Probability of getting the third ace in the j'th draw

Question

So the problem is the following:

Given a complete deck of 52 cards, 9 cards are drawn one by one with reposition. What is the probability that in a hand (of 9 cards) in which 3 aces come out, the 3rd will come out at the j'th draw?

My answer:

I was able to reach the correct answer (according to the solutions to the exercise) for the 9'th draw, which is

$$\binom{8}{2}\left(\frac{4}{52}\right)^3\left(\frac{48}{52}\right)^6$$

My thinking being: Probability of getting 2 aces in the first 8 cards times the probability of getting an ace in the 9'th (and last) card.

For the j'th draw I ended up with the following answer:

$$\binom{j-1}{2}\left(\frac{4}{52}\right)^3\left(\frac{48}{52}\right)^6$$

The thinking being very similar to the one I used before. According to the solution this is wrong. The correct answer should be:

$$\binom{j-1}{2}\left(\frac{4}{52}\right)^3\left(\frac{48}{52}\right)^{j-3}$$

Where did I go wrong? Did they actually mean "at least 3 aces"?

Answer 1

Since there is exactly 3 aces in the 9 cards you can calculate the number of different positions for the aces, i.e.

AAA------  // ace in position 1, 2, 3
AA-A-----  // ace in position 1, 2, 4
A-AA-----  // ace in position 1, 3, 4
-AAA-----  // ace in position 2, 3, 4
AA--A----  // ace in position 1, 2, 5
... etc ...

That is $$\binom{9}{3} = 84$$

combinations in total.

The last ace can be in position 3, 4, ... 9

If the last ace is in position 3 there need to be exactly 2 aces in the 2 positions before position 3. Now calculate in how many ways that can be done:

$$\binom{2}{2} = 1$$

If the last ace is in position 4 there need to be exactly 2 aces in the 3 positions before position 4. Now calculate in how many ways that can be done:

$$\binom{3}{2} = 3$$

If the last ace is in position 5 there need to be exactly 2 aces in the 4 positions before position 5. Now calculate in how many ways that can be done:

$$\binom{4}{2} = 6$$

And so on...

This gives

Last ace in position 3 :  1 combination  --> probabilty  1/84
Last ace in position 4 :  3 combinations --> probabilty  3/84
Last ace in position 5 :  6 combinations --> probabilty  6/84
Last ace in position 6 : 10 combinations --> probabilty 10/84
Last ace in position 7 : 15 combinations --> probabilty 15/84
Last ace in position 8 : 21 combinations --> probabilty 21/84
Last ace in position 9 : 28 combinations --> probabilty 28/84

So in general

$$\binom{j - 1}{2} / 84$$