Can a randomized system cause unfair dice to make a fair result?

Question

Say I have 6 dice with 6 faces in a bag that are all weighted unfairly so that each dice have the same unfair probability of landing on a different side. (i.e. the probability that one of the unfair dice has a higher chance on landing on the one's side has the same probability as another unfair dice to land on the twos side that another die has the same probability on the threes side ect)

I then grab a die at random from the bag and I roll get a number then put it back into the bag. I then randomly grab another dice.

Dose the resulting number have the same chance as a regular 6 sided dice ($\frac{1}{6}$)?

Rather then doing the math I have tried using logic and I believe that it will cause the resulting number to be fair but I probability is not strong suit and I would love to hear someone else's opinion.

Answer 1

Yes, this is fair. Think of the bag as containing $36$ possible faces. Your hypothesis about how the weighting is distributed means that each of the six possible values has the same probability when you choose a die and roll it: $1/6$.

The hypothesis is very strong. The only way I can imagine satisfying it in a physical situation is to have six blank dice equally unbalanced in the same known way. Then mark the faces as specified.

Answer 2

For each die, there is probability $p$ to land on the biased face. Then, it has probability $\frac{1-p}{5}$ to land on any of unbiased faces. Since each number is biased on $1$ die and unbiased on $5$ dice, we have \begin{align*} P(\text{rolling a 1}) &= P(\text{rolling a biased 1}) + P(\text{rolling an unbiased 1}) \\ &= \frac{1}{6} \times p + \frac{5}{6} \times \frac{1-p}{5}\\ &= \frac{1}{6} \times p + \frac{1}{6} \times (1-p) \\ &= \frac{1}{6} \end{align*}

Therefore, this process replicates rolling an unbiased 6-sided die.