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.