Unusually Common Dice: Find the Unexpected Ordinary

[RandallB]

Here's another set, it has 2 dice in common with yours:

1, 1, 1, 2, 2, 2
0, 0, 0, 3, 3, 3
1, 1, 2, 2, 3, 3
0, 0, 2, 2, 4, 4

I was just dealing with the SIX sided dice.
Your above set of four dice is what gives the last column odds results that dosn’t match normal dice.
Since the less than six sided dice you show are a simple reduction of the set that dosn’t work I’d expect it would have fewer permutations but the same wrong odds.

For dice of less than six sides (Or call them coins and three sided rods), I’d reduce “Randall’s Dice”
FROM:
0, 0, 1, 1, 2, 2
0, 0, 0, 3, 3, 3
1, 1, 2, 2, 3, 3
1, 1, 1, 4, 4, 4

and
1, 2, 2, 3, 3, 4
1, 1, 3, 3, 5, 5
0, 0, 0, 3, 3, 3

TO:
0, 1, 2
0, 3
1, 2, 3
1, 4

and
1, 2, 2, 3, 3, 4
1, 3, 5
0, 3

These should all give correct odds to match a normal pair of dice.

 

[shmoe]

I was just dealing with the SIX sided dice.
Your above set of four dice is what gives the last column odds results that dosn’t match normal dice.
Since the less than six sided dice you show are a simple reduction of the set that dosn’t work I’d expect it would have fewer permutations but the same wrong odds.

In that case I disagree. Let's just look at the number of ways you can get a "3" from the dice:

1, 1, 1, 2, 2, 2
0, 0, 0, 3, 3, 3
1, 1, 2, 2, 3, 3
0, 0, 2, 2, 4, 4

The answer should be 72 but you claim the above give 108 ways. The 1st and 3rd dice give at least a total of 2, so the 2nd and 4th must both roll 0, there are 3*2 ways of doing this. So we just need a 3 total from the 1st and 2nd. This is either a 1 from the first and a 2 from the second, 3*2 ways here, or a 2 from the first and a 1 from the second, so 3*2 ways. The total is therefore:3*2*(3*2+3*2)=72.

Or just multiply out the corresponding polynomials:

(3x+3x^2)(3+3x^3)(2x+2x^2+2x^3)(2+2x^2+2x^4)

or factor them and compare my last post.

Alternatively, your 4 dice and my 4 dice had 2 dice in common. Toss them out and just compare the outcomes of the remaining dice, you get the same answer: 6 ways each for a 1, 2, 3, 4, 5, or 6. So if one set is correct, the other is.

I just realize that I completely misinterpreted your 3 dice set, I had thought it was the start of a 4 dice set. So example in my last post doesn't make sense in that regard, but it still illustrates how to use unique factorization to find 4 dice (modifying to any number would be similar).

 

[RandallB]

In that case I disagree.

I agree you should disagree.
Don't know what I did wrong when I tested your numbers on the spread sheet - must have been a typo.
You number set came up as I was looking for more variartions of the working number sets. Over a dozen so far.

Here's a good one with a die with no 0 or 1.

000111
000333
001122
224466

Also on the three dice sets best one has only one blank or zero surface in the set as:

011223
113355
111444

Also found a bit of info on Sicherman Dice:
Looks like they were first described by George Sicherman of Buffalo, NY in a 1978 Scientific American column on Mathematical Games by Martin Gardner, who later reprinted it in his book “Penrose Tiles to Trapdoor Ciphers”.

My guess is we won't be able to do this with 5 dice.

RB

 

[RandallB]

Also shmoe spliting your set of four into two pair thus:
002244
111222

and

001122
111444

Both pair act as a single die.

SO doubling either set like;
001122
111444
001122
111444
as a set of 4 will give same results as a normal pair.

 

[DaveC426913]

Personally, I am fond of the sequences that go above 6.

 

[geniusprahar_21]

sum of numbers on opp. sides are equal( also true for normal dice.)