined with this. You can just make a 3x3x3 table if all the possibilities

 

[FallenApple]

Oh, it is possible. That is interesting!

So there is nothing undefined with this. You can just make a 3x3x3 table if all the possibilities

Got it. Though we have the tables, we wouldn't find convergence to a fixed probability if we were to repeat the experiment many times if we were interested in the random variable M=max(A,B,C).

 

[Dale]

Got it. Though we have the tables, we wouldn't find convergence to a fixed probability if we were to repeat the experiment many times if we were interested in the random variable M=max(A,B,C).

Sure you would, you could easily get an exact PMF. There are only 27 possibilities and the max is well defined for each of them.

 

[FallenApple]

Sure you would, you could easily get an exact PMF. There are only 27 possibilities and the max is well defined for each of them.

Oh right, that is true. They are independent. I could just multiply them. (2/3)*(2/3)*(2/3)=8/27

Hmm. But there's still something about this problem there runs counter to intuition. Well, in the long run, I think we still wouldn't know which is the most frequent. So the max is just a black box solution.

 

[Dale]

Well, in the long run, I think we still wouldn't know which is the most frequent

What are you referring to by "which"?

Each of the 27 possibilities is equally probable if the dice are fair. If the dice are not fair then you could weight the probabilities accordingly. Most importantly, the three rolls are independent.

 

[FallenApple]

What are you referring to by "which"?

Each of the 27 possibilities is equally probable if the dice are fair. If the dice are not fair then you could weight the probabilities accordingly. Most importantly, the three rolls are independent.

I am referring to the dice that would take precedence with regards to frequency after many trials when each trial is rolling all three dice at once.

On the first roll, the A>B with probability 4/9, B>C with probability 4/9 and C>A with probability 4/9, which does not suggest that any of them would win among the three because there is no transitivity in probability when considering all three simultaneously. On the second roll, due to independence between trials, the same thing happens and so on. This means the long run effect is the same as well with regards to non transitivity.

So in the long run, A<B , B<C and C<A but nothing can be said globally. Which seems to go against frequentist probability because we are totaling the amounts so that A<B , B<C and C<A, which can't happen on the one dimensional number line.

 

[Dale]

So in the long run, A<B , B<C and C<A but nothing can be said globally.

Sure you can. You can calculate the exact probability by building the table I described. I don't know for sure what the result would be, but it would clearly be well defined.

 

[Dale]

So in the long run, A<B , B<C and C<A but nothing can be said globally.

If you build the table and simply count then you find that A is the greatest in 10 of 27 cases, B is the greatest in 10 of 27 cases, and C is the greatest in 7 of 27 cases.

 

[FallenApple]

If you build the table and simply count then you find that A is the greatest in 10 of 27 cases, B is the greatest in 10 of 27 cases, and C is the greatest in 7 of 27 cases.

This is what I ended up with when I used tables. I didn't get the same results. C is most common, B second and A last.

 

[FallenApple]

With different data I got the following table. It follows the non transitive property and got me a uniform distribution.

 

[FactChecker]

What are the probabilities you have listed at the bottom? Are they like P(A largest)?
For A ∈ {2,6,7}, B ∈ {1,5,9}, C ∈ {3,4,8}, I get P(A largest) = 8/27. The combinations for that are:
(A, B, C) with A largest = combinations of (A ∈ {6,7}, B ∈ {1,5}, C ∈ {3,4} )
(6,1,3)
(6,1,4)
(6,5,3)
(6,5,4)
(7,1,3)
(7,1,4)
(7,5,3)
(7,5,4)

 

[Dale]

This is what I ended up with when I used tables.

These tables are not correct. You are rolling all 3 dice simultaneously so each element of the table should have the result for each of the three dice. You should never have an entry with just results for A and B but without C because you are rolling all three.

 

[FallenApple]

These tables are not correct. You are rolling all 3 dice simultaneously so each element of the table should have the result for each of the three dice. You should never have an entry with just results for A and B but without C because you are rolling all three.

