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

• Mar 18th 2013, 01:44 PM
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!
• Mar 18th 2013, 02:02 PM
Plato
Re: roulette number repeat...two workings dif answers..1wrong-why?
Quote:

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 $\displaystyle \mathcal{T}=\binom{12+38-1}{12}$ total of outcomes.

There are $\displaystyle \mathcal{D}=\binom{38}{12}$ ways to have all different outcomes.

The probability of all different outcomes is $\displaystyle \frac{\mathcal{D}}{\mathcal{T}}$.

So the probability of at least one repeated outcome is $\displaystyle 1-\frac{\mathcal{D}}{\mathcal{T}}$.

Now I am not sure that is actually what you mean by "any number repeated".
• Mar 18th 2013, 02:29 PM
Re: roulette number repeat...two workings dif answers..1wrong-why?
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?
• Mar 18th 2013, 03:13 PM
Plato
Re: roulette number repeat...two workings dif answers..1wrong-why?
Quote:

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.
• Mar 19th 2013, 03:29 AM