# Sunday Times Teaser 2739 – Funny Dice

### by Michael Fletcher

#### Published: 22 March 2015 (link)

I have two cube-shaped dice, one red and one blue, with a positive whole number on each face. When I throw the dice and add up the two numbers, the most likely total is 7. The next most likely totals are 6 and 8, the next are 5 and 9, the next are 4 and 10, the next are 3 and 11, and the least likely are 2 and 12. However, my dice are not standard: indeed, the total of the six numbers on the red dice is higher than the total of those on the blue dice.

What are the six numbers on the red dice?

1. 2. I’m personally of the opinion that “non-standard dice” are too well-known for this teaser to pass the “originality” test.

3. A bit easy – how about if the dice are twelve sided – what are the numbers on each dice – or even if the dice have 2N sides what are the numbers on each dice?

• I suspect that is only easy if you have experienced it (or something like it) before – we can see from my site for manual solvers (https://sites.google.com/site/sundaytimesteasers/teaser-index-1/2015-q1/teaser-2739) that it created quite a bit of discussion.

I have written to Victor Bryant about ‘originality’ before now and his view is that there are very few truly original teasers so it would make life hard if he insisted on using this as a criterion.

For me it opened up an area I had not delved into before (I had looked at non-transitive dice but not non-standard ones). So I certainly enjoyed this one both intellectually and as a programming exercise – well done!

For those who are interested, here is a link covering the non-standard dice https://en.wikipedia.org/wiki/Sicherman_dice.

4. Assume the use of two dodecahedral dice numbered 1 to 12. The same probability distribution can then also be obtained by the use of any of the following seven combinations:

(1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7) and (1, 3, 5, 7, 7, 9, 9, 11, 11, 13, 15, 17);

(1, 2, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9) and (1, 2, 3, 7, 7, 8, 8, 9, 9, 13, 14, 15);

(1, 2, 4, 5, 5, 6, 8, 9, 9, 10, 12, 13) and (1, 2, 3, 3, 4, 5, 7, 8, 9, 9, 10, 11);

(1, 2, 2, 3, 5, 6, 6, 7, 9, 10, 10, 11) and (1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13);

(1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6) and (1, 4, 5, 7, 8, 9, 10, 11, 12, 14, 15, 18);

(1, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8) and (1, 2, 5, 6, 7, 8, 9, 10, 11, 12, 15, 16);

(1, 2, 2, 3, 3, 4, 7, 8, 8, 9, 9, 10) and (1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14).

I worked this out by hand (and spreadsheet) after work and before dinner.

I believe it is exhaustive; it exhausted me.

But thank you. “It was a tough assignment.”

5. Peter Hayes has alerted me to a neat way of tackling non-standard dice using a mathematical method based on generating functions. The idea is that the six faces on a die can be represented by the polynomial: $x+x^2+x^3+x^4+x^5+x^6$where the powers of $$x$$ represent the face values on the die and the polynomial coefficients represent the number of times each face value occurs. The above polynomial hence represents a standard die. If we now throw two dice, we can find the result by squaring the above polynomial to give:$x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^{10}+2x^{11}+x^{12}$Now the powers represent the sum of the two dice and the coefficients give the number of times each sum will occur. So sums of 2 and 12 each occur just once, while a sum of 7 occurs 6 times (1+6, 2+5, 3+4, 4+3, 5+2, 6+1).
It turns out that the generating polynomial for a standard six sided can be factored:$x+x^2+x^3+x^4+x^5+x^6=x(x+1)(x^2-x+1)(x^2+x+1)$ and this allows the result of throwing two standard dice to be represented by the polynomial:$x^2(x+1)^2(x^2-x+1)^2(x^2+x+1)^2$ Although for two standard dice we select the same factors for each die, we can produce the same result from other selections of the two subsets. For example we might chose the two subsets: $x(x+1)(x^2-x+1)=x+x^4$$x(x+1)(x^2+x+1)^2(x^2-x+1)=x+2x^2+3x^3+3x^4+3x^5+3x^6+2x^7+x^8$These two polynomials give a two sided die (1, 4) and an 18 sided one (1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 8). In addition to different numbers of faces, other choices can also involve negative face values but just one: $x(x+1)(x^2+x+1)=x+2x^2+2x^3+x^4$ $x(x+1)(x^2+x+1)(x^2-x+1)^2=x+x^3+x^4+x^5+x^6+x^8$ produces a pair of six-sided dice. It is these dice with face values (1, 2, 2, 3, 3, 4) and (1, 3, 4, 5, 6, 8) that solve this teaser. They were first discovered by George Sicherman and are named after him (see Sicherman Dice for more details).