Following your suggestion, I ended up with 3 cases. Two of them are the same as the one you got. When I collapse on B, it's different. So it seems no unique 3x3x3 table exists.

Though there is a weighting towards cases 2 and 3.

So maybe this problem is inherently 4 dimensional? Since there are 3 levels of 3d arrays, producing a unique 4 dimensional table. Of course with extra variables, case1, 2, 3 that needs to be codified somehow. Tangent: this reminds me of quaternions, though I don't know exactly how. This whole situation seems somewhat analogous to rotation in 3space where the order matters.

 

[FallenApple]

What are the probabilities you have listed at the bottom? Are they like P(A largest)?
For A ∈ {2,6,7}, B ∈ {1,5,9}, C ∈ {3,4,8}, I get P(A largest) = 8/27. The combinations for that are:
(A, B, C) with A largest = combinations of (A ∈ {6,7}, B ∈ {1,5}, C ∈ {3,4} )
(6,1,3)
(6,1,4)
(6,5,3)
(6,5,4)
(7,1,3)
(7,1,4)
(7,5,3)
(7,5,4)

They are the proportions in which the in which one is larger than the other, comparing the variables pairwise. I got the same 8/27 when I was looking at 3 different pairwise tables. Though it was probably just luck. I should have looked at a 3x3x3 table instead. I posted a solution using 3x3x3 on a different data set using collapsed tables.

 

[Dale]

Following your suggestion, I ended up with 3 cases. Two of them are the same as the one you got. When I collapse on B, it's different. So it seems no unique 3x3x3 table exists.

In case 1 you mistakenly have 7 winning over 9 in three places. All of your cases are the same thing and give the same result, just written down differently.

 

[FallenApple]

In case 1 you mistakenly have 7 winning over 9 in three places. All of your cases are the same thing and give the same result, just written down differently.

Ok got it. Fixed it. Yes, it makes sense. There is a unique table.

I guess the non intuitive part is where the first two wins out compared to the third. I'll build the table for a different set of data that satisfies same type of pairwise intransitivity to see if it shifts.

 

[Dale]

You might want to do it programmatically so that it is easier to make a large number of comparisons.

 

[FallenApple]

You might want to do it programmatically so that it is easier to make a large number of comparisons.

So basically I fix the probability and then sample by iteration under the probability constraints?

 

[Dale]

So basically I fix the probability and then sample by iteration under the probability constraints?

I would fix the dice numbers and then generate the table and calculate the probabilities.

 

[mfb]

I don’t know if it is possible at all (an interesting question), but if you want each die to have the maximum in 1/3 of the cases you need at least 6 sides on each die (without having three pairs of sides as in the discussed example).

Edit: It is impossible with the set of most unfair balanced dice with 6 sides (list). It might be possible with a less unfair set.

 

[mfb]

Forget my previous post (now deleted), I had a bug where rounding errors accumulated too much. I looked for unique combinations of six-sided intransitive dice (using unique labels, from 0 to 17) where, rolled together, each die has 1/3 chance to have the highest value.

In total there are 2,858,856 unique possible sets of dice. Out of these, 10705 are (strictly) intransitive. Out of these, 1910 give the same winning chances for A over B, B over C and C over A ("fair intransitive"). Out of these, 7 give each die a 1/3 chance to have the highest value.

While 21/36, the best chance requiring intransitive dice only, is not possible, 20/36 = 5/9 is possible.

And here are the combinations:

