6
$\begingroup$

Question: What is the probability you draw Jack of Hearts in a hand of $5$ cards?

Assume you have a deck with with $52$ cards ($4$ suits of $13$ cards: numbers $1\ldots 9$, and faces J, Q, K).

My way of thinking is the following:

$$\frac{\left(\dfrac{1\cdot51\cdot50\cdot49\cdot48}{4!}\right)}{ \left(\dfrac{52\cdot51\cdot50\cdot49\cdot48}{5!}\right)}$$

$1$ is for the Jack of Hearts being drawn, and then $51\ldots48$ for the rest of the $4$ cards

$\endgroup$

4 Answers 4

Reset to default
6
$\begingroup$

Your thinking is correct, though let me provide another way of looking at the problem that might make its structure clearer:

  • There are ${51\choose 4}$ ways to pick a hand that include the Jack of Hearts (because once we've picked the Jack, we can choose 4 other cards from the remaining 51)
  • There are ${52\choose 5}$ ways to pick a hand, with no restrictions.

Therefore the probability of getting a hand with the Jack of Hearts is

$$\frac{51\choose 4}{52\choose 5} = \frac{51!5!47!}{4!47!52!} = \frac{5}{52}$$

You can check that the obvious generalization is, in fact, true: the probability of drawing a particular card in a hand of $m$ cards with a deck of size $n$ is $m/n$.

$\endgroup$
6
$\begingroup$

Alternatively, there are $\binom{51}{5}$ ways of picking a hand that does not have the Jack of Hearts. There are a total of $\binom{52}{5}$ ways of picking $5$ cards, so the probability of choosing a hand with the Jack of Hearts is: $$1 - \frac{\binom{51}{5}}{\binom{52}{5}} = 1-\frac{47}{52} = \frac{5}{52}$$

$\endgroup$
1
  • 1
    $\begingroup$ Although a bit more complicated, this is valid (+1) $\endgroup$
    – robjohn ♦
    Mar 5, 2014 at 23:42
4
$\begingroup$

The probability you draw the jack of hearts is the same as the probability of drawing any other particular card. Since you draw 5 cards, the 52 individual probabilities have to add up to 5, so each probability is 5/52. In particular, the probability of drawing the jack of hearts is 5/52.

$\endgroup$
0
$\begingroup$ 
5/52 seems wrong to me.  

I suggest the slightly higher probability of:
                          n = (1/52 + 1/51 + 1/50 + 1/49 + 1/48)

Which approximates to:
                          n = 5/50

Each time a cards is picked the deck gets smaller, and the probability of 
picking the "good" card the next round increases.


5/52 would be the probability ONLY if after each time you drew one card, 
you replaced it into the deck before the next draw.
$\endgroup$
3
  • 1
    $\begingroup$ With replacement, the probability would be $1-\left(\frac{51}{52}\right)^5$ $\endgroup$
    – robjohn ♦
    Mar 5, 2014 at 23:39
  • $\begingroup$ No, I think the answer is right, as using hypergeometric distribution, which is the way you solve these sorts of problems, you get the answer 5/52 $\endgroup$
    – R3FL3CT
    Jan 5, 2022 at 16:28
  • $\begingroup$ Also, you're forgetting to account for the probability that you don't get the card, so it would actually be n = ( 1/52 + ( 51/52 * 1/51 ) + ( 51/52 * 50/51 * 49/50 ) ) etc. So it actually is 5/52 $\endgroup$
    – R3FL3CT
    Jan 6, 2022 at 13:46

You must log in to answer this question.

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