Sicherman dice

[Mister_JWO]

Hello Everyone

I want to try as exercice to proof the sicherman dices without using polonomials.

Are there properties I can use to proof the case with 7 as biggest number on one die and 5 as biggest number on the other die? The case with max 8 on a die I have proved taking all the cases that were relevant, but with 7 as maximum on a die it are just too much. I think there has to be an interesting propertie I can't find.

Thank you very much.

Kind Regards
Mister JWO

 

[blamocur]

Hello Everyone

I want to try as exercice to proof the sicherman dices without using polonomials.

Are there properties I can use to proof the case with 7 as biggest number on one die and 5 as biggest number on the other die? The case with max 8 on a die I have proved taking all the cases that were relevant, but with 7 as maximum on a die it are just too much. I think there has to be an interesting propertie I can't find.

Thank you very much.

Kind Regards
Mister JWO

From Wikipedia: "They are notable as the only pair of 6-sided dice that are not normal dice, bear only positive integers, and have the same probability distribution for the sum as normal dice."

 

[Mister_JWO]

Yes, I know that, but I'm interested in it and I want to proof it (as a kind of exercise) in another way than the proof with polonomials.

 

[pka]

I want to try as exercice to proof the sicherman dices without using polonomials.
Are there properties I can use to proof the case with 7 as biggest number on one die and 5 as biggest number on the other die? The case with max 8 on a die I have proved taking all the cases that were relevant, but with 7 as maximum on a die it are just too much. I think there has to be an interesting propertie I can't find.

I think that any us that have done any teaching/research in this will ask you "why reinvent a wheel"?
Suppose that [imath]\left\{ {\boxed1{\kern 1pt} \boxed2\boxed2{\kern 1pt} \boxed3{\kern 1pt} \boxed4{\kern 1pt} \boxed4\quad \boxed5{\kern 1pt} \boxed7\boxed7{\kern 1pt} \boxed8{\kern 1pt} \boxed8{\kern 1pt} \boxed9} \right\}[/imath] is a set of two sets of such dice.
Now consider the generating polynomial: [imath](x+x^2+x^2+x^3+x^4+x^4)\cdot (x^5+x^7+x^7+x^8+x^8+x^9)[/imath]
See the expansion HERE the term [imath]7x^{10}[/imath] in that expansion tells us that there are seven ways to toss a total of ten.

 

[Steven G]

I understand that you want to prove that there is exactly one other pair of 6-sided dice that gives the same probability for a given sum as a pair of regular dice does.

Then you lose me completely. I know the (polynomial) proof of these dice very well.

The Sicherman dice are numbered {1,2,2,3,3,4} and {1,3,4,5,6,8} so I have no idea at all what you are trying to do with having a 7 on one side. It is not possible to get the correct probabilities with a 7 on a die.

You can argue that to get a sum of 12, the other die must have exactly one 5 on it. To get a sum of 2 exactly one way you need to have a 1 on each die. To get a sum of 3 you need to have two 2's--either both on one die or one on each die. Just continue and you'll get a contradiction eventually if you are trying to prove that a die can't have a 7 on it.

May I ask why you don't want to use polynomials?