In my last blog post, I explained how to quickly and easily work out, say, the number of ways to get a 10 on three dice, just by multiplying together some polynomials. It doesn’t have to be real dice of course. The trick works just as well for spinners, balls in a hat, or computer random number generators.

The trick is like this –

• Represent the three dice by a polynomial. An ordinary dice gives the polynomial x + x² + x³ + x⁴ + x⁵ + x⁶, with one power of x for each of its faces.
• Multiply the polynomials together. Three dice gives  x³ + 3x⁴ + 6x⁵ + 10x⁶ + 15x⁷ + 21x⁸ + 25x⁹ + 27x¹⁰ + 27x¹¹ + 25x¹² + 21x¹³ + 15x¹⁴ + 10x¹⁵ + 6x¹⁶ + 3x¹⁷ + x¹⁸ .
• Then, just read off the number of combinations for each number. The coefficient of x¹⁰ is 27, so there are 27 ways to get a 10 on three dice. (Don’t believe me? Count them and leave a comment below!)

This is a lot of fun – if you know this trick, you can do certain types of probability sums much faster than using any other method. The real fun, however, comes no from multiplying the polynomials for a die or set of dice, but factorising it.

The polynomial for a normal six-sided die is x + x² + x³ + x⁴ + x⁵ + x⁶. I can partially factorise this as (1 + x) (x + x³ + x⁵). This tells me that, using the trick of the previous post,

• I’ll take a coin, with one side blank and a dot on the other side.
• I’ll also take a spinner, with the numbers one, three and five.
• Their polynomials are 1+x and x + x³ + x⁵.
• If I multiply these together, I get x + x² + x³ + x⁴ + x⁵ + x⁶.

So with my coin and my spinner, there’s exactly one way to get 1, exactly one way to get 2, exactly one way to get 3, and so on. I’ve made a six-sided die out of a coin and a triangular spinner!

Another way to factorise x + x² + x³ + x⁴ + x⁵ + x⁶ is as (1 + x³) (x + x² + x³). So, I can imitate a six-sided die as

• a spinner with the numbers 1, 2 and 3, giving polynomial x + x² + x³, and
• a coin, one side with no dots, the other side with 3 dots, giving polynomial 1 + x³.

We could find all possible substitutes for a normal die by listing all the different ways to factorise its polynomial. Unfortunately, all other factorisations ask us to put certain numbers on negative (or worse, complex) numbers of faces of our spinner.

So, for more entertainment, we turn to pairs of normal dice. The polynomial for a pair of normal dice is (x + x² + x³ + x⁴ + x⁵ + x⁶) (x + x² + x³ + x⁴ + x⁵ + x⁶), or (x + x² + x³ + x⁴ + x⁵ + x⁶)². If I factorise this fully, I get  x² (1 + x)² (1 + x + x²)² (1 – x + x²)². I want to choose some of these 8 factors (count them), multiply them together, to get a plan for a spinner or fancy die.  The remaining factors will give the plan for another, and these two dice or spinners will simulate a pair of normal 6-sided dice.

Useful fact – if I set x=1, I get the total number of faces on my spinner or dice. A normal die has 1 + 1² + 1³ + 1⁴ + 1⁵ + 1⁶ = 6 faces. A pair of dice, with polynomial (x + x² + x³ + x⁴ + x⁵ + x⁶)², can be simulated with a spinner with (1 + 1² + 1³ + 1⁴ + 1⁵ + 1⁶)² = 36 sides.

If I put x=1 into the four different factors of (x + x² + x³ + x⁴ + x⁵ + x⁶)², I get

• 1 for x
• 2 for 1+x
• 3 for 1+x+x²
• 1 for 1-x + x².

This makes it easy to choose, in advance, how many faces on each of our spinners or dice. If I want to make a polynomial for a 6-sided dice, using these factors, I need include one of 1+x and one of 1+x+x² as factors. For good measure, I’ll also include 1 of x. Here’s one possibility for my die and its partner :

• x (1 + x) (1 + x + x²) (1 – x + x²), and x (1 + x) (1 + x + x²) (1 – x + x²)

this is what I get if I make the die and its partner the same. The polynomials multiply together to give x² (1 + x)² (1 + x + x²)² (1 – x + x²)², or (x + x² + x³ + x⁴ + x⁵ + x⁶)², just like they should. Unfortunately, that’s because each individual one multiplies out to give x + x² + x³ + x⁴ + x⁵ + x⁶. That’s not very interesting – it means my two dice are just normal dice. We already know that two normal dice gives the same numbers of combinations as two normal dice.

So, let’s move one of the factors from one die to the other, and use

• x (1 + x) (1 + x + x²) (1 – x + x²)², and x (1 + x) (1 + x + x²)

These polynomials still multiply together to give x² (1 + x)² (1 + x + x²)² (1 – x + x²)², or (x + x² + x³ + x⁴ + x⁵ + x⁶)². But they are different polynomials, so they must give different dice! So –

• Multiply out the polynomials x (1 + x) (1 + x + x²) (1 – x + x²)², and x (1 + x) (1 + x + x²)
• This gives x + x³ + x⁴ + x⁵ + x⁶ + x⁸ and x + 2x² + 2x³ + x⁴ .
• The first of these represents a die with faces 1, 3, 4, 5, 6 and 8
• The second represents a die with faces 1, 2, 2, 3, 3 and 4.

So, if we make these two dice, the probabilities of getting any particular number when you roll them will be exactly the same as if you had rolled two normal dice! These amazing nonstandard dice are called Sicherman Dice and were discovered in the 1970’s. You can, of course, get them from Amazon.com, but it’s probably more fun to make your own pair!

Even more fun, still, is to use the tricks here to find other pairs of strange or crazy dice – can you find, for example, a pair of nonstandard four-sided dice that gives the same probabilities as the standard pair? Or 12-sided or 20-sided dice? Finding four-sided dice will be easier than the larger sizes, so if you’re keen to start, start there.

If you really want to get into this in-depth, try googling “cyclotomic polynomials” – they turn out to be very useful for this kind of puzzle – but many of the explanations on the web are (be warned) quite technical!

(1 – x + x²)