Probability of having 4 aces after taking turns to pick cards I've started to learn probability and I get stuck with the following problem:
My friend and I are playing a card game with 36 unique cards. There are four suits (diamonds, heart, clubs and spades), each of them having cards numbered from 1 to 9 (cards numbered with 1 are defined to be the aces). In this game, each player grabs 9 cards from the top of a shuffled deck, one at a time, taking turns. So I grab a card first, then my friend grabs a card, and this continues until we each have 9 cards in our hand. What are the odds that I choose all four aces?
After thinking about how to solve this problem I’ve came up with 2 different approaches that give me different results. I would like to know if I'm wrong with one of them or both.
Option 1: Trying to use the conditional probability.
A total of 18 cards will be picked. I want to find the probability of 4 aces to appear among these cards.
1) Total # of outcomes: It is just the number of different 9-card hands I can make with 36 cards.
2) # of ways it can happen: # of ways of having 4 aces among 18 cards.
$P(A): \frac{C(18,4)}{C(36,9)} = \frac{9}{276892}$
Now, given that 4 aces are among those 18 cards, I want to find the probability of having those 4 aces in my hand:
1) Total # of outcomes: # of ways of making 9-card hands with 18 cards.
2) # of ways it can happen:  # of ways of having 4 aces in one 9-card hand.
$P(B|A): \frac{C(9,4)}{C(18,9)} = \frac{63}{24310}$
And finally:
$P(A\cap$$B) = P(B|A) * P(A) = \frac{63}{24310}*\frac{9}{276892} \approx 8.42*10^{-8}$
Option 2: Multiplying probabilities.
Since I start picking one card, then my friend picks the next and so on, in this situation I’m going to pick the card #36, #34, #32, # 30. I hope to find an ace in each of these turns, so:
$P(B) = \frac{4}{36} * \frac{3}{34} * \frac{2}{32} * \frac{1}{30} = \frac{1}{48960}$
Thanks !!
 A: Note that there are $\binom{36}{9}$ nine-card hands, all equally likely.
There are $\binom{32}{5}$ hands that have all four Aces.  For we must choose $5$ non-Aces from the $32$  non-Aces.  If you wish, you can multiply this by $\binom{4}{4}$, the number of ways to choose $4$ Aces from the $4$ available. But since $\binom{4}{4}=1$ that makes no numerical difference. 
A: Not sure if this is right but we want the number of ways to choose 4 non-distinct cards out of 36, divided by the total number of ways to choose 9 cards of out 36.
So $P(A) =  \binom{36}{4} / \binom{36}{9} \approx 0.000625$.
A: Far less elegant, but you can view the string of $18$ cards as an ordered list. Let $$P(n,k)=\frac{n!}{(n-k)!}.$$ Then there are $P(36,18)$ possible orders in which the $18$ cards may be taken ($36$ choices for the first card, $35$ for the second, and so on...). In any of these possible outcomes, we are interested in having $4$ aces and $14$ non-aces. There are $P(32,14)$ ways of picking the non-aces and $P(4,4)$ ways of picking the aces, respecting the order in which they are picked. Finally, there are $\binom{9}{4}$ positions in which the aces may lie, because they can be anywhere in the first player's hand. Our answer is therefore $$\binom{9}{4}\frac{P(4,4)P(32,14)}{P(36,18)}.$$ I'm not suggesting this is a particularly good method, but it is interesting to compare it to the others.
Note The $P$ in $P(n,k)$ is supposed to stand for the word 'pick' (as opposed to 'choose', which disregards the order of the selection) and is not to be confused with the probability.
