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So if I roll a fair six sided die twice, I can expect the probability of getting 2 6’s to be 1 in 36.

I want to start with a probability of say 1 in 100 and calculate how many times I’d need to roll a die to match that.

I get it wouldn’t be an integer (1 in 100 would be between 2 and 3), but what specifically would it be

By extension, how could I calculate the same thing but for 5’s and 6’s. My thinking in that case would be to treat it in the equation as a 3-sided die. Would that be accurate?

Scott Page
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1 Answers1

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We're treating each die roll as independent, so to get the final probability we just multiply the probability of success at each step.

With the repeated 1-in-6 rolls: The probability of success in one step is $p = \frac 1 6$, so the probability of success $n$ times in a row is $p^n$. If you wanted to solve for $n$ to get a final probability of $\frac{1}{100}$, you could use $n = \log_p \left(\frac{1}{100}\right) = -\log_p(100)$. With $p=\frac 1 6$ that comes out to $\approx 2.57$. (Note this is similar to your guess in the comments but not exactly the same.)

If you used 2-in-6 rolls instead (success on 5 or 6) then the math would be the same but with $p = \frac 1 3$ instead of $p = \frac 1 6$.

Finally - what if you really want to get an exactly 1-in-100 result using just your 6-sided die? You can do this using a trick. First, roll the die 3 times. Let $a, b, c$ be those three results. Then define a number $X$ by $X = 36(a-1) + 6(b-1) + (c-1)$. Now $X$ is a uniformly random number from $0$ to $6^3-1 = 215$.

If $X <= 199$, we can use the last two digits of $X$ as the result from a $100$-sided die. For example, $X=157$ would count as rolling $57$ on the d100. To get the $\frac{1}{100}$ probability we wanted, we can just define success as $X=0$ (or $100$) and failure as any other value.

If $X \ge 200$, the process has failed and we need to restart. Roll 3 more times and compute a new $X$. In theory this process could go on forever (if we keep getting $X \ge 200$ but in practice we'll usually be fine even on the first set of rolls, since the probability of getting $X \ge 200$ is only $\frac{8}{216} \approx 7\%$.

You could extend this process to get any probability you want. To get $\frac{1}{N}$, just roll enough d6 so that the total number of outcomes is $\ge N$, then break the die roll outcomes into $N$ different equal groups. The leftover outcomes always mean failure and you'll need to restart the process in those cases.

Further reading if you're interested: 1, 2.

David Clyde
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