Probability of picking m numbers from a subset of n numbers [Bingo game] I have one problem that I'm not able to solve. Let's say we have set A of 50 integers (1 to 50). We choose randomly 30 numbers (one by one, no repeat) from set A and let's name this new set, set B. Now, let's create third set C of 6 integers (possible numbers are from set A). This is some sort of a bingo game, so whats the probability that in n-th picking from set A (where we create set B) that we hit all 6 numbers from set C. 
I'm looking for a formula in terms of A, B and C where this represents size of these set and n which indicates n-th picking.
Hope I was clear enough. Thanks in advance :)
EDIT: I wasn't clear enough.
I'm picking 30 times from set A. My question is what's the probability that in n-th picking I'll hit all 6 numbers I choosed in set C. I understand that in first 5 picks the probability is 0. 
 A: I hope I've understood your question correctly. I'll use $a=|A|$ and $c=|C|$.
Disregarding order, the number of ways to get all $c$ numbers in your first $n$ numbers for $B$ is the number of ways to choose the other $n-c$ numbers from the other $a-c\;$ non-$C$ numbers available. This equals $\binom{a-c}{n-c}$.
The total number of ways to choose $n$ numbers from $a$ available is $\binom{a}{n}$.
So the probability of your $n$ numbers having all $c$ numbers is $$\binom{a-c}{n-c}\bigg/\binom{a}{n}.$$
However, it sounds like you only want the cases where the $n^{th}$ number belongs to set $C$ so that you haven't got all $c$ numbers until you choose your $n^{th}$ number.
The fraction of the $\binom{a-c}{n-c}$ ways of having all $c$ numbers that have the $n^{th}$ number belonging to set $C$ is just $\dfrac{c}{n}$ because the $n^{th}$ number is equally likely to be any of the $n$ numbers selected.
So with this additional requirement, the probability is $$\dfrac{c}{n}\binom{a-c}{n-c}\bigg/\binom{a}{n}.$$
A: Let
$$
S(k, n)
$$
denote the probability that after $n$ picks, you have selected exactly $k$ of the $c = |C|$ numbers in $C$. You're wondering about $S(c, n)$. I can't tell you a formula for that, but I can at least write down a recurrence, for general $k$ and $n$, based on the idea that the only ways you can have $k$ good numbers on the $n$th pick are (a) to have $k$ on the $n-1$th pick, and pick a bad number, or (b) to have $k-1$ on the $n-1$th pick, and pick a good number. The result is therefore in general
$$
S(k, n) = pS(k-1, n-1) + q S(k, n-1)
$$
where $0 < k \le c$ and $1 \le n \le a$, and $p$ is the chance of picking a "good" number (one that's in the set $A$, but has not yet been picked in the first n-1 picks) from the set of all numbers not yet picked, and $q$ is the change of picking a "bad" number (one not in $A$), given that $n-1$ numbers have been picked. In the left-hand part, the number of "good" numbers remaining, under the assumption that $k-1$ have been picked, is $c - (k-1) = 1 + c - k$. 
The total count of numbers remaining in $A$ is $a - (k-1) = 1 + a - k$. The probability $p$ is therefore
$$
p = \frac{1 + c - k}{1 + a - k}.
$$
Similarly, for the right-hand term, given that $k$ "good" numbers have already been picked, there are $c - k$ numbers remaining in $C$, and $a - k$ numbers remaining in $a$, and we want to pick one of the non-C numbers; thus
$$
q = \frac{(a-k) - (c - k)}{a-k} = \frac{a - c}{a - k}.
$$
Given these, we now have to think about the corner cases. 


*

*What about when $n = 1$? To use the recurrence, we need a formula for $S(k, 0)$ for every $k$. That's not tough: $S(0,0) = 1; S(k, 0) = 0$ for all $k > 0$.

*What about when $k = c+1$? You can't every pick $c+1$ of the $c$ elements in $C$, so $S(k, n) = 0$ for every $k > c$.   

*What about when $n > a$? You can't possibly pick more than $a$ items from a set of $a$ items, so $S(n, k) = 0$ for every $n > a$. 
Now we can write down a general recurrence:
\begin{align}
S(k, n) &= \begin{cases}
1 & k = 0, n = 0 \\
0 & k > 0, n = 0 \\
0 & k > c\\
0 & n > a \\
\frac{1 + c - k}{1 + a - k}S(k-1, n-1) + \frac{a - c}{a - k} S(k, n-1) & \text{else}
\end{cases}
\end{align}
Using dynamic programming or memoization (yeah, that's a word!), you can now compute $S(k, n)$ for any $k,n$ pair in a number of operations proportional to $nk$, which for practical matters isn't too bad. 
As for an explicit formula -- I doubt that's easy to find, but perhaps some MSE person can prove me wrong. I'd love to see it!
