There's a deck of cards (52 cards total). When drawing two cards from the deck, what is the probability of both the cards not being kings?

So, here's my line of thought. The chance of NOT getting a king is $\frac{48}{52}$ (because there are 4 kings). Then, the chance of, when removing 2 cards, not getting any king, would be $\frac{48}{52} \times \frac{48}{52}$, because of this rule:

$$P(A\rm{\,and\,}B)=P(A\cap B)=P(A)P(B),$$

I got this exercise on a book and on the Answers, it says it's $\frac{47}{221}$. That's $\sim21%$, is there a chance the answers are wrong, because when I think of it, the probability should be much higher (like the one I got, $\sim85%$).

Notes: The first card is not replaced. There are 4 kings, the rest of the cards shouldn't really matter.

  • $\begingroup$ Both what cards? All you've said you have is a deck of cards. $\endgroup$ Oct 19 '12 at 18:01
  • $\begingroup$ I edited it, although I had already said there were 4 kings :) $\endgroup$ Oct 19 '12 at 18:03
  • $\begingroup$ The first card's not a king with probability $12/13$. Given that the first card wasn't a king, then the second card's not a king with probability $47/51$. The product is $188/221=0.8506787...$. $\endgroup$
    – mjqxxxx
    Oct 19 '12 at 18:03
  • $\begingroup$ The bit about the card not being replaced is important. Also, the rule $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ is only true if the events in question are independent, which they're not here - if you've removed a king from the packet, then it's going to be less likely that the next card you take out will be a king. $\endgroup$ Oct 19 '12 at 18:05
  • 2
    $\begingroup$ @user996056: You haven't in any way addressed my question. You have a deck of cards, then you're suddenly talking about "both the cards". What cards are these? Are they two cards drawn at random from the deck with replacement? Without replacement? Something else entirely? You haven't even said explicitly they're drawn randomly. You need to tell us. $\endgroup$ Oct 19 '12 at 18:06

Since the draws are made without replacement, the probability that the first card is not a king is, as you say, $\frac{48}{52}$. However, the probability that the second card is not a king, given that the first card is not a king, is $\frac{47}{51}$: there are only $51$ cards left in the deck, of which $47$ are not kings. The correct answer is therefore $$\frac{48}{52}\cdot\frac{47}{51}=\frac{12}{13}\cdot\frac{47}{51}=\frac4{13}\cdot\frac{47}{17}=\frac{188}{221}\approx 0.85068\;.$$

Alternatively, you can look at it like this: there are $\binom{52}2$ pairs of cards, of which $\binom{48}2$ contain no king, so the probability of drawing a pair with no king is


exactly as before.

  • $\begingroup$ Downvoting without leaving an explanation is singular unhelpful $-$ especially when the answer is correct. $\endgroup$ Oct 20 '12 at 9:19

The rule:

$$\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$$

is in fact incorrect. It only gives the right answer when $A$ and $B$ are independent. The correct relation is

$$\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B|A)$$

Where $\mathbb{P}(B|A)$ is the probability that $B$ happens, given that $A$ has already happened.

So, to start off with, you are drawing one card, which is not a king with probability $\frac{48}{52}$.

But now you want to find the probability that you don't get a king, given that you didn't get a king the first time. If you've already taken out a card, then there are now $51$ cards, $4$ of which are kings. So the probability is $\frac{47}{51}$. So the answer is $\frac{48}{52}\times\frac{47}{51}$.


Hint: Use the hypergeometric distribution to find the probability of getting two kings without replacement. Then the probability of NOT $\ldots$


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