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?

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.
