# Nonstandard 6-Die

9 ianuarie 2012 • All, Miscellaneous

Since 2004, more often than not, while teaching with Dofin ASE I kept assigning the class homework involving a nonstandard 6-die problem. The reason could be related to a remembrance since my times with Oxford University when one of my teachers realized that Romania did not have a border with Germany.

The real reason is that in order to understand finance one needs to span many mathematical branches including probability and combinatorics. “Playing” with an n-die is a good starting point.

A standard 6-die refers to a fair, six-sided die, whose faces are numbered 1 to 6.

The homework asks to find an alternative pair of 6-dice, with faces labeled with positive integers different from the standard 6-dice. The new pair should have the same odds for throwing every number as a normal pair of 6-dice.

Moreover it must be proved that there is precisely one pair of nonstandard 6-dice with the same probability as a pair of standard 6-dice.

The answer is known as Sicherman Dice and is represented by a pair of 6-dice labeled with (1,2,2,3,3,4) and (1,3,4,5,6,8).

For unicity one can check Proposition 9, here.

One of the first instances when the problem was mentioned was in M. Gardner, Penrose Tiles to Trapdoor Ciphers.

### Abonare la articole

Poti introduce adresa de email mai jos daca vrei sa fii la curent cu ultimele articole postate pe LucianIsar.com

### 2 Comentarii

Sergiu
18-01-2012 la ora 11:35 pm

Nice problem. For those who intend to read the paper from Caltech I have some observations:
1. There is a typing error at Question 4. The product obviously should be made by Ai(x) power series
2. The proofs of DEM 5&6 require solid abstract algebra knowledge such as Galois theory. However, the exemples of cyclotomic polynomial are intuitive. Another helpful observation is that the degree of phi[n](x) is given by Euler’s number phi(n).
3. Based on the list related to DEM 7 one might speculate that all coefficients of all cyclotomic polynomials are either +1, -1, or 0. But this is not true. It is true for n prime and for n having at most two distinct prime divisor. The smallest n where phi[n]() has an exotic coefficient is n=3*5*7=105. Also, the proof is not trivial.

Lucian
31-08-2018 la ora 7:40 am

Multumesc Sergiu