2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ There is nothing wrong with your logic. Are you sure that you copied the problem correctly? Are you sure that the proposed solution is $\frac{1}{33}$ rather than $\frac{1}{17}$ which is after all equal to $\frac{3}{51}$? $\endgroup$ – JMoravitz Jan 24 at 20:44
  • $\begingroup$ Also worth clarifying, are you drawing two cards in total or might you be intending to talk about higher numbers of cards being drawn? The problem of asking "what is the probability that when drawing two cards that both are aces given that the first was an ace" is a very different problem than asking "what is the probability that when drawing five cards that there are two aces given that the first was an ace" which is further different than asking "what is the probability that when drawing two cards that there are two aces given at least one of them (not necessarily first) is an ace?" $\endgroup$ – JMoravitz Jan 24 at 20:49
  • $\begingroup$ I have edited the question, hope it makes more sense now. Thank you. $\endgroup$ – OkThen Jan 24 at 20:52
6
$\begingroup$

The answer of $\frac{1}{33}$ which matches a problem which is worded similarly to yours is the answer to the following problem:

What is the probability that when drawing two cards from a deck without replacement that both cards are aces given that at least one of them (not necessarily the first one) is an ace?

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:

$$\dfrac{2\times \frac{1\times 3}{52\times 51}}{\frac{1}{52}+\frac{1}{52}-0}=\frac{1}{17}$$

$\endgroup$
3
$\begingroup$

Your own text says "Given that the first draw is an ace..." But the problem is "Given that at least one of the two cards is an Ace..." Can you see the difference?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.