2
$\begingroup$

You draw at random five cards from a standard deck of 52 cards. What is the probability that there is an ace among the five cards and a king or queen?

I want to do this by counting. There are 3 mutually exclusive events that need to be summed: ace and king with no queen, ace and queen with no king, ace and both.

Ace and king with no queen:

$$4 * 4 * {46 \choose 3} $$

One of 4 aces, one of 4 kings, 3 from the 46 other possible cards. Same thing for ace with queen and no kings.

Ace with both:

$$4 * 4 * 4 * {49 \choose 2} $$

One of 4 aces, one of 4 kings, one of 4 queens, 2 from the 49 cards remaining.

There are a total of $52 \choose 5$ ways to draw hands of 5 cards, so the answer is:

$$ \frac{2 * 4 * 4 * {46 \choose 3} + 4 * 4 * 4 * {49 \choose 2}}{52 \choose 5} = 0.215$$

...which is wrong. The answer is $0.1765$, which was obtained by taking $1 - P(A\cup B)$, where $A$ is no aces and $B$ is no king or queen. Where did I go wrong?

$\endgroup$
4
  • $\begingroup$ There is some overcounting in each computation. For example, for ace and king with no queen, it is still possible to draw more aces in the $\binom{46}{3}$ term. So that computation counts some hands (with multiple aces) several times. $\endgroup$
    – angryavian
    Commented Feb 17, 2019 at 4:19
  • $\begingroup$ Maybe I don't understand the question then, is more than one ace not acceptable? The answer seems to allow that possibility. $\endgroup$
    – badmax
    Commented Feb 17, 2019 at 4:21
  • 1
    $\begingroup$ @badmax: yes, a hand like AAK23 is acceptable, but you have counted it twice: once when the first ace is the required one and the second is one of the extra cards and a second time when the second ace is the required one. $\endgroup$
    – Ross Millikan
    Commented Feb 17, 2019 at 4:23
  • 2
    $\begingroup$ Yes more than one ace is acceptable. I'm just saying you count such hands multiple times. For example, "A of hearts, A of spades, K of hearts, 1 of diamonds, 2 of spades" is a single hand, but you count it twice in the computation $4 \cdot 4 \cdot \binom{46}{3}$: once when the first "4" chooses the A of hearts, and once when the first "4" chooses the A of spades $\endgroup$
    – angryavian
    Commented Feb 17, 2019 at 4:23

1 Answer 1

Reset to default
1
$\begingroup$

As users have pointed out in the comments, counting might complicate matters. You would have to first determine all the cases and then be very careful in your computations to avoid over-counting.

Here I present an alternate approach which, in my opinion, makes this question easier to tackle.

Consider the three events $A$,$B$,$C$ defined as follows:

$A-$ The $5$ cards picked contain at least one Ace

$B-$ The $5$ cards picked contain at least one King

$C-$ The $5$ cards picked contain at least one Queen

We are interested in determining $P(A \cap(B\cup C)) $.

Now, using De Morgan's Law: $$(A \cap(B\cup C))^c =A^c \cup(B\cup C)^c $$

Therefore, $$P(A \cap(B\cup C)) $$$$=1-P(A^c \cup(B\cup C)^c)$$

$$=1-\bigg(P(A^c)+P((B\cup C)^c)-P(A^c\cap(B\cup C)^c)\bigg)$$

$$=1-\bigg(\frac{{48 \choose 5}}{52\choose 5}+\frac{{44 \choose 5}}{52\choose 5}-\frac{{40 \choose 5}}{52\choose 5}\bigg)$$$$=0.1765$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .