# 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: