Exploring this idea for twelve-sided dice, in how many different ways can one construct an (unordered) pair of 12-sided dice which
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Write D n for a uniform distribution on { 0 , 1 , … , n − 1 } . Then we are looking for different ways to rewrite 2 + D 1 2 + D 1 2 . Now we can split D 1 2 into at most three parts, featuring D 3 once and D 2 twice: D 1 2 = D 3 + 6 D 2 + 3 D 2 = 2 D 3 + 6 D 2 + D 2 = 4 D 3 + 2 D 2 + D 2 .
New twelve-sided dice are created by swapping out D 2 terms. We explore the six possible decompositions of 2 + D 1 2 + D 1 2 :
( 1 + D 3 + 6 D 2 + 3 D 2 ) + ( 1 + D 3 + 6 D 2 + 3 D 2 ) . There is one possible swap: D A = 1 + D 3 + 6 D 2 + 6 D 2 = [ 1 , 2 , 3 , 7 , 7 , 8 , 8 , 9 , 9 , 1 3 , 1 4 , 1 5 ] ; D B = 1 + D 3 + 3 D 2 + 3 D 2 = [ 1 , 2 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 , 8 , 9 ] .
( 1 + 2 D 3 + 6 D 2 + D 2 ) + ( 1 + 2 D 3 + 6 D 2 + D 2 ) . Again, one possible swap: D A = 1 + 2 D 3 + 6 D 2 + 6 D 2 = [ 1 , 3 , 5 , 7 , 7 , 9 , 9 , 1 1 , 1 1 , 1 3 , 1 5 , 1 7 ] ; D B = 1 + 2 D 3 + D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 ] .
( 1 + 4 D 3 + 2 D 2 + D 2 ) + ( 1 + 4 D 3 + 2 D 2 + D 2 ) . Another single swap here: D A = 1 + 4 D 3 + 2 D 2 + 2 D 2 = [ 1 , 3 , 3 , 5 , 5 , 7 , 7 , 9 , 9 , 1 1 , 1 1 , 1 3 ] ; D B = 1 + 4 D 3 + D 2 + D 2 = [ 1 , 2 , 2 , 3 , 5 , 6 , 6 , 7 , 9 , 1 0 , 1 0 , 1 1 ] .
( 1 + D 3 + 6 D 2 + 3 D 2 ) + ( 1 + 2 D 3 + 6 D 2 + D 2 ) . Again, one possible swap: D A = 1 + D 3 + 6 D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 4 , 7 , 8 , 8 , 9 , 9 , 1 0 ] ; D B = 1 + 2 D 3 + 6 D 2 + 3 D 2 = [ 1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 1 0 , 1 1 , 1 2 , 1 4 ] .
( 1 + 2 D 3 + 6 D 2 + D 2 ) + ( 1 + 4 D 3 + 2 D 2 + D 2 ) . One swap: D A = 1 + 2 D 3 + 2 D 2 + D 2 = [ 1 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 , 8 ] ; D B = 1 + 4 D 3 + 6 D 2 + D 2 = [ 1 , 2 , 5 , 6 , 7 , 8 , 9 , 1 0 , 1 1 , 1 2 , 1 5 , 1 6 ] .
( 1 + 4 D 3 + 2 D 2 + D 2 ) + ( 1 + D 3 + 6 D 2 + 3 D 2 ) . This is the most interesting case, because there are four different D 2 terms involved, allowing for six possible pairings. Thus there are five possible swaps:
D A = 1 + 4 D 3 + 2 D 2 + 3 D 2 = [ 1 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 1 0 , 1 1 , 1 2 , 1 4 ] ; D B = 1 + D 3 + 6 D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 4 , 7 , 8 , 8 , 9 , 9 , 1 0 ] . But we must discard this because it is equivalent to one we found before! We continue:
D A = 1 + 4 D 3 + 2 D 2 + 6 D 2 = [ 1 , 3 , 5 , 7 , 7 , 9 , 9 , 1 1 , 1 1 , 1 3 , 1 5 , 1 7 ] ; D B = 1 + D 3 + 3 D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 ] . Again, this only doubles a result we found earlier.
D A = 1 + 4 D 3 + 3 D 2 + D 2 = [ 1 , 2 , 4 , 5 , 5 , 6 , 8 , 9 , 9 , 1 0 , 1 2 , 1 3 ] ; D B = 1 + D 3 + 6 D 2 + 2 D 2 = [ 1 , 2 , 3 , 3 , 4 , 5 , 7 , 8 , 9 , 9 , 1 0 , 1 1 ] .
D A = 1 + 4 D 3 + 6 D 2 + D 2 = [ 1 , 2 , 5 , 6 , 7 , 8 , 9 , 1 0 , 1 1 , 1 2 , 1 5 , 1 6 ] ; D B = 1 + D 3 + 2 D 2 + 3 D 2 = [ 1 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 , 8 ] . This, too, is a result we found earlier.
D A = 1 + 4 D 3 + 3 D 2 + 6 D 2 = [ 1 , 4 , 5 , 7 , 8 , 9 , 1 0 , 1 1 , 1 2 , 1 4 , 1 5 , 1 8 ] ; D B = 1 + D 3 + 2 D 2 + D 2 = [ 1 , 2 , 2 , 3 , 3 , 3 , 4 , 4 , 4 , 5 , 5 , 6 ] .
In total, then 7 different ways to construct the new dice.
The "different ways" would have to be clarified further. If I swap the pair around, does that constitute a different way?
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I believe that question is sufficiently answered by the case of the Sicherman dice, which are unique, up to swapping the pair around.
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Would you mind if it stated "a (unordered) pair of 12-sided dice"?
Comparing this solution and that of @Mark Hennings , an interesting question is how different the solutions really are. A partial translation, at least, is possible. Define d n ( x ) = x n − 1 + ⋯ + x 2 + x + 1 . Then my " D " notation above can be translated to generating function polynomials as follows:
1 ↦ t ,
+ ↦ × ,
D n ↦ d n ( t ) ,
a D n ↦ d n ( t a ) .
Thus D A = 1 + 4 D 3 + 3 D 2 + 6 D 2 ↦ t ( t 8 + t 4 + 1 ) ( t 3 + 1 ) ( t 6 + 1 ) . Factorize these factors further, and we find Mark's solution. In this case, for instance, t 3 + 1 = ( t 2 − t + 1 ) ( t + 1 ) , t 6 + 1 = ( t 4 − t 2 + 1 ) ( t 2 + 1 ) , t 8 + t 4 + 1 = ( t 2 + t + 1 ) ( t 2 − t + 1 ) ( t 4 − t 2 + 1 ) . This further factorization introduces negative coefficients and therefore non-solutions. On the other hand, it results a unique factorization, while my D notation has non-unique decomposition and therefore results in generating identical solutions. D 1 2 = ⎩ ⎪ ⎨ ⎪ ⎧ 4 D 3 + 2 D 2 + D 2 6 D 2 + 2 D 3 + D 2 6 D 2 + 3 D 2 + D 3 ↦ t 1 1 + ⋯ + t + 1 = ⎩ ⎪ ⎨ ⎪ ⎧ ( t 8 + t 4 + 1 ) ( t 2 + 1 ) ( t + 1 ) ( t 6 + 1 ) ( t 4 + t 2 + 1 ) ( t + 1 ) ( t 6 + 1 ) ( t 3 + 1 ) ( t 2 + t + 1 ) = ( t 