Problem Statement

Let our dice select numbers on their faces with equal probability, i.e. fair dice. Dice may have more or less than six faces. (The possibility of there being a 3D physical shape that has that many "faces" that allow them to be fair dice, is ignored for this task - a die with 3 or 33 defined sides is defined by the number of faces and the numbers on each face). Throwing dice will randomly select a face on each die with equal probability. To show which die of dice thrown multiple times is more likely to win over the others:

If two dice X and Y are thrown against each other then X likely to: win, lose, or break-even against Y can be shown as: X > Y, X < Y, or X = Y respectively.

If X is the three sided die with 1, 3, 6 on its faces and Y has 2, 3, 4 on its faces then the equal possibility outcomes from throwing both, and the winners is: Both die will have the same statistical probability of winning, i.e.their comparison can be written as X = Y In mathematics transitivity are rules like: If, for example, the op, (for operator), is the familiar less than, <, and it's applied to integers we get the familiar if a < b and b < c then a < c These are an ordered list of dice where the '>' operation between successive dice pairs applies but a comparison between the first and last of the list yields the opposite result, '<'. (Similarly '<' successive list comparisons with a final '>' between first and last is also non-transitive).

Three dice S, T, U with appropriate face values could satisfy To be non-transitive.

Find all the ordered lists of three non-transitive dice S, T, U of the form S < T, T < U and yet S > U; where the dice are selected from all four-faced die , (unique w.r.t the notes), possible by having selections from the integers one to four on any dies face. Solution can be found by generating all possble individual die then testing all possible permutations, (permutations are ordered), of three dice for non-transitivity. Find lists of four non-transitive dice selected from the same possible dice from the non-stretch goal.

Show the results here, on this page.

Let's start with the solution:

// Non-transitive dice. Nigel Galloway: August 9th., 2020
let die=[for n0 in [1..4] do for n1 in [n0..4] do for n2 in [n1..4] do for n3 in [n2..4]->[n0;n1;n2;n3]]
let N=seq{for n in die->(n,[for g in die do if (seq{for n in n do for g in g->compare n g}|>Seq.sum<0) then yield g])}|>Map.ofSeq 
let n3=seq{for d1 in die do for d2 in N.[d1] do for d3 in N.[d2] do if List.contains d1 N.[d3] then yield (d1,d2,d3)}
let n4=seq{for d1 in die do for d2 in N.[d1] do for d3 in N.[d2] do for d4 in N.[d3] do if List.contains d1 N.[d4] then yield (d1,d2,d3,d4)}
n3|>Seq.iter(fun(d1,d2,d3)->printfn "%A<%A; %A<%A; %A>%A\n" d1 d2 d2 d3 d1 d3)
n4|>Seq.iter(fun(d1,d2,d3,d4)->printfn "%A<%A; %A<%A; %A<%A; %A>%A" d1 d2 d2 d3 d3 d4 d1 d4)


// Non-transitive diceNigel Galloway: August 9th., 2020
let die=[for n0 in [1..6] do for n1 in [n0..6] do for n2 in [n1..6] do for n3 in [n2..6] do for n4 in [n3..6] do for n5 in [n4..6]->[n0;n1;n2;n3;n4;n5]]
let N=seq{for n in die->(n,[for g in die do if (seq{for n in n do for g in g->compare n g}|>Seq.sum<0) then yield g])}|>Map.ofSeq 
let n3=[for d1 in die do for d2 in N.[d1] do for d3 in N.[d2] do if List.contains d1 N.[d3] then yield (d1,d2,d3)]
let d1,d2,d3=List.last n3 in printfn "Solutions found = %d\nLast solution found is %A<%A; %A<%A; %A>%A" (n3.Length) d1 d2 d2 d3 d1 d3


// Non-transitive diceNigel Galloway: August 9th., 2020
let die=[for n0 in [1..7] do for n1 in [n0..7] do for n2 in [n1..7] do for n3 in [n2..7] do for n4 in [n3..7] do for n5 in [n4..7]->[n0;n1;n2;n3;n4;n5]]
let N=seq{for n in die->(n,[for g in die do if (seq{for n in n do for g in g->compare n g}|>Seq.sum<0) then yield g])}|>Map.ofSeq 
let n3=[for d1 in die do for d2 in N.[d1] do for d3 in N.[d2] do if List.contains d1 N.[d3] then yield (d1,d2,d3)]
let d1,d2,d3=List.last n3 in printfn "Solutions found = %d\nLast solution found is %A<%A; %A<%A; %A>%A" (n3.Length) d1 d2 d2 d3 d1 d3