# Unusually common dice

1. May 3, 2005

### 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.

Last edited: May 3, 2005
2. May 3, 2005

### 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??

3. May 3, 2005

### 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.

4. May 3, 2005

### ceptimus

Code (Text):
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

Last edited: May 3, 2005
5. May 3, 2005

### Jimmy Snyder

Nicely displayed ceptimus.

Last edited: May 3, 2005
6. May 3, 2005

### 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

7. May 3, 2005

### 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 :)

8. May 4, 2005

### 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.

9. May 4, 2005

### 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

10. May 4, 2005

### Jimmy Snyder

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.

11. May 4, 2005

### minger

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

12. May 4, 2005

### DaveC426913

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?

13. May 4, 2005

### Jimmy Snyder

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

14. May 4, 2005

### 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...

15. May 4, 2005

### DaveC426913

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?

16. May 5, 2005

### Jimmy Snyder

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).

Last edited: May 5, 2005
17. May 5, 2005

### 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.

18. May 5, 2005

### DaveC426913

:uhh: :grumpy: :yuck: :surprised
And in English, the answer would be ... :grumpy:

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)

19. May 5, 2005

### 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 lets 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 wierd 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).

20. May 5, 2005

### DaveC426913

Code (Text):

[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.
.

Last edited: May 5, 2005