Finding the probability that two sets of numbers chosen randomly from the same set have the same numbers Background : At my local bingo hall, there is a game called Casino. They give you 2 cards per dollar, and the goal of the game is to get a full card ("blackout") in 53 balls or less for 500 dollars, or 54 balls or more for 100 dollars.
I am trying to find the probability of having a full card for $n$ amount of ball pulls.
In the easiest sense I would want to explain it, you have set A, B, and Z. At the beginning, A and B are empty. Z consists of all integers between 1 and 75. $c$ amount of elements from set Z are randomly selected and allocated to set A without any repeating elements. Then, $d$ amount of elements from set Z are randomly selected and allocated to set B. Assuming $d \geq c $, what is the probability that all elements in set A are contained in set B?
If possible, I would like the answer to be formatted to a graphable equation, with :
$x$ is the amount of balls, $y$ is probability of blackout
Although, for the full card number of possibilities, $nCr(75, 24)$ isn't right, the proper amount of possibilities is $nCr(15, 5)^{4}nCr(15, 4)$. Although, the equation $y=\frac{nCr(15, 5)^{4}nCr(15, 4)}{nCr(75, x)} $ provides weird results. How do I get the correct results?
 A: $$
\frac{\binom{75-c}{d-c}}{\binom{75}{d}}
$$
Explanation: As far as we're concerned, the set $A$ is fixed. There are $\binom{75-c}{d-c}$ ways to complete the set $A$ into a set consisting of $d$ numbers. There are $\binom{75}{d}$ ways of selecting the set $B$.
A: First, let's do the probabilities for one card, because (unless you would get paid double if both cards score on the 53rd or 54th pull) you chances of winning with two cards depends (slightly) on how many numbers the two cards have in common.  (You can see this by noting that if the two cards are identical, having the second card gives no improvement at all.)
OK, then, the probability of filling one card on 53 pulls is 
$$
\frac{\binom{51}{29}}{\binom{75}{53}} \approx 0.00003
$$
since you have no choice but to include all $24$ of your numbers, and there are $28$ more numbers to be had.  
For 54 pulls it is 
$$
\frac{\binom{51}{28}}{\binom{75}{54}} \approx 0.000094
$$
and assuming you would get paid double for two cards both winning, the expected payout is about $4.9$ cents(!)
Are you sure the payout is not \$5,000 and \$1,000?
As stated, the game still favors the house if you allow 57 pulls for the big win and 58 for the small win.
