TITLE: New to probability - Is this true?
QUESTION [26 upvotes]: 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?

REPLY [4 votes]: 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.