Friday 12 March 2021

Dice

Imagine you try to simulate a single six-sided die roll by rolling a large number of die, using the following system: 1. Roll a die 2. Check what number appears the most frequent 3. If there is a single number, take that number 4. If there is a tie, scrap both and continue.

For example, if you roll

132 ⚀’s, 144 ⚁’s, 120 ⚂’s, 97 ⚃’s, 101 ⚄’s and 118 ⚅’s

Then your result is ⚁ cause 144 is the highest.

If you roll 144 ⚀’s, 144 ⚁’s, 120 ⚂’s, 97 ⚃’s, 101 ⚄’s and 118 ⚅’s

Then ⚀ and ⚁ tie, so they are scrapped, so the highest remaining is ⚂.

The question now is the following:

For what number of dice (n) is this method: 1. Well defined 2. Does it produce the numbers 1 to 6 with equal chance

and does this work for dice with other than 6 sides?

First off, the fact that it produces all number with equal chance can be explained by a simple isomorphism argument.

The only thing that is needed to proof this is well defined, is to make sure that not all numbers are tied. There are different ways in which the frequencies can all tie, namely

2-2-2 tie, for example 0x⚀ 0x⚁ 1x⚂ 1x⚃ 3x⚄ 3x⚅ (roll: ⚂⚃⚄⚄⚄⚅⚅⚅ for 8 dice) 2-4 tie, for example 1x⚄ 1x⚅ (roll:⚄⚅ for 2 dice) 3-3 tie for example 1x⚃,1x⚄,1x⚅ (roll: ⚃⚄⚅ for 3 dice)

But for the first two ties, this can only happen if the number of die is a multiple of 2, and for the last die, this can only happen if this is a multiple of 3.

This means that if the number of die is not a multiple of 2 or 3, the frequencies cannot tie.

The argument in the other direction is similar. If n is a multiple of 2, then n/2 fives and n/2 sixes will tie. Same if n is a multiple of 3, then n/3 fours, fives and sixes will tie.

For all other n

n = 2,3 or 4 has a similar argument.

And for 5, and all numbers starting from 7, there is a tie of the form

2-3-x with x a positive integer that is not one (if it was one, that one die would not be tied).

And if we assign a frequency of zero to the x, this turns into the question, can be find for each positive number n two positive integers p and q such that 2p + 3q = n. Ans this is obviously true if n is not one.