# Unusually common dice

shmoe said:
Bah, I can't seem to view your image. So you made some dice? Out of what?
Don't know why but if your logged in - many times your cannot see a jpeg etc.
And bingo - you can see the images.

System must need to block some access function that affects viewing them; that it does not need to block when your not logged in.
RB

Randall's Dice

DaveC426913 said:
Took me a while
Took me an Even longer while for me to get this one.

Jimmy
Do you think working the “polynomial linear factors” might find these four dice?
I’m calling them “Randall’s Dice” unless I find someone somewhere has made them already.

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

Opposite faces would total 2, 3, 4, 5 respectively.
I’m sure you can tell from the Minimum and Maximum possible rolls what the objective here is.

I’ll allow first shot at showing just how ‘close’ I got to the objective,
to you guys and shmoe plus anyone that would like to.

Now if I can just find someplace that can actually cut a set of these for me.

RB

shmoe
Homework Helper
RandallB said:
Do you think working the “polynomial linear factors” might find these four dice?
Yes it would. It will be the same as the 2 die case except you have an extra 2^2 and a 3^2 to distribute as well (going by how I arranged it). How did you find them?

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

There's not many options on how you break the dice up, and I didn't want to include a non-random one like 2,2,2,2,2,2.

It might also be amusing to allow different sided die and rearrange like:

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

shmoe said:
except you have an extra 2^2 and a 3^2 to distribute
Not sure what you mean here if you have an extra it mean there'd be a short somewhere - did you use the polynomial method to figure it? - from my method of checking mine are spot on for the odds of all 11 possible results.
How did you find them?
I used some educated guesses.
Didn’t take long to see that 20 out 24 numbers were required to be used for the high and low numbers in order to get the 2 and 12 odds right.
Then it was picking how to distribute the high numbers and balance the interior 4 values to get the other odds to come out right.
Used a Excel spread sheet to produce all 1296 possible outcomes and the Data Analysis - Histogram tool to collect and check results.
I put your alternative 4 dice set in – it misses on 3, 4, 9, & 10; yours is the last column below.

Stat results:
PHP:
 2    36	36
3    72	108
4	108	144
5	144	144
6	180	180
7	216	216
8	180	180
9	144	108
10   108	72
11   72	72
12   36	36
– And I just got lucky with three dice, was beginning to think it couldn’t be done, but solving for four helped me work on the three.

1, 2, 2, 3, 3, 4
1, 1, 3, 3, 5, 5
0, 0, 0, 3, 3, 3
Should work.
RB

shmoe
Homework Helper
shmoe said:
It might also be amusing to allow different sided die and rearrange like:

1, 2
0, 0, 3, 3
1, 1, 2, 2, 3, 3
0, 0, 2, 2, 4, 4
This set you're saying is wrong? That conclusion doesn't suprise me when I look back at my horribly ambiguous wording. "different sided die" means "dice with different number of sides". The list above represents a 2-sided dice, a 4- sided dice and two 6 sided dice. If my first set works, so does this one as the probabilities of the corresponding dice are the same (meaning my 2 sided 1,2 die has the same outcomes as a 6-sided 1,1,1,2,2,2 die).

For 2-six sided dice our polynomial had 36 possible outcomes (viewing each face on each die as distinct). If you divide the relevant polynomial by 36 you get the probabilities of the outcomes as coefficients:$$(x^1+x^2+x^3+x^4+x^5+x^6)^2/36=x^2/36+x^3/18,\ etc$$

For 4 dice we will have 1296 outcomes so we'll be dividing our polynomial p(x) by 1296 and we should get the same probabilities as above so

$$p(x)/1296=(x^1+x^2+x^3+x^4+x^5+x^6)^2/36$$

and the polynomial representing our 4 dice will be

$$36(x^1+x^2+x^3+x^4+x^5+x^6)^2=2^{2}3^{2}x^2(x+1)^2(x^2+x+1)^2( x^2-x+1)^2$$

It's now a matter of splitting this into 4 polynomials, each with positive coefficients that sum to 6.

For the 3 you've started, they correspond to:

$$x+2x^2+2x^3+x^4=x(x+1)(x^2+x+1)$$
$$2x+2x^3+2x^5=2x(x^2+x+1)(x^2-x+1)$$
$$3+3x^3=3(x+1)(x^2-x+1)$$

so the 4th die must make up the rest:

$$6(1)$$

and you have a die with 6 zero sides,

0, 0, 0, 0, 0, 0

shmoe said:
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
Homework Helper
RandallB said:
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).

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shmoe said:
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

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

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