Chance of getting $20$ three times in a row on a $20$-sided die.

Question

My friend and I are having an argument and I would like some help proving I'm right or perhaps proving me wrong. My friend and I where playing dungeons and dragons the other night when discussing the chances of rolling a $20$-sided die and getting a $20$ three times in a row. I believe the answer should $\frac1{8000}$. He is saying the chances are $\frac1{8420}$ why is that? He says it has something to do with recurrence.

Answer 1

Assuming that the dice is fair and that the rolls are independent, which is what one would expect, the probabilities multiply. So the probability is indeed $(1/20)^3 = 1/8000$.

Answer 2

The easiest way to think about this would be to say that we have a 20 sided die, it is fair (every face is equally likely to be landed on) and then we know that when we roll the die each time we are equally likely to land on the same side. Now, what we usually do here is something called a tree diagram and you just follow your branch.

Okay, now I said easiest because it truly is. Each time you roll you have a $\frac{1}{20}$ chance of rolling any side, then the second time you roll it's a $\frac{1}{20}$ chance you land on the same side, on the 3rd time it's a $\frac{1}{20}$ chance you land on the same side as the first time. Now, when doing this you multiply all of your rolls together:

$(\frac{1}{20})(\frac{1}{20})(\frac{1}{20}) = \frac{1}{20^{3}} = \frac{1}{8000}$

That's the probability you will roll the same side (face - I use them interchangeably, sorry!) 3 times in a row on your fair die.

Answer 3

It's curious why your friend thinks the probability should be $\frac{1}{8420}.$ "Something about recursion" does not explain much.

It's true that if all you know is that a die is rolled at least once and not more than three times, there are $20$ possible outcomes of a single roll, $400$ possible outcomes of two rolls, and $8000$ possible outcomes of three rolls, so if we can have "stop after one roll" and "stop after two rolls" as possible sets of outcomes, there are $20+400+8000 = 8420$ possible outcomes altogether. On the other hand,

• There's no reason to think the one-roll outcomes or two-roll outcomes are equally as likely as each of the three-roll outcomes.

• There aren't any "stop after one" or "stop after two" outcomes to consider, because if you keep getting $20$ you're going to roll the die three times!

Now, if the deal is that you roll the die and roll again only if it comes up $20,$ and roll a third time only if the second roll also is a $20,$ then you do have one-roll and two-roll events, but you still do not have $8420$ equally-likely outcomes. What you have is

• $19$ one-roll outcomes each with probability $\frac{1}{20},$
• $19$ two-roll outcomes consisting of a $20$ followed by something other than $20,$ each with probability $\frac{1}{40}$ ($\frac{1}{20}$ probability of rolling the first $20,$ and each possible second roll has $\frac{1}{20}$ of that probability), and
• $20$ two-roll outcomes consisting of two consecutive $20$s followed by anything from $1$ through $20,$ each with probability $\frac{1}{8000}$ ($\frac{1}{400}$ probability of rolling the two consectuive $20$s, and each possible third roll has $\frac{1}{20}$ of that probability).

The total of all these probabilities is $$ 19\left(\frac{1}{20}\right) + 19\left(\frac{1}{400}\right) + 20\left(\frac{1}{8000}\right) = 1, $$ showing that we have covered all the outcomes.

Answer 4

The probability of rolling a $20$-sided die three times and getting three $20$s is $$ \left(\frac1{20}\right)^3=\frac1{8000} $$ If we were to roll the die three times and compare the triple rolled with $(20,20,20)$, it would, on average, take $8000$ of these triples to get the triple $(20,20,20)$.

However, if we were to simply roll the die until three $20$s occurred in succession, it would take, on average, $8420$ rolls to get three $20$s in a row. Here are a couple of ways to compute this expected duration.

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Generating Functions

The generating function for the probability of getting three $20$s in a row in exactly $n$ rolls is $$ \begin{align} f(x) &=\sum_{k=0}^\infty\left(\vphantom{\frac{19}{20}}\right.\overbrace{\ \ \ \frac{19}{20}x\ \ \ }^\text{non-$20$}+\overbrace{\ \frac{19}{400}x^2\ }^{\substack{\text{non-$20$}\\\text{one $20$}}}+\overbrace{\frac{19}{8000}x^3}^{\substack{\text{non-$20$}\\\text{two $20$s}}}\left.\vphantom{\frac{19}{20}}\right)^k\ \ \overbrace{\ \ \frac{x^3}{8000}\ \ }^{\text{three $20$s}}\\[6pt] &=\frac{x^3}{8000-7600x-380x^2-19x^3} \end{align} $$ The probability of getting three $20$s in a row eventually is $$ f(1)=1 $$ The expected number of rolls to get three $20$s in a row is the derivative at $1$. $$ f'(x)=\frac{24000x^2-15200x^3-380x^4}{\left(8000-7600x-380x^2-19x^3\right)^2} $$ The expected number of rolls to get three $20$s in a row is $$ f'(1)=8420 $$

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Recursion

Let $E$ be the expected number of rolls to get three $20$s in a row. $$ \begin{align} E &=\overbrace{\frac1{8000}\cdot3}^{\text{three $20$s}}+\overbrace{\frac{19}{8000}(3+E)}^{\substack{\text{two $20$s}\\\text{non-$20$}}}+\overbrace{\frac{19}{400}(2+E)}^{\substack{\text{one $20$}\\\text{non $20$}}}+\overbrace{\frac{19}{20}(1+E)}^\text{non $20$}\tag1\\ &=\frac3{8000}+\frac{57}{8000}+\frac{38}{400}+\frac{19}{20}+\left(\frac{19}{8000}+\frac{19}{400}+\frac{19}{20}\right)E\tag2\\ &=\frac{421}{400}+\frac{7999}{8000}E\tag3\\[6pt] &=8420\tag4 \end{align} $$ Explanation:
$(1)$: collect the contributions to the expected duration
$(2)$: separate constants and coefficients of $E$
$(3)$: simplify
$(4)$: $8000\cdot(3)-7999\cdot(1)_\text{left}$