4 − t 2 + 1 ) ( t 2 + t + 1 ) ( t 2 − t + 1 ) ( t 2 + 1 ) ( t + 1 ) . Reconciling the two methods could be accomplished by introducing the pseudo-die D 3 ⋆ = { 2 , 1 ⋆ , 0 } with the rules D 3 + D 3 ∗ = { 4 , 3 , 3 ⋆ , 2 , 2 , 2 ⋆ , 1 , 1 ⋆ , 0 } = { 4 , 2 , 0 } = 2 D 3 , D 2 + D 3 ∗ = { 3 , 2 , 2 ⋆ , 1 , 1 ⋆ , 0 } = { 3 , 0 } = 3 D 2 . where the ∗ operator indicates removal from the set. Now we get D 1 2 = 2 D 3 ⋆ + D 3 + D 3 ⋆ + 2 D 2 + D 2 , and the former ambiguity is resolved: D 1 2 = ⎩ ⎪ ⎨ ⎪ ⎧ 4 D 3 + 2 D 2 + D 2 = ( 2 D 3 ⋆ + D 3 + D 3 ⋆ ) + 2 D 2 + D 2 6 D 2 + 2 D 3 + D 2 = ( 2 D 3 ⋆ + 2 D 2 ) + ( D 3 ⋆ + D 3 ) + D 2 6 D 2 + 3 D 2 + D 3 = ( 2 D 3 ⋆ + 2 D 2 ) + ( D 3 ⋆ + D 2 ) + D 3 . The problem becomes a redistribution of the ten D -terms in D 1 2 + D 1 2 = D A + D B in two groups in such a way, that
each group has one non-starred D 3 term, and two D 2 terms,
each D 3 ⋆ remains combined with D 2 or D 3 , and each 2 D 3 ⋆ remains combined with 2 D 2 or 2 D 3 = D 3 + D 3 ⋆ .
The eight solutions (or fifteen possible dice) are easily reproduced by considering where the four starred terms go:
D 3 + 2 D 3 ⋆ + 2 D 3 ⋆ + D 3 ⋆ + D 3 ⋆ + 2 D 2 + D 2 ; die with maximum value = 18.
D 3 + 2 D 3 ⋆ + 2 D 3 ⋆ + D 3 ⋆ + 2 D 2 + 2 D 2 ; maximum value = 17.
D 3 + 2 D 3 ⋆ + 2 D 3 ⋆ + D 3 ⋆ + 2 D 2 + D 2 ; maximum value = 16.
D 3 + 2 D 3 ⋆ + 2 D 3 ⋆ + 2 D 2 + 2 D 2 ; maximum value = 15.
D 3 + 2 D 3 ⋆ + D 3 ⋆ + D 3 ⋆ + 2 D 2 + D 2 ; maximum value = 14.
D 3 + 2 D 3 ⋆ + D 3 ⋆ + D 3 ⋆ + D 2 + D 2 ; maximum value = 13.
D 3 + 2 D 3 ⋆ + D 3 ⋆ + 2 D 2 + 2 D 2 ; maximum value = 13.
D 3 + 2 D 3 ⋆ + D 3 ⋆ + 2 D 2 + D 2 ; maximum value = 12 (the "normal" die).
D 3 + 2 D 3 ⋆ + D 3 ⋆ + D 2 + D 2 ; maximum value = 11.
D 3 + 2 D 3 ⋆ + 2 D 2 + 2 D 2 ; maximum value = 11.
D 3 + 2 D 3 ⋆ + 2 D 2 + D 2 ; maximum value = 10.
D 3 + D 3 ⋆ + D ⋆ 3 + 2 D 2 + D 2 ; but the other die would be wrong
D 3 + D 3 ⋆ + D ⋆ 3 + D 2 + D 2 ; maximum value = 9.
D 3 + D 3 ⋆ + 2 D 2 + 2 D 2 ; but the other die would be wrong
D 3 + D 3 ⋆ + 2 D 2 + D 2 ; maximum value = 8.
D 3 + D 3 ⋆ + D 2 + D 2 ; maximum value = 7.
D 3 + 2 D 2 + 2 D 2 ; but the other die would be wrong.
D 3 + 2 D 2 + D 2 ; maximum value = 6.
D 3 + D 2 + D 2 ; but the other die would be wrong.