(2, 3, 7, 10, 12, 17)
(1, 5, 6, 9, 14, 16)
(0, 4, 8, 11, 13, 15)
winning chance: 19/36
--
(2, 4, 5, 10, 14, 16)
(1, 3, 7, 12, 13, 15)
(0, 6, 8, 9, 11, 17)
winning chance: 19/36
--
(2, 4, 6, 9, 13, 17)
(1, 3, 8, 11, 12, 16)
(0, 5, 7, 10, 14, 15)
winning chance: 19/36
--
(3, 4, 5, 8, 15, 16)
(1, 2, 9, 12, 13, 14)
(0, 6, 7, 10, 11, 17)
winning chance: 20/36
--
(3, 4, 5, 8, 15, 16)
(1, 2, 10, 11, 13, 14)
(0, 6, 7, 9, 12, 17)
winning chance: 20/36
--
(3, 4, 6, 7, 15, 16)
(1, 2, 9, 12, 13, 14)
(0, 5, 8, 10, 11, 17)
winning chance: 20/36
--
(3, 4, 6, 7, 15, 16)
(1, 2, 10, 11, 13, 14)
(0, 5, 8, 9, 12, 17)
winning chance: 20/36Edit: I had a look at 9-sided die. There are 37,978,905,250 combinations, and the code can just study ~25,000 per second. Way too many to check all.

It is easy to find hundreds of thousands of fair intransitive dice just by going through the list, but finding triples with the 1/3 max condition is more challenging.

I restricted the analysis to dice with a mean between 12.5 and 14.5 and sampled the first die at random to avoid starting from silly border cases (e. g. 0,1,2,3,4 on the same die). There are many dice here as well. Some examples:

(2, 5, 6, 9, 11, 17, 20, 22, 25)
(1, 3, 7, 10, 15, 16, 18, 23, 24)
(0, 4, 8, 12, 13, 14, 19, 21, 26)
winning chance: 0.506172839506 <- 41/81
--
(2, 5, 6, 9, 11, 17, 20, 22, 25)
(1, 3, 7, 12, 13, 16, 18, 21, 26)
(0, 4, 8, 10, 14, 15, 19, 23, 24)
winning chance: 0.493827160494 <- or 40/81. This means die 3 wins over die 2, 2 wins over 1, 1 wins over 3

(2, 4, 7, 9, 11, 15, 22, 23, 24)
(1, 6, 8, 10, 12, 13, 16, 25, 26)
(0, 3, 5, 14, 17, 18, 19, 20, 21)
winning chance: 0.469135802469 <- 38/81

Some dice are exceptionally good in producing these triples:

Spoiler

(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 10, 13, 17, 18, 22, 26)
(0, 2, 11, 12, 14, 15, 16, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 10, 14, 16, 18, 22, 26)
(0, 2, 11, 12, 13, 15, 17, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 10, 15, 16, 17, 22, 26)
(0, 2, 11, 12, 13, 14, 18, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 11, 12, 17, 18, 22, 26)
(0, 2, 10, 13, 14, 15, 16, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 11, 13, 16, 18, 22, 26)
(0, 2, 10, 12, 14, 15, 17, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 11, 14, 15, 18, 22, 26)
(0, 2, 10, 12, 13, 16, 17, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 11, 14, 16, 17, 22, 26)
(0, 2, 10, 12, 13, 15, 18, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 12, 13, 15, 18, 22, 26)
(0, 2, 10, 11, 14, 16, 17, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 12, 13, 16, 17, 22, 26)
(0, 2, 10, 11, 14, 15, 18, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 12, 14, 15, 17, 22, 26)
(0, 2, 10, 11, 13, 16, 18, 23, 24)
winning chance: 41.0/81
--
(3, 5, 7, 8, 9, 19, 20, 21, 25)
(1, 4, 6, 13, 14, 15, 16, 22, 26)
(0, 2, 10, 11, 12, 17, 18, 23, 24)
winning chance: 41.0/81
--

The mean of all these is exactly 13, I limited the search to dice with 13 as mean, that sped up the discovery a lot. The code now spends most of the time on the maximum probabilities as finding fair intransitive dice is so easy. Some more with 38/81 or the equivalent 43/81 (found >20 now):

