<s>[INST] Probability of getting the same number three times. If I have a set of numbers $\{1 \ldots n\}$ where $n \ge 1$ and I pick $3$ numbers from the set independently and uniformly. Whats the probability I'll get all $2$'s, the probability I get all the same numbers and the probability that my first two numbers are the same.
I've I compared this to rolling a fair die three times. Wouldn't probability of getting three $2$'s  and getting all the same numbers be $\left(\dfrac 16 \right)^3$ since its uniform. And the probability of getting two of the same numbers be $\left(\dfrac 16 \right)^2 \left(1-\dfrac{1}{6} \right)$??
Any help would be appreciated.
 [/INST] *

*You're right, the probability of all 2s is
$$P(222) = P(2)P(2)P(2) = \frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6} = \left(\frac{1}{6}\right)^3= \frac{1}{216},$$
where use the product by independence.

*When calculating probability that they are all the same is, you have to account for all the choices you have for a match, which is $\binom{6}{1}$. So, 
$$P(\text{All match}) = \binom{6}{1}\left(\frac{1}{6}\right)^3 =\frac{1}{36}.$$

*For the first two to match you have again $\binom{6}{1} = 6$ choices for this pair. After, you will only have $\binom{5}{1} = 5$ choices for the remaining single. Lastly, you need to account for the order of the first two since for example $221$ is the same as if I switched the first two $221$, which is $\binom{3}{2,1} = 3$. So the probability of interest is
$$\binom{3}{2,1}\binom{6}{1}\left(\frac{1}{6}\right)^2\binom{5}{1}\left(\frac{1}{6}\right) = 90\left(\frac{1}{6}\right)^3 = \frac{5}{12}.$$
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