A card counting problem in which order matters

I can't seem to figure out the correct solution to this problem I'm having. Out of a deck of 40 cards, 4 are selected. What is the probability that the third card is the first ace chosen?

Note that the deck has 40 cards, 4 suits, and 10 denominations (A-10)

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What can be the first card? Anything but ace: 36 good cases out of 40 possibilities.

The 2nd one? Anything but ace, and neither the first one: 35 good cases out of 39 possibilities.

The 3rd one? Ace: 4 good cases out of 38 possibilities [cards left]

The 4th one? all the 37 possibilities are good.

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This is what I thought. And it turns out this is right. I was just looking at the wrong question number in the answer key. My bad. But thanks for the help – user979616 Sep 23 '12 at 20:43
@user979616: That’s the right answer; why do you think that it’s wrong? – Brian M. Scott Sep 23 '12 at 20:45

Take the event $A$ to be an Ace is chosen on the third card, $C_1,C_2$ to be something other than an ace for card $1$ and $2$. You seek $\mathbb{P}(A\cap C_1\cap C_2)$. Using conditionnal probabilities you find \begin{align} \mathbb{P}(A\cap C_1\cap C_2)&=\mathbb{P}(C_1)\mathbb{P}(C_2 | C_1)\mathbb{P}(A|C_1\cap C_2)\\ & =\frac{36}{40}\times\frac{35}{39}\times \frac{4}{38}\\ &=\frac{21}{247} \end{align} Note that the fourth card is not important here

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I assume that the first ace chosen means that the third card is the first ace.

We pick the first card. 36 from 40 are not an ace. We pick the next card. 35 from 39 are not an ace. We pick the third card. 4 cards from 38 are an ace. The fourth card doesn't matter. P = 36/40 * 35/39 * 4/38 = 5040/59280 = 0.085 = 8.5%

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