Probability of drawing certain hand of cards There is a deck of 32 cards (not regular playing cards). There are four cards labeled 0, four cards labeled 1, four cards labeled 2, ..., and four cards labeled 7. So there are essentially eight types of cards, with four of each type in the deck. 
If you deal these cards in the normal manner (cycling through players) to $n$ players until each player has four cards, what is the probability that any of the $n$ players has four cards all of the same label?
(Note $n \leq 8$, otherwise not all players could get 4 cards.)
What I have tried:
I thought the answer might be
$$
\frac{8n}{{32 \choose 4}}
$$
because there are ${32 \choose 4}$ possible hands, and there are 8 types of cards to potentially get all of, and any of the $n$ players could get the 4-of-a-kind.
I simulated this using python
import random

NTRIALS = 2000000

def sim(n):
    deck = [0] * 32
    for i in range(32):
        deck[i] = i % 8

    random.shuffle(deck)
    players = [[] for x in range(n)]
    for i in range(4 * n):
        players[i % n].append(deck.pop())

    counts = [0] * n
    for i in range(n):
        for j in range(1, 4):
            if players[i][j] == players[i][0]:
                counts[i] += 1

    if any(c == 3 for c in counts):
        return 1
    else:
        return 0

def main():
    count = [0] * 5
    for n in [4, 5, 6, 7, 8]:
        for i in range(NTRIALS):
            count[n - 4] += sim(n)

    for i in range(5):
        print("n =", i + 4, ":", count[i] / NTRIALS)

if __name__ == '__main__':
    main()

with result
n = 4 : 0.000889
n = 5 : 0.001087
n = 6 : 0.001288
n = 7 : 0.001569
n = 8 : 0.0017525

The values predicted by my formula are
n = 4 : 0.0008898
n = 5 : 0.0011124
n = 6 : 0.0013348
n = 7 : 0.0015573
n = 8 : 0.0017798

All these values are fairly close, but I'm not sure if my formula is correct. I'd appreciate any help in understanding how to calculate this probability.
 A: You have to consider the fact that the events are dependent. Simply multiplying by 8n won't lead you to the answer.
For the sake of giving an example, consider n=2.
After the first person takes his cards, what the next person gets will change based on what the first person gets. For example, if the first person gets 1,2,3, and 4, the next person will be less likely to get 4 of the same because there are only 4/(28 C 4) ways he can get what he wants. You can see how this could quickly branch out into a very difficult to calculate problem. 
Another issue is that, if the first person picks 1,1,1, and 1, the second person should not even be considered, but your formula will multiply in the second probability anyways.
As more cards are taken, the probabilities will be more and more dependent on others. This explains why the error increases as n increases. For n = 1, there would be no error.
That said, I believe that your solution gives you the correct value for the expected value of the answer. This is based off of the idea that probability is linear - https://brilliant.org/wiki/linearity-of-expectation/ explains this idea very well. However, the expected value of the answer is not what you are looking for - you are looking for the number of times at least one has it, and therefore, if two or more people have it, you are overcalcuating them. 
Knowing this, I can easily write out a formula for n=2, simply avoiding doublecounting any cases where both win -
P(2) = (8*2)/(32 C 4) - 8/(32 C 4) * 7/(28 C 4)
However, as n increases, this will very quickly get very messy. I do not know if there is a clean method to solve this problem accurately for larger values of n.
