I read this somewhere and can't find it to verify:

"If you don't have a dice with 12 sides, but do have one with 20 sides and you need to make rolls for 12 sides, you can use the numbers 1-12 on the 20-sided die, ignoring any other numbers if they come up." (not necessarily a direct quote, but as good as I can remember)

Assuming that each side is numbered from 1 to n and that the dice are balanced, so each side has an equal probability of occurring:

In my mind, a 20-sided die has a 1/20 chance that any given number will come up and a 12/20 (3/5) chance that a number between 1 and 12 will come up, although I believe this 3/5 probability becomes a 1/1 probability if you ignore any numbers between 13 and 20.

Assuming that you ignore numbers 13-20, does the probability of numbers 1-12 occurring become 1/12 (i.e. the same as a 12-sided die)? Or is it more complicated than that?

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    $\begingroup$ The key here is that each roll of the die is independent of all other rolls. So rolling a 1, then a 13, then a 6 is 'no different' (in this context) from rolling a 1 then a 6, as rolling 13 doesn't effect the probability of then getting a 6. If 13 comes up, pretend it didn't and roll again. $\endgroup$ – Daniel Freedman Oct 31 '11 at 13:00
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    $\begingroup$ FYI, the general principle behind this example is called rejection sampling. $\endgroup$ – FelixCQ Oct 31 '11 at 13:16
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    $\begingroup$ The only "catch" is that there is a non-zero chance that you will have to wait an arbitrarily long time before you will get a number in the correct range. (In practice, after two or three rolls you'll almost certainly get at least one good one) $\endgroup$ – Random832 Oct 31 '11 at 13:57
  • $\begingroup$ Also, see this related interview question. $\endgroup$ – FelixCQ Oct 31 '11 at 15:38
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    $\begingroup$ 20-sided die (and also 12-sided die) are notoriously non-random. Because the angles at the vertices and edges are so obtuse, even without deliberate modification, they are very easy to wear to the point where some sides are favored rather than others. A better physical device to get 1 through 12 would be to roll a 6-sided side and flip a coin: add 6 for tails and 0 for heads. But of course theoretically, leaving out 13 through 20 works just fine. $\endgroup$ – Mitch Oct 31 '11 at 20:23

Let's find the probability of, say, rolling a 5. You could roll a 5 right away with probability $\frac{1}{20}$, roll a number over 12 and then roll again and get a 5 with probability $\frac{8}{20} \frac{1}{20}$, roll two numbers over 12 and then roll a 5 with probability $\frac{8}{20} \frac{8}{20} \frac{1}{20}$, etc. All these ways of rolling a 5 are mutually exclusive so their probabilities sum up. In total the probability of eventually rolling a 5 is given by the geometric series below, which I evaluated using the usual geometric series sum formula.

$$S = \frac{1}{20} + \frac{8}{20} \frac{1}{20} + \frac{8}{20} \frac{8}{20} \frac{1}{20} + \ ... \ = \sum_{i=0}^{\infty} \frac{1}{20} \left(\frac{8}{20}\right)^i = \frac{\frac{1}{20}}{1-\frac{8}{20}} = \frac{1}{12}$$

The fact that when you reduce the sample space there are only 12 of the 20 equally likely possible outcomes remaining, one of which must occur eventually, is a nice informal way to find the sum of this infinite series.

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    $\begingroup$ Nice observation! You did not simply use the GP, but pointed out that the intuitively clear answer of $\frac{1}{12}$ gives us the sum of the series. $\endgroup$ – André Nicolas Oct 31 '11 at 15:33
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    $\begingroup$ +1. Another "intuitive" way is to consider getting 13+ on a 20-sided dice to a 12-sided dice rolling of the table (whatever probability that is...). It will not count... $\endgroup$ – Macke Oct 31 '11 at 16:06
  • $\begingroup$ Thanks, @BrendanCordy, good clear answer! $\endgroup$ – oliver-clare Nov 1 '11 at 10:21
  • $\begingroup$ This answer is not very good - it is complicating a non complicated problem. All that is needed is to observe the symmetry as @AMPerrine is doing. $\endgroup$ – hpekristiansen Jan 24 '12 at 16:29

What you are looking for is the notion of "conditional probability". The process which you are describing can be thought of "sampling the dice" conditioned on the event that the result is between $1$ to $12$.

