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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 !!

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    $\begingroup$ You can simplify this problem greatly if you ignore your friend altogether. The probability you ask about equals the probability that, just yourself drawing 9 cards, you get all 4 aces. $\endgroup$ – Lee Mosher Jul 31 '14 at 18:51
  • $\begingroup$ @LeeMosher That's 2/935. I'd thought that, because it's like if I continue picking cards after having made one 9-card hand (since the deck doesn't know if just one person is playing or there are more). But I still can't make sense the 'impact' other player has over the final probability: Does it mean that if there are 2 player or more, all of them have a 2/935 of getting 4 aces in their own hand, right? $\endgroup$ – user163665 Jul 31 '14 at 21:57
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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.

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  • $\begingroup$ Yep. In fact in that way I obtain the same answer as if I were playing cards alone and I'd want to find this probability in this situation, which makes sense since the deck of card doesn't know if the person's hand picking cards is the same all the time or not. Thanks !! $\endgroup$ – user163665 Aug 1 '14 at 14:38
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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.

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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 $P(A) = \binom{36}{4} / \binom{36}{9} \approx 0.000625$.

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