Unusually Common Dice: Find the Unexpected Ordinary

[Jimmy Snyder]

The number of spots on the faces of a normal die are 1, 2, 3, 4, 5, 6.

Here is an unusual pair of dice. One has the following number of spots on its faces:

1, 3, 4, 5, 6, 8

The other has:

1, 2, 2, 3, 3, 4

Obviously, these are not ordinary dice. The problem is to find out what is unexpectedly ordinary about them.

 

[minger]

It's just a guess, but a somewhat educated guess as I'm not a big fan of doing statistics but:

The odds of any roll 2-12 is the same as on a regular pair of dice??

 

[Jimmy Snyder]

It's a good guess. I am willing to give you 90%. But I don't think it should count as the answer unless you actually show that it is true so that you can say it with more confidence.

 

[ceptimus]

Total  Regular dice                  Funny dice                      Odds
  2    1-1                           1-1                             1/36
  3    1-2, 2-1                      1-2, 1-2                        2/36
  4    1-3, 2-2, 3-1                 1-3, 1-3, 3-1                   3/36
  5    1-4, 2-3, 3-2, 4-1            1-4, 3-2, 3-2, 4-1              4/36
  6    1-5, 2-4, 3-3, 4-2, 5-1       3-3, 3-3, 4-2, 4-2, 5-1         5/36
  7    1-6, 2-5, 3-4, 4-3, 5-2, 6-1  3-4, 4-3, 4-3, 5-2, 5-2, 6-1    6/36
  8    2-6, 3-5, 4-4, 5-3, 6-2       4-4, 5-3, 5-3, 6-2, 6-2         5/36
  9    3-6, 4-5, 5-4, 6-3            5-4, 6-3, 6-3, 8-1              4/36
 10    4-6, 5-5, 6-4                 6-4, 8-2, 8-2                   3/36
 11    5-6, 6-5                      8-3, 8-3                        2/36
 12    6-6                           8-4                             1/36

 

[Jimmy Snyder]

Nicely displayed ceptimus.

 

[minger]

eh 90% is good enough for an A or at least A-. For the work I actually did, I'm pretty happy with my score

 

[T@P]

actually this brings up a funny question, which is if you take a standard monopoly dice, and roll it, the probabilities arent the same for getting a 2 or a 3. (because its not an 'ideal' dice). the question is which way are they slanted?

idk if it has anything to do with the previous one, but its a good question :)

 

[Jimmy Snyder]

T@P,

'Monopoly dice' aren't standard. There is no standard way of tossing them either. Also there is no standard surface upon which to toss them, or a standard environment (temperature, air motion, etc.) But if these standards could be established, then you could take a pair of standard dice and toss them in the standard way enough times to establish the probabilities of each outcome. Then, in order to take advantage of this knowledge, you would have to make sure that those conditions prevailed while you were playing the game. However, you would not be able to make sure that your opponents were tossing in the standard way. Also, you would be ethically in the wrong if you knowingly used dice that weren't 'fair' unless your opponents were also aware of it.

 

[ArielGenesis]

just a rough guess but i think that the sum of the opposite side is always equal

7=1+6=2+5=3+4
9=1+8=3+6=4+5
5=1+4=2+3=2+3

 

[Jimmy Snyder]

just a rough guess but i think that the sum of the opposite side is always equal

7=1+6=2+5=3+4
9=1+8=3+6=4+5
5=1+4=2+3=2+3

It is possible to arrange things so that the number of spots on two opposite faces is always 9 on one die and always 5 on the other. The arrangement would be as you indicated in the two lines above that start with 9= and 5=. I'm not sure what the 7= line refers to. However, it was not stipulated that they would be arranged that way.

 

[minger]

actually this brings up a funny question, which is if you take a standard monopoly dice, and roll it, the probabilities arent the same for getting a 2 or a 3. (because its not an 'ideal' dice). the question is which way are they slanted?

idk if it has anything to do with the previous one, but its a good question :)

Um...not 100% sure what you're talking about, but the odds are heavily not in favor of rolling a 2 because there is only 1 possible outcome that results in a 2 (1 and 1 obviously). The dice are naturally "slanted" towards 7's, then 6's and 8's. so on and so on

 

[DaveC426913]

Total  Regular dice                  Funny dice                      Odds
  2    1-1                           1-1                             1/36
  3    1-2, 2-1                      1-2, 1-2                        2/36
  4    1-3, 2-2, 3-1                 1-3, 1-3, 3-1                   3/36
  5    1-4, 2-3, 3-2, 4-1            1-4, 3-2, 3-2, 4-1              4/36
  6    1-5, 2-4, 3-3, 4-2, 5-1       3-3, 3-3, 4-2, 4-2, 5-1         5/36
  7    1-6, 2-5, 3-4, 4-3, 5-2, 6-1  3-4, 4-3, 4-3, 5-2, 5-2, 6-1    6/36
  8    2-6, 3-5, 4-4, 5-3, 6-2       4-4, 5-3, 5-3, 6-2, 6-2         5/36
  9    3-6, 4-5, 5-4, 6-3            5-4, 6-3, 6-3, 8-1              4/36
 10    4-6, 5-5, 6-4                 6-4, 8-2, 8-2                   3/36
 11    5-6, 6-5                      8-3, 8-3                        2/36
 12    6-6                           8-4                             1/36

This is fascinating! I MUST make a pair of these and use them in the next game I play. Watch em wail!

