<p>This is a hypergeometric distribution problem.  Luckily, calculators exist that let you just plug in the numbers and get an answer.  I used <a href="http://stattrek.com/online-calculator/hypergeometric.aspx" rel="nofollow">this one</a> with</p>

<p>Population size = 24 (the size of the deck)</p>

<p>Number of successes in population = 4 (the total number of kings)</p>

<p>Sample size = 5 (the number of cards in your hand)</p>

<p>Number of successes in sample = 3 (the number of kings you are interested in)</p>

<p>The results show that the chances of getting exactly 3 kings is 1.79%.  If you also include hands with 4 kings, it goes up to 1.84%.  </p>

CC BY-SA 3.0
2016-10-10T16:00:44.267Z
2016-10-10T16:00:44.267Z
bwarner
<p>You could also think of it like this, using combinations. Recall that a combination C(x, n) tells you the number of ways you could arrange n objects from a set of size x where order doesn't matter. So if you have 10 books, C(10, 3) tells you the number of ways you could select 3 of them.</p>

<p>There are 4 Kings in a Euchre deck, and 24 total cards. So there are C(4, 3) = 4 ways of choosing 3. Now since you need 2 other cards to finish out your hand, and you have already chosen from the 4 kings to leave 20 remaining cards, there are C(20, 2) = 190 ways to choose the remaining cards. </p>

<p>So there are 4 * 190 ways to form a hand of 3 Kings (and 2 other cards). There are C(24, 5) = 42504 total possible Euchre hands. So your probability is:</p>

<p>P = (4 * 190) / 42504 = 780 / 42504 = .01788 ~= 1.788% chance. </p>

CC BY-SA 3.0
2017-12-24T03:25:46.327Z
24-4=20. You count the fourth king twice. The probability is then 1.788% in agreement with the existing answer using a calculator.
Nij
CC BY-SA 3.0
2017-12-24T01:47:38.087Z
2017-12-24T03:59:49.407Z
2017-12-24T03:59:49.407Z
nsv
<p>In the game of euchre, the deck consists of the nine, ten, jack, queen, king and ace of each suit. Players are dealt a five card hand.</p>

<p>What is the probability that a player is dealt 3 kings in that hand of five cards?</p>

<probability><euchre>
What is the probability that a player is dealt 3 kings in a single euchre hand?
daniele
# What is the probability that a player is dealt 3 kings in a single euchre hand?

(Asked by: username_0 on 2016-10-10)

In the game of euchre, the deck consists of the nine, ten, jack, queen, king and ace of each suit. Players are dealt a five card hand.

What is the probability that a player is dealt 3 kings in that hand of five cards?

**Tags:** probability, euchre


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## 2 answers

### Answer by username_1 on 2016-10-10. Score: 3

This is a hypergeometric distribution problem. Luckily, calculators exist that let you just plug in the numbers and get an answer. I used [this one](http://stattrek.com/online-calculator/hypergeometric.aspx) with

Population size = 24 (the size of the deck)

Number of successes in population = 4 (the total number of kings)

Sample size = 5 (the number of cards in your hand)

Number of successes in sample = 3 (the number of kings you are interested in)

The results show that the chances of getting exactly 3 kings is 1.79%. If you also include hands with 4 kings, it goes up to 1.84%.


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### Answer by username_2 on 2017-12-24. Score: 2

You could also think of it like this, using combinations. Recall that a combination C(x, n) tells you the number of ways you could arrange n objects from a set of size x where order doesn't matter. So if you have 10 books, C(10, 3) tells you the number of ways you could select 3 of them.

There are 4 Kings in a Euchre deck, and 24 total cards. So there are C(4, 3) = 4 ways of choosing 3. Now since you need 2 other cards to finish out your hand, and you have already chosen from the 4 kings to leave 20 remaining cards, there are C(20, 2) = 190 ways to choose the remaining cards.

So there are 4 \* 190 ways to form a hand of 3 Kings (and 2 other cards). There are C(24, 5) = 42504 total possible Euchre hands. So your probability is:

P = (4 \* 190) / 42504 = 780 / 42504 = .01788 ~= 1.788% chance.

> **Comments:**
> **username_3**: 24-4=20. You count the fourth king twice. The probability is then 1.788% in agreement with the existing answer using a calculator.  

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