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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?

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    $\begingroup$ For starters, barrel dice can easily have sixteen sides, so if you really wanted to you could manufacture a fair sixteen sided die just fine. As for the mathematics question, assuming that you don't get frustrated by the time it takes to stop rolling $E$ over and over, yes the final probabilities will be the same between the sixteen sided die and the twentysided die after any potential rerolls. The downside to the twentysided though is that you could potentially have to reroll thousands of times before getting a result (however unlikely) $\endgroup$ – JMoravitz Jun 14 '17 at 19:41
  • $\begingroup$ Thank you so much. This has been extremely helpful. $\endgroup$ – user3803238 Jun 14 '17 at 20:03
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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.

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  • $\begingroup$ Thank you so much. The heavy math is lost on me. Congratulations on being really clever and a nice enough person to answer silly questions of your own back. People like you will eventually take us to the stars. $\endgroup$ – user3803238 Jun 14 '17 at 20:04
  • $\begingroup$ No problemo ;) If you found this answer helpful, consider upvoting and or accepting. $\endgroup$ – Saketh Malyala Jun 14 '17 at 20:49
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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.

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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.

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  • $\begingroup$ Thank you so much. The heavy math is lost on me. Congratulations on being really clever and a nice enough person to answer silly questions of your own back. People like you will eventually take us to the stars. $\endgroup$ – user3803238 Jun 14 '17 at 20:04
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    $\begingroup$ @user3803238 My pleasure, and welcome to MathSE. If this or any other answer was of help to you, please consider upvoting/accepting. $\endgroup$ – jvdhooft Jun 14 '17 at 20:05

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