Found these, using a Python script: A=[1,4,5,7,8,9,10,11,12,14,15,18] ; B=[1,2,2,3,3,3,4,4,4,5,5,6] A=[1,3,5,7,7,9,9,11,11,13,15,17] ; B=[1,2,2,3,3,4,4,5,5,6,6,7] A=[1,3,4,5,6,7,8,9,10,11,12,14] ; B=[1,2,2,3,3,4,7,8,8,9,9,10] A=[1,3,3,5,5,7,7,9,9,11,11,13] ; B=[1,2,2,3,5,6,6,7,9,10,10,11] A=[1,2,5,6,7,8,9,10,11,12,15,16] ; B=[1,2,3,3,4,4,5,5,6,6,7,8] A=[1,2,4,5,5,6,8,9,9,10,12,13] ; B=[1,2,3,3,4,5,7,8,9,9,10,11] A=[1,2,3,7,7,8,8,9,9,13,14,15] ; B=[1,2,3,4,4,5,5,6,6,7,8,9]
For dice with other number of sides 1-15: 4: A=[1,3,3,5] ; B=[1,2,2,3] 6: A=[1,3,4,5,6,8] ; B=[1,2,2,3,3,4] 8: A=[1,3,5,5,7,7,9,11] ; B=[1,2,2,3,3,4,4,5], A=[1,3,3,5,5,7,7,9] ; B=[1,2,2,3,5,6,6,7], A=[1,2,5,5,6,6,9,10] ; B=[1,2,3,3,4,4,5,6] 9: A=[1,4,4,7,7,7,10,10,13] ; B=[1,2,2,3,3,3,4,4,5] 10: A=[1,3,5,6,7,8,9,10,12,14] ; B=[1,2,2,3,3,4,4,5,5,6] 14: A=[1,3,5,7,8,9,10,11,12,13,14,16,18,20] ; B=[1,2,2,3,3,4,4,5,5,6,6,7,7,8] 15: A=[1,4,6,7,9,10,11,12,13,14,15,17,18,20,23] ; B=[1,2,2,3,3,3,4,4,4,5,5,5,6,6,7]
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We want to factorise the probability generating function 1 4 4 1 ( t + t 2 + t 3 + ⋯ + t 1 1 + t 1 2 ) 2 = 1 4 4 1 t 2 ( t + 1 ) 2 ( t 2 + t + 1 ) 2 ( t 2 + 1 ) 2 ( t 2 − t + 1 ) 2 ( t 4 − t 2 + 1 ) 2 of the random variable of the sum of two normal 1 2 -sided dice as the product P ( t ) Q ( t ) of two polynomials where
There are a number of possible candidates:
P ( t ) = P a , b ( t ) = 1 2 1 t ( t 2 + t + 1 ) ( t + 1 ) 2 ( t 2 − t + 1 ) a ( t 4 − t 2 + 1 ) b and Q ( t ) = Q a , b ( t ) = 1 2 1 t ( t 2 + t + 1 ) ( t 2 + 1 ) ( t 2 − t + 1 ) 2 − a ( t 4 − t 2 + 1 ) 2 − b for some integers 0 ≤ a , b ≤ 2 . All of these possibilities satisfy the condition P ( 1 ) = Q ( 1 ) = 1 , but not all of these options give nonnegative coefficients. The cases that do give nonnegative coefficients are ( a , b ) = ( 1 , 0 ) , ( 1 , 1 ) , ( 2 , 0 ) , ( 2 , 1 ) . This gives us 4 possible dice pairs.
P ( t ) = R a , b ( t ) = 1 2 1 t ( t 2 + t + 1 ) ( t + 1 ) ( t 2 + 1 ) ( t 2 − t + 1 ) a ( t 4 − t 2 + 1 ) b and Q ( t ) = R 2 − a , 2 − b ( t ) for some integers 0 ≤ a , b ≤ 2 . The ( a , b ) = ( 0 , 2 ) case yields some negative coefficients. The cases ( 0 , 0 ) and ( 2 , 2 ) give the same pair of dice, as do the cases ( 0 , 1 ) and ( 2 , 1 ) and also the cases ( 1 , 0 ) and ( 1 , 2 ) . The case ( 1 , 1 ) gives the normal pair of dice. Thus there are 3 possible new pairs of dice.
That makes 7 pairs of dice in total.