Math Help - roulette number repeat...two workings dif answers..1wrong-why?

Hi,
Suppose a roulette (with 38 numbers) is spun 12 times, and the probability of any number repeated is required.

So:
Working 1: 38C1*12C2*(1/38)^2*(37/38)^10.......gives a value greater than 1 so obviously wrong. (success occuring twice for a single number, then accounting for 38 separate no.s)
Working 2: No repeat= 38*37*36*..*27/(38)^12. So at least one repeat=1-no repeat=1-.143=.857 probability...looks better, and sufficient for my task.

Problem is...how do I find out the probability of any number occuring thrice exactly, or at least probability of any number occuring twice exactly, as the working 1 doesn't work, and working 2 is only applicable for the special case and no more Why just when I think I have learnt something about probability it eludes me.

Heeeelp!

Originally Posted by ssadi

Hi,
Suppose a roulette (with 38 numbers) is spun 12 times, and the probability of any number repeated is required.

There are total of outcomes.

There are ways to have all different outcomes.

The probability of all different outcomes is .

So the probability of at least one repeated outcome is .

Now I am not sure that is actually what you mean by "any number repeated".

no, I don't think I meant what you think I meant
You are saying the probability of a particular set of number of appearing
I am saying that say..2 appears at least twice, or 15 appears exactly 3 times... in the 12 trials
How to get that probability?

Originally Posted by ssadi

no, I don't think I meant what you think I meant
You are saying the probability of a particular set of number of appearing
I am saying that say..2 appears at least twice, or 15 appears exactly 3 times... in the 12 trials
How to get that probability?

In that case, you need to give a full and complete detailed description of the setup.

When you use the word any it does not mean a particular set.

for example
12 trials

any number in the set (00,0...36) repeats itself

within the 12 trials

I don't know how i can be more specific

similarly, say

any number in the set (00, 0...36) appears exactly 3 times in the 12 trials