8
$\begingroup$

I got this question and answered it incorrectly. I haven't yet seen the correct answer. The possible answers were:

  1. $\frac{4}{52}$
  2. $\frac{16}{221}$ (my answer)
  3. $\frac{2}{52}$
  4. $\frac{32}{221}$

My reasoning is the following:

Event A: Card is not an ace.

Event B: Card is an ace.

$$P(A)\times P(B|A)=\frac{4}{52}\times \frac{48}{51}=\frac{16}{221}$$

This is assuming the cards are drawn sequentially. Drawing exactly one ace from a single draw is $4/52$ but to ensure that only a single ace was drawn one should consider the probability of not getting a second ace.

How am I wrong?

$\endgroup$
  • 4
    $\begingroup$ What if the first card is not an ace but the second is? $\endgroup$ – Ethan Bolker Dec 25 '17 at 16:21
  • 3
    $\begingroup$ If you are Lemmy, then in probability 1 you will draw the ace of spades. $\endgroup$ – Asaf Karagila Dec 26 '17 at 8:18
  • $\begingroup$ You can also do it in another way (as a check). The probability of two aces is $\frac{4}{52}\cdot\frac{3}{51}$. The probability of zero aces (that is two non-aces), is $\frac{48}{52}\cdot\frac{47}{51}$. So the answer should be $$1-\frac{4}{52}\cdot\frac{3}{51}-\frac{48}{52}\cdot\frac{47}{51}$$ $\endgroup$ – Jeppe Stig Nielsen Dec 26 '17 at 13:44
13
$\begingroup$

You forgot to assume $P(B)\times P(A|B)=\frac{48}{52}\times\frac4{51}=\frac{16}{221}$.

So $P(A)\times P(B|A)+P(B)\times P(A|B)= \frac{16}{221}+ \frac{16}{221}= \frac{32}{221}$.

The correct answer is $4$ , i.e. $\frac{32}{221}$

$\endgroup$
  • $\begingroup$ Why? Isn't it the same thing, just reversed? $\endgroup$ – Rivasa Dec 25 '17 at 21:02
  • $\begingroup$ @Annabelle, there are two possible ways you can get it, so you account for both. Thinking about it intuitively, because there aren’t many aces, you would expect your chances to be better by taking two cards than by taking one. Otherwise, if you kept taking more and more, your chances wouldn’t change, which would be strange when you got to 50 cards and are guaranteed to get two aces... $\endgroup$ – Jeff Dec 25 '17 at 21:12
  • 1
    $\begingroup$ @Annabelle: Consider that you have a deck of 3 cards, 2 aces. You draw two cards. What's the probability of drawing exactly one ace? Well, the event "drawing one" is equivalent to "an ace remains in the deck after drawing two cards". What is the probability of that? Clearly 2/3 as there are two aces in a deck of three if you imagine randomly picking a card to remain in the deck after drawing. But computing it as the OP did here you would end with P=1/3 because of mistakenly not considering that both drawing an ace first and drawing an ace second must be taken into account. $\endgroup$ – Jon Dec 25 '17 at 21:58
  • 1
    $\begingroup$ I think I get it now... @Jon $\endgroup$ – Rivasa Dec 26 '17 at 2:46
14
$\begingroup$

We have: $$P_{\text{ only one ace }} = P_{\text{ ace }1} P_{\text{ non-ace }2} + P_{\text{ non-ace }1} P_{\text{ ace }2} = \frac{4}{52}\times \frac{48}{51} + \frac{48}{52}\times \frac{4}{51} = \frac{32}{221}$$

$\endgroup$
  • $\begingroup$ Why the downvote? $\endgroup$ – Rohan Dec 26 '17 at 13:40
6
$\begingroup$

Answer 4 seems correct to me because:

$$P=\frac {\binom4 1\binom {48} 1}{\binom {52} 2}=\frac {32}{221}$$

$\endgroup$
3
$\begingroup$

You can figure this out without any knowledge of conditional probabilities:

There are 52 choices for the first pick and 51 choice for the second pick. So, in total, there are $52 \times 51 = 2652$ possible "hands" that consist of two cards.

Now, the question is, how many of these 2652 contain a single ace. Let's do some counting of the possible one-ace hands:

There are $4 \times 48 = 192$ hands that have an ace as their first pick and a non-ace as the second pick (because there are 4 aces and 48 non-aces). Similarly, there are $48 \times 4 = 192$ hands that have a non-ace as their first pick and an ace as the second pick. So the total number of hands that have a single ace is $192+192=384$.

So, the probability of a one-ace hand is $384/2652$. Dividing top and bottom by $12$ gives us $384/2652 = 32/221$.

$\endgroup$
  • $\begingroup$ Does the downvoter care to explain his or her reasoning? $\endgroup$ – bubba Dec 27 '17 at 2:51

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.