Spoiler

(2, 5, 8, 10, 11, 12, 20, 24, 25)
(1, 3, 6, 13, 14, 15, 18, 21, 26)
(0, 4, 7, 9, 16, 17, 19, 22, 23)
winning chance: 38.0/81
--
(2, 5, 8, 10, 11, 12, 20, 24, 25)
(1, 3, 6, 13, 14, 16, 17, 21, 26)
(0, 4, 7, 9, 15, 18, 19, 22, 23)
winning chance: 38.0/81
--
(2, 5, 6, 7, 13, 15, 21, 23, 25)
(1, 3, 4, 14, 16, 18, 19, 20, 22)
(0, 8, 9, 10, 11, 12, 17, 24, 26)
winning chance: 43.0/81
--
(3, 4, 5, 10, 11, 18, 19, 21, 26)
(1, 6, 7, 9, 12, 13, 22, 23, 24)
(0, 2, 8, 14, 15, 16, 17, 20, 25)
winning chance: 38.0/81

Found one with 44/81, or 0.5432. The best one so far, but still worse than the best 6-sided dice. 45/81 would make them equal.
(2, 4, 8, 11, 12, 13, 16, 25, 26)
(1, 6, 7, 9, 10, 15, 22, 23, 24)
(0, 3, 5, 14, 17, 18, 19, 20, 21)
winning chance: 44.0/81

Spoiler: More with 44/81

(5, 6, 7, 8, 9, 12, 21, 23, 26)
(2, 3, 4, 11, 15, 18, 19, 20, 25)
(0, 1, 10, 13, 14, 16, 17, 22, 24)
winning chance: 44.0/81
--
(5, 6, 7, 8, 9, 12, 21, 23, 26)
(2, 3, 4, 11, 16, 17, 19, 20, 25)
(0, 1, 10, 13, 14, 15, 18, 22, 24)
winning chance: 44.0/81
--
(2, 4, 9, 10, 12, 13, 16, 25, 26)
(1, 6, 7, 8, 11, 15, 22, 23, 24)
(0, 3, 5, 14, 17, 18, 19, 20, 21)
winning chance: 44.0/81
--

Two more hours and nothing better than 44/81 popped up. If that exists it has to be extremely rare. Interestingly, the 5 6 7 8 9 12 21 23 26 sets appeared 4 times. Makes me wonder how good the random number generator is, there are 2 million choices for the first die and many of them should have an average of 13.

Tuned the program even more, it is now able to search through all avg=13 dice now so I switched back to a sorted search. 17,495,345 fair intransitive dice combinations found, out of these have 5077 a fair maximum, out of these only 4 triples have the "maximal intransitivity" of 44/81, all of them have been found in the random search already.

If I drop the 1/3 max condition (but keep the average of 13 as condition), the best sets of intransitive dice get 48/81=0.593.. (30 times) and 33/81 (12 times). They are slightly better than the best 6-sided intransitive dice.

I'll keep updating this post, although the final version will probably be cleaned up and posted as separate thread or insights article.

 

[Zafa Pi]

Following your suggestion, I ended up with 3 cases. Two of them are the same as the one you got. When I collapse on B, it's different. So it seems no unique 3x3x3 table exists.

Though there is a weighting towards cases 2 and 3.

So maybe this problem is inherently 4 dimensional? Since there are 3 levels of 3d arrays, producing a unique 4 dimensional table. Of course with extra variables, case1, 2, 3 that needs to be codified somehow. Tangent: this reminds me of quaternions, though I don't know exactly how. This whole situation seems somewhat analogous to rotation in 3space where the order matters.

View attachment 213229

There are 3 tice (3 sided dice) A, B, and C. As @Dale said, rolling all 3 there are 27 equally likely out comes due to uniformity and independence.
In 10 of these A gets the highest roll, in 10 B gets the highest roll, and in 7 C gets the highest roll.
In 15 of these A beats B, in 15 B beats C, and in 15 C beats A.
The table is a 3x3x3 cube.
What is your problem with this?

 

[haruspex]

So A is more likely than B and B is more likely than C, so it's common to assume that A is more likely than C. But it's not since C is more likely than A. So it's non transitive

It does not mean anything to say that A is more likely than B. They are not events, they are random variables.

If you try to fix that by saying A is "better than" B then you need to define the criterion. If the criterion is tending to produce a higher value on average then you can expect transitivity (and in this case they all have the same average, so no paradox). If the criterion is tending to win a one-on-one roll then there is no good reason to expect transitivity.