Probability of dice roll affected by reroll. Say I wanted a 16-sided die with either A, B, C or D on each side, with the following probabilities:
A. 1/16
B. 7/16
C. 3/16
D. 5/16
I could roll the die and get either A, B, C or D with above probabilities. However, I read that 16-sided dice are not isohaedral: not every side can have the same size. 
Now consider a 20-sided die (one of the Platonic solids) with the following probabilities:
A. 1/20
B. 7/20
C. 3/20
D. 5/20
E (null reroll). 4/20
Four sides now have a value of $E$, which is void and results in a reroll until either A, B, C or D are obtained. Does this have implications on the probability of getting each of the valid results? Would using the 20-sided die be similar to using the 16-sided die with equal size sides?
 A: Notice that the probabilities are the same regardless of rerolls.
The probability of rolling A is still $\frac{1}{16}$, B still $\frac{7}{16}$, etc.
Let's consider side A.
Say I roll A with probability $\frac{1}{20}$, or maybe I get a null and reroll A, or I get a null and another null and reroll A, or i get a null.....(x) times and roll A.
What this looks like is $\frac{1}{20}$(roll) + $\frac{1}{20}*(\frac{1}{5})$(chance to null & reroll) + $\frac{1}{20}*(\frac{1}{25})$(chance to roll null twice & reroll)...$\frac{1}{20}(\frac{1}{5^n})$
The chance of rolling A first then would be $\frac{1}{20}+\frac{1}{100}+\frac{1}{500}+\frac{1}{2500}$.... it's a geometric series with sum $\frac{1}{16}$.
Similar logic for B, C, D. 
This is because you can't actually end with E, so the probabilities are still in the ratio of A:B:C:D.
A: Those two dice will be identical in terms of probabilities.
Consider the 20-sided die. The expected outcome is:
$$E=A\frac{1}{20}+B\frac{7}{20}+C\frac{3}{20}+D\frac{5}{20}+E\frac{4}{20}$$
Rerolling results in $E$ because $E$ is the expected outcome. Now modify this equality.
$$E\frac{16}{20}=A\frac{1}{20}+B\frac{7}{20}+C\frac{3}{20}+D\frac{5}{20}$$
$$E=A\frac{1}{16}+B\frac{7}{16}+C\frac{3}{16}+D\frac{5}{16}+E\frac{4}{16}$$
which is exactly the same with 16-sided die's expectation equation.
A: Indeed, using the proposed 20-sided die does not affect the probability of resulting in $A$, $B$, $C$ or $D$. Let $A_n$ denote the event that, starting from the $n^{th}$ roll, we end up with an $A$. Then we get:
$$P[A_1] = P[A] + P[E_1] P[A_2] = \frac{1}{20} + \frac{4}{20} P[A_2]$$
However, the turn in which we're in does not affect the probability of, eventually, ending up with $A$ - it is a memoryless process. We can thus state that $P[A_1] = P[A_2]$, and we get:
$$P[A_1] = \frac{1}{20} + \frac{4}{20} P[A_1] \iff \frac{16}{20} P[A_1] = \frac{1}{20} \iff P[A_1] = \frac{1}{16}$$
You can also look at it this way: since we always reroll the die after we hit $E$, it is impossible to end the series with $E$. As such, we can only consider the events of hitting $A$, $B$, $C$ and $D$:
$$P[A_1] = \frac{P[A]}{P[A] + P[B] + P[C] + P[D]} = \frac{\frac{1}{20}}{\frac{16}{20}} = \frac{1}{16}$$
As mentioned in the comments, however, one disadvantage to this approach is that you could potentially end up with a high number of rerolls before you finally hit a valid side of the die.
