simulating a fair six with a four equal sector spinner

Question

Whist teaching basic probability I needed a group to use a fair four sector spinner but I'd none left. I gave them a die asking them to disregard 5,6 should they arise. The problem got me thinking about what I could do if I needed a die and only had a spinner.

I have tried to formulate some rules for the puzzle:

1. You can relabel the spinner after a spin. For example $1234$ means the four equally likely sectors were labelled $1,2,3$ and $4$. Similarly $1123$ means $2$ of the sectors were labelled $1$.
2. It is ok to label $5$ and $6$ e.g. $3456$
3. You cannot choose a sector to ignore essentially reducing the spinner to three equally likely sectors.
4. You can spin and re-label as many times as you like.

Is it possible under these rules to simulate a fair six sided die with such a spinner?

I have adopted a notation of writing the state of the spinner for each spin underneath each other.

$$\begin{array}{c} 1122 \\ 3344 \\1123 \\1223 \\1233 \end{array}$$

The above example an attempt of mine.

Let the outcomes of successive spinners be called a branch (thinking like a probability tree) The branch$ (1,4,2,2,3)$ represents the outcome of $1$ on the first spin $4$ on the second etc. I chose this labelling because there would be $2\times2\times3\times3\times3=108$ different branches and $108$ is divisible by $6$

I know the branches are not equally likely but I was hoping to assign sets of branches to the six outcomes of the die e.g. $\{(1,4,2,2,3),(1,3,2,2,1)...\} \mapsto 1$ So far Ive got lost because the tree diagram gets out of hand.

I'm not sure this problem can be solved as no power of $4$ is divisible by $6$

This is the first problem I've invented by myself so I apologize if there is something I have overlooked that makes the problem trivial.

Thanks in advance and I hope you find it an interesting puzzle if it turns out easier than I've thought.

Answer 1

Well, one way to do this is to express 1/6, 2/6, 3/6, 4/6 and 5/6 in base 4.

Then construct a base 4 representation of a real number between 0 and 1, by using the stream of random numbers in base 4 - till the number can be unambigously classified as falling in one of the 6 intervals (i/6, (i+1)/6).

---

To make it precise - here are the boundaries -

0. 000000...
1. 022222...
2. 111111...
3. 200000...
4. 222222...
5. 311111...
6. 333333...

Keep getting the random spin of numbers from 0 to 3, and construct a sequence. The moment the sequence constructed so far is unambigously between any two of the above, say (n) and (n+1), stop. Then classify the draw as a n. In that case n is an uniform random number between 0 to 5.

---

Some examples -

"110" -> 1
"01" -> 0
"22223" -> 4
"33" -> 5

etc.

Answer 2

Essentially, your attempt runs afoul of unique factorization. You cannot define a procedure with a bounded number of spins that generates each of six different outcomes with equal probability, because $6$ is never a factor of $4^n$ for any positive integer $n$.

Practically, this is of no import. There are any number of procedures one can follow to generate values between $1$ and $6$, inclusive, with equal probability, but because of the above, they will always admit infinite sequences of spins that fail to generate any valid value (such as, for instance, the sequences mentioned by KalEl). However, because such sequences collectively have zero probability, these procedures will "almost surely" complete.