Using two coins, simulate a fair dice with 7 sides, A, B, C ,D ,E, F, G.
Coin 1:
Head 6/7
Tail 1/7
Coin 2:
Head 50%
Tail 50%
My approach
Step 1:
------------- Coin1 -----------------
/ \
/ \
6/7 / \ 1/7
/ \
/ \
Step2 G
Step 2:
A new problem, simulate a fair dice using a fair coin and the biased coin (coin1)
I can do this only using a fair coin, flipping 3 times, 8 possibilities, assigning 1-6 to A, B, C, D, E, F.
For 0 and 7, repeat the process. Since termination possiblity is 3/4, it will stop at some point.
But, can it be done in a finite number of steps? perhaps the biased coin could be used again?
I think it cannot and the problem is the following. Let there be $N-n$ throws of the fair coin and $n$ throws of the unfair coin (hence $N$ total trows). Then the probability of any configuration happening is $$ \frac{1}{2^{N-n}}\frac{6^k}{7^{n}} $$ wherre $0\leq k\leq n$.
Now in order to simulate a fair dice you want to associate to some subset of configuration each face, and you want such a subset to have probability $1/7$. In other words you want to divide the configurations with probability given by the expression at the beginning into groups whit probability 1/7. Then, for instance, for the first face you will have to find integers $k_1,k_2,\dots,k_i$ such that $$ \frac{1}{7}=\sum_{j=1}^i\frac{1}{2^{N-n}}\frac{6^{k_i}}{7}=\frac{1}{2^{N-n}}\frac{6^{\sum_i k_i}}{7^n}. $$
Then
$$ 7\times 6^{\sum_i k_i}=2^{N-n}7^n $$
This is definitely impossible since the left hand side is divisble by 3 while the right hand side is not.
Then it is impossible to simulate the fair dice with a fixed finite number of tries.