Let $A$ be the event that the dice falls between $1$ to $12$. This event has probability $12/20$. Now, denote by $B_i$ the event that the dice falls on the number $i$, where $i$ is between $1$ and $12$. Each $B_i$ has probability $1/20$. Then, you have that:

$P(B_i \mid A) = P(B_i\cap A)/P(A) = \frac{1/20}{12/20} = \frac{1}{12}$,

which is indeed uniform over the numbers $1$ to $12$.


Yes. The probabilities of rolling any given number are still equal, so when you discard the extras you are left with twelve numbers with equal probability. Therefore the probability of any single number must be $\frac{1}{12}$.


It is that simple. You start with 1/20 for each side. If you agree to ignore 13-20, you are saying you will roll the die as many times as necessary to get a number in the range 1-12. You can then scale up the probabilities to sum to 1, which is a factor 20/12, and you get 1/12 for each side 1-12.

Another way to get there is to ask what is the chance that I accept a 1? I can either get a 1 on the first roll (probability 1/20), get something higher than 12 on the first roll and a 1 on the second roll (probability (8/20)*(1/20)), or ..., leading to $\sum_{i=0}^{\infty}\frac{1}{20}(\frac{8}{20})^i=\frac{1}{20}\frac{1}{1-\frac{8}{20}}=\frac{1}{12}$


A $20$-sided die is an icosahedron and has $12$ vertices. So if you want to roll only once and don't mind a little extra effort in interpreting the result, you could identify the vertices (for instance by remembering or writing down for each vertex the numbers on three faces adjacent to it) and associate them with the numbers $1$ through $12$. Then you could roll the die so that it lands near some straight line, e.g. the edge of the table, and use the vertex on the top face that's nearest to the line.

You can avoid having to identify the vertices by utilizing the duality with the dodecahedron in a slightly more subtle way: Roll the die so that it lands near some straight line. There are three faces adjacent to the top face. Determine the one closest to the line, and determine the rank of its number among the three numbers on the three faces. This is uniformly distributed between $1$ and $3$. You can combine it with the number on the top face, which is uniformly distributed between $1$ and $20$, to get a number uniformly distributed between $1$ and $60$. Then taking the remainder modulo $12$ (with $0\equiv12$) yields the desired number uniformly distributed between $1$ and $12$. The reason this works is that the number of faces times the number of faces adjacent to a face is twice the number of edges, and since the dual has the same number of edges this is also the number of vertices times the number of vertices adjacent to a vertex, and thus a multiple of the number of vertices.


Just for fun, you can improve on that method. If you roll 1-12 take that result. Otherwise, you have a number 13-20; interpret that number as
13-14: 0
15-16: 3
17-18: 6
19-20: 9
and roll again. On the second and subsequent rolls, if you get a 19 or 20, reroll. Otherwise, interpret the number as
1-6: 1
7-12: 2
13-18: 3

and add the two numbers to get your result.

In practice this takes longer than just rolling until you get a good result.* But instead of taking 5/3 = 1.66... rolls on average, this takes only 13/9 = 1.44... rolls on average.

.* Actually you can improve on the practical method by rolling two dice, say a red one and a blue one. If the red one is 1-12 take its result, otherwise look at the blue one. If neither are 1-12, reroll both.


"Assuming that you ignore numbers $13$-$20$, does the probability of numbers $1$-$12$ occurring become $1/12$ (i.e. the same as a $12$-sided die)? Or is it more complicated than that?"

It is nearly as simple as that. As @RossMillikan and others have noted, you are going to repeatedly roll the fair $20$-sided die until one of the $12$ numbers $1$-$12$ shows up, and stop right then. So, on the roll after which you stopped (and so you know the roll resulted in one of $1$-$12$), what are the probabilities that you got $1$,$2$, $\ldots, 12$ respectively? The $12$ probabilities clearly sum to $1$, and they must all equal $1/12$ because any supposed "proof" that $P(i) > 1/12$ (which implies $P(j) < 1/12$ for some other $j$) can be modified (by interchanging $i$ and $j$ everywhere) into a proof that it is $P(j)$ that is greater than $1/12$ and $P(i)$ that is smaller than $1/12$.


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