I would like to know why my reasoning is wrong on this problem.
Given that the first draw is an ace, I think there would be 3/51 chances of getting two aces after I know that the first one is an ace.
It seems that the probability of this event is 1/33.
Where is my reasoning incorrect?
Edit: The problem is as follows. I have a standard deck of cards and I draw $2$ cards at random. I would like to compute the probability $
P( \text{Having two aces} | \text{ Have one ace} )$ and $P( \text{Having two aces} | \text{Having one ace of spades} )$. It seems, based on these lectures from Harvard: https://www.youtube.com/watch?v=JzDvVgNDxo8&list=PL2SOU6wwxB0uwwH80KTQ6ht66KWxbzTIo&index=5 at $9:30$ mark, that
$$
P( \text{Having two aces} | \text{ Have one ace} ) = 1/33
$$
while
$$
P( \text{Having two aces} | \text{Having one ace of spades} ) = 1/17
$$
I do not understand why my combinatorics is incorrect, even though I think I understand his solution.
Best Answer
The answer of $\frac{1}{33}$ which matches a problem which is worded similarly to yours is the answer to the following problem:
This is not the same problem as asking what the probability is that both are aces given that very specifically the first card drawn is an ace. The differences between these two problems are discussed at great length on the page for the Boy-Girl Paradox. This is just a different flavor of the same Boy-Girl problem described in the link, using cards and ranks rather than children and genders.
The answer to the problem where we asked for the probability given that the first card was an ace will indeed be $\frac{1}{17}$
The answer to the intended question with the answer of $\frac{1}{33}$ is found by direct calculation using definitions of conditional probability and will wind up being:
$Pr(\text{Two Aces}\mid \text{At least one ace in either position}) = \frac{Pr(\text{Two Aces})}{Pr(\text{At least one ace in either position})}$ (note how two aces is a subevent of having at least one ace in either position, so the intersection can be left out for the numerator)
Calculating $Pr(\text{Two aces})$ is straightforward as being $Pr(\text{First card is an ace})\times Pr(\text{Second card is an ace}\mid \text{First card is an ace}) = \frac{4}{52}\times\frac{3}{51}$
Calculating $Pr(\text{At least one ace})$ expands as $Pr(\text{First card is an ace})+Pr(\text{Second card is an ace}) - Pr(\text{Both cards are aces}) = \frac{4}{52}+\frac{4}{52}-\frac{4\times 3}{52\times 51}$
Putting these together we have:
$$\dfrac{\frac{4\times 3}{52\times 51}}{\frac{4}{52}+\frac{4}{52}-\frac{4\times 3}{52\times 51}}=\frac{1}{33}$$
As for the problem of $Pr(\text{Two aces}\mid \text{Ace of spades})$, we approach similarly as above, noting that the ace of spades might be in the first or the second position, yielding the final calculation of:
$$\dfrac{2\times \frac{1\times 3}{52\times 51}}{\frac{1}{52}+\frac{1}{52}-0}=\frac{1}{17}$$