There are 10,000 possible numbers, since 10^4 is 10,000.
10 possible quads
10/10,000 -> 1/1,000
Anon wins!

Ok then, if you don't count 0000, then 9/10,000

no, because the number set would have been 1111,2222,...,9999 (nine numbers) and the whole set 0001,0002,9999 (including quads, 9,999 numbers, exculding quads, 9,990 numbers) So we were both wrong. It was 9/9990 or 1/1110.

Actually, the wholes set is 10,000 numbers because you are excluding 0000 from the set, even though it could be rolled.
# of 1 digit numbers (0-9):-------------------10 or 10^1 or 10
# of 2 digit numbers (00-99):--------------100 or 10^2 or 10*10
# of 3 digit numbers (000-999):--------1,000 or 10^3 or 10*10*10
# of 4 digit numbers (0000-9999):---10,000 or 10^4 or 10*10*10*10

Or, thanks to probability, you could think of it as this:
Possible outcomes for the 1st digit:----10 (0-9)
Possible outcomes for the 2nd digit:---10 (0-9)
Possible outcomes for the 3rd digit:----10 (0-9)
Possible outcomes for the 4th digit:----10 (0-9)
Multiply the possible outcomes: 10 * 10 * 10 * 10 = 10,000 (0000-9999)

Now, we know for sure that there are 10,000 possible outcomes. When finding the probability of selecting one at random, you put the number of outcomes you want over the number of total possible outcomes.

If you count 0000 as a quad:_______________________________________
Wanted outcomes(quads, including 0000):-----------10 (0000, 1111, 2222, etc.)
Total possible outcomes:-----------------------------10,000 (0000, 0001, 0002, etc.)
Chance of rolling quads (including 0000):---10/10,000 or 1/1000 0r .001

If you DON'T count 0000 as a quad:_________________________________
Wanted outcomes(quads, except 0000):-----------------9 (1111, 2222, 3333, etc.)
Total possible outcomes(doesn't change):------10,000 (0000, 0001, 0002, etc.)
Chance of rolling quads (excluding 0000):----9/10,000 or .0009

Sorry for all the math, but I like explaining this stuff to people.

**psykobear rolls 5,491**