Question involving probability Think of a card game. cards can either be "big" or other cards
we consider a game with 2 players
each player has a deck of 40 cards
each player shuffles their deck, deals 7 cards which is their hand
assume that player 1 has 10 big cards in their deck and player 2 has 20
I would appreciate if you could answer the simpler question first, and post your answer immediately after figuring it out.
Simpler question: What is the probability that player 2 has exactly 3 big cards in their hand?
Tricker question: What is the probability that player 2 has more big cards in their hand than player one?
After working for 3 hours on the trickier question, I got the answer wrong. I do not know whether my answer to the easier question is correct, however I think my issue may come from getting the easier question wrong.
Here's my approach for the easier problem:
Imagine that player two pulls out 4 "non-big" cards in a row, then pulls out 3 "big cards."
The probability of this occurring is ((20/40)(19/39)(18/38)(17/37)(20/36)(19/35)(18/34)).
If player two has exactly 3 big cards in their hand, there are 35 ways that all those 7 cards can be arranged (7 choose 3), all of which are equally likely.
Multiplying 35 by the long product computed before, I get the answer of 0.296, or 29.6%.
Is that correct? If not, where did I go wrong?
 A: I don't think the second problem is particularly tricky, but it's long.  All we can hope to do is to arrange the calculations efficiently.
The probability that play $1$ has exactly $k$ big cards, where $k=0,1,2,\cdots,6$ is $$p_k:=\frac{\binom{10}{k}\binom{30}{7-k}}{\binom{40}7}$$  Now we need to know the probability $q_{k+1}$ that player $2$ has at least $k+1$ big cards.  When $k=0$ we need $q_1$, the probability that he has at least $1$ big card.  This is the complement of the probability that he has no big cards:$$q_1=1-\frac{\binom{30}{7}}{\binom{40}{7}}$$
Once we have computed $q_k$, it's easy to computed $q_{k+1}$.  The probability that player $2$ has at least $k+1$ big cards is the probability that he has at least $k$ big cards, minus the probability that he has exactly $k$ big cards: $$q_{k+1}=q_k-\frac{\binom{10}{k}\binom{30}{7-k}}{\binom{40}7}$$
Once we've computed all these probabilities, the probability that player $2$ has more big cards than player $1$ is $$\sum_{k=0}^6p_kq_{k+1}$$