Say, can this be worked out for 3 dice? Any takers?

 

[Jimmy Snyder]

I MUST make a pair of these and use them in the next game I play.

And if you tell 'em these dice don't change the odds any, you'll be dead right.

 

[T@P]

well no... the answer to my question is more physics ish...
look at a reglular dice. the numbers are indicated by makine little holes and painting them black... so if there are more holes on the 6 side than the 1, its lighter on that side...
*light little lightbulb on top of your head*

 

[DaveC426913]

Say, can this be worked out for 3 dice? Any takers?

I'm really interested in doing this. I can't seem to figure out an algorithm to solve for any other possibilities though. And then transposing it to three dice.

jimmy, where did you find this?

 

[Jimmy Snyder]

I'm really interested in doing this. I can't seem to figure out an algorithm to solve for any other possibilities though. And then transposing it to three dice.

jimmy, where did you find this?

This problem was posed to me 30 years ago by Professor Donald Newman of Temple University when I was taking a course of his on recreational mathematics. He noted the following fact:

The coefficents of the powers of x in this polynomial

$$(x + x^2 +x^3 + x^4 + x^5 + x^6)/6$$

are all 1/6, the probability of rolling the exponent with a single die.

He then pointed out that the coefficients of the powers of x in the expansion of this polynomial

$$(x + x^2 +x^3 + x^4 + x^5 + x^6)^2/36$$

are the probabilities of rolling the exponent with a pair of dice.

He challenged us to factor the polynomial into linear factors (factors of the form x + c) and rearrange the factors into a second pair of polynomials whose coefficients add up to 1 and that produce the same product. The result is

$$(x + 2x^2 + 2x^3 + x^4)/6$$
and
$$(x + x^3 + x^4 + x^5 + x^6 + x^8)/6$$

From these polynomials, you can read off the number of spots needed on each face of a new pair of dice. I do not remember the linear factors of the polynomial and I am too lazy to figure them out again, but the first two are easy

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

Therefore, if there is a solution for three dice, it would involve finding the rest of the factors and rearranging them into three other polynomials whose coefficients add to 1 and whose product is

$$(x + x^2 +x^3 + x^4 + x^5 + x^6)^3/216$$

By the way, Professor Newman was also my source for the plane colorization problem(s).

 

[Jimmy Snyder]

It is trivially easy to extract a second pair of linear factors from the polynomial so here is the ammended equation:

$$(x + x^2 + x^3 + x^4 + x^5 + x^6)^2/36 = (x + 0)^2(x + 1)^2(1 + x^2 + x^4)^2/36$$

Reducing the higher degree polynomial from 5 to 4 is important because there is a formula for factoring fourth degree polynomials, but not fifth degree.

One more point, I don't remember if it is necessary to completely factor the polynomial into linear factors. It may be that the problem can be solved by merely factoring the 4th degree polynomial into a pair of 2nd degree factors. The same goes for the three dice problem.

Four funny dice can be produced by simply taking two pairs of the dice I have already described. Indeed, the only interesting problems are for prime numbers of dice. Once the polynomial is factored, all of these problems can be addressed one at a time, or perhaps in bulk.

 

[DaveC426913]

It is trivially easy to extract a second pair of linear factors from the polynomial so here is the ammended equation:

$$(x + x^2 + x^3 + x^4 + x^5 + x^6)^2/36 = (x + 0)^2(x + 1)^2(1 + x^2 + x^4)^2/36$$

Reducing the higher degree polynomial from 5 to 4 is important because there is a formula for factoring fourth degree polynomials, but not fifth degree.

One more point, I don't remember if it is necessary to completely factor the polynomial into linear factors. It may be that the problem can be solved by merely factoring the 4th degree polynomial into a pair of 2nd degree factors. The same goes for the three dice problem.

Four funny dice can be produced by simply taking two pairs of the dice I have already described. Indeed, the only interesting problems are for prime numbers of dice. Once the polynomial is factored, all of these problems can be addressed one at a time, or perhaps in bulk.

And in English, the answer would be ...





I'm going to have to do this with brute force...

Ideally, I'd like:
- 3 dice
- none "normal" (I can make a 3 dice solution by merely adding one normal die to the 2 dice solution)
- each one different (solutions with 2 identical dice are less interesting)
- each one clearly 'crazy' (I hope to find a set where one die has a 7 or 8, and another die has a zero)

 

[shmoe]

I've seen these called Sicherman dice before, so if you want to google that might help.


