What is the chance of 3 particular numbers, occurring in 6 randomly generated numbers? Let's say I generate $6$ numbers: $X_1, \ldots, X_6$
where $X_i$ can be any integer between $1$ and $59$ with equal probability (inclusive).
For example, the following could have been randomly generated:
$$1, 2, 3, 4, 5, 59$$
$$1, 1, 2, 3, 4, 59$$
$$\text{etc}\ldots$$
What is the chance that $1,2,3$ is a  subset of the $6$ numbers generated?
What is the chance that $1,1,3$ is a  subset of the $6$ numbers generated?
And in general,
What is the chance that $x,y,z$ is a  subset of the $6$ numbers generated?
Edit: As per the definition of "subset", order does not matter. So $1,2,3$ is a subset of $1,2,3,4,5,6$ as well as $1,6,3,5,2,4$
 A: Assuming that ${a, a, b}$ is not the same as ${a, b},$ we can solve the problem using basic combinations.
In a set with six elements, $S = \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\},$ there are $\dbinom{6}{3} = 20$ "subsets" of $S$ with three elements in it. If there are $59$ numbers to choose from, we can construct a total of $59 + 2 \cdot \dbinom{59}{2} = 3481$ distinct sets of three numbers (that are not necessarily different).
The probability is simply $\boxed{\frac{20}{3481}},$ which is about $0.57\%.$
A: $$Pr(1,2,3\text{ in set})=Pr(\text{three numbers in }\{1,2,3\})Pr(1,2,3 \text{ all included})\\
+Pr(\text{four numbers in }\{1,2,3\})Pr(1,2,3 \text{ all included})+...\\
={6\choose3}(\frac3{59})^3(\frac{56}{59})^3\frac6{27}+
{6\choose4}(\frac3{59})^4(\frac{56}{59})^2\frac{36}{81}+...$$
By Inclusion-Exclusion, Pr(1,2,3 all in five)=$(3^5-3.2^5+3.1^5)/3^5$, and
Pr(1,2,3 all in six)=$(3^6-3*2^6+3*1^6)/3^6$.  I get about 0.00054
The odds will be different for $1,1,2$.  3/59 becomes 2/59, and Pr(1,1,2 all included) is different.  I get about 0.000275