It's enough to factor the polynomial as $$x^1+x^2+x^3+x^4+x^5+x^6=x(x+1)(x^2+x+1)(x^2-x+1)$$, into irreducible factors (irreducible over Z), since the polynomial representing your dice in the end will be over Z. (I'm removing the 1/6 factor found in jimmysnyder's post, we can manage without it)

So suppose we have three dice whose behavior matches 3 normal dice, then the product of their corresponding polynomials (call them p(x),q(x),r(x)) will be $$(x^1+x^2+x^3+x^4+x^5+x^6)^3=x^3(x+1)^3(x^2+x+1)^3(x^2-x+1)^3$$.

Concentrating on just p(x) for now, $$p(x)=x^a(x+1)^b(x^2+x+1)^c(x^2-x+1)^d$$ (this follows from unique factorization of polynomials with integer coefficients). Now the condition that this die has 6 faces is the same as requiring the sum of the coefficients in the polynomial to be 6. This is the same as requiring p(1)=6. Sticking this into the above form we see $$6=p(1)=1^a(1+1)^b(1^2+1+1)^c(1^2-1+1)^d=2^b3^c$$, so we must have b=c=1 (by unique factorization of another kind;)). This holds for q and r as well.

Since you wanted an interesting answer, for the moment let's allow 0 as a valid face for the die (reading again I see you actually desired this!). What remains is to assign a and d. Let me relabel slightly, and say up to this point we know:

$$p(x)=x^{a_p}(x+1)(x^2+x+1)(x^2-x+1)^{d_p}$$
$$q(x)=x^{a_q}(x+1)(x^2+x+1)(x^2-x+1)^{d_q}$$
$$r(x)=x^{a_r}(x+1)(x^2+x+1)(x^2-x+1)^{d_r}$$
$$p(x)q(x)r(x)=x^3(x+1)^3(x^2+x+1)^3(x^2-x+1)^3$$

so all that remains is to choose (non-negative) integers so that:

$$a_p+a_q+a_r=3$$ and
$$d_p+d_q+d_r=3$$

There are a few way to do this and get 'funny' results. One result is:
$$p(x)=x^{1}(x+1)(x^2+x+1)(x^2-x+1)^{0}=x^4+2x^3+2x^2+x$$
$$q(x)=x^{0}(x+1)(x^2+x+1)(x^2-x+1)^{1}=x^5+x^4+x^3+x^2+x+1$$
$$r(x)=x^{2}(x+1)(x^2+x+1)(x^2-x+1)^{2}=x^9+x^7+x^6+x^5+x^4+x^2$$

Giving dice 1,2,2,3,3,4, one with 0,1,2,3,4,5 and one with 2,4,5,6,7,9

If you were to not allow 0 as a face, this condition is equivalent to p(0)=0 so that a>=1 and consequently a=1 (since all die will need at least one on this exponent) and you end up not being able to meet all your criterea- the only possibilities are 3 normal dice or 1 normal and the two weird ones from the original post.


This same technique will show that the only way to mimic the behavior of 2 normal dice is the funny ones in the original post (assuming we don't allow 0 as a face).

 

[DaveC426913]

[B]
1,2,3,4,5,6		1,2,2,3,3,4
1,2,3,4,5,6		0,1,2,3,4,5
1,2,3,4,5,6		2,4,5,6,7,9
[/B]
3  1			3  1
4  3			4  3
5  6			5  6
6  10			6  10
7  15			7  15
8  21			8  21
9  25			9  25
10 27			10 27
11 27			11 27
12 25			12 25
13 21			13 21
14 15			14 15
15 10			15 10
16 6			16 6
17 3			17 3
18 1			18 1

You are a supergenius!


I am looking forward to rolling 0's and 9's on 3D6 at my next game.
.

 

[shmoe]

You are a supergenius!
.

Thanks, but it's just a demonstration of how fantastic unique factorization can be.


The original funny dice of this thread may give the same probability of a 2, 3 etc. as the usual dice, but if you are playing monopoly why aren't they equivalent? (this requires some knowledge of monopoly)

 

[Jimmy Snyder]

The probability of doubles (1-1, 2-2, 3-3, 4-4, 5-5 and 6-6) are not the same. Backgammon is another game that would be changed by introducing these funny dice. And there are many others I'm sure.

 

[DaveC426913]

And another property of dice is their handedness. Most dice are right handed, it's rare to find a left-handed die.

No, I'm not talknig about how you throw them!

If you place the six face up, and the five toward you, most dice wil have the 3 on the right face. A very few (hard to find) dice will have the 3 on the left.

This property has no known effect on any game though.

(The assignation of left vs. right is purely arbitrary. Left is called left because it is relatively rarer of the two.)

 

[DaveC426913]

Took me a while*, but here are my alternate 3D6s:
1 2 2 3 3 4
0 1 2 3 4 5
2 4 5 6 7 9

*(a lot longer than my stupidhead friends appreciate... )

 

[shmoe]

Bah, I can't seem to view your image. So you made some dice? Out of what?

 

[RandallB]

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.
Log out - Select return to last view.
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

 

[RandallB]

Randall's Dice

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]

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

 

[RandallB]

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:

 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]

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 surprise 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

 

[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.