Probability and Statistics > Probability > Probability of picking from a deck of cards

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## Probability of picking from a deck of cards: Overview

Questions about how to figure out the probability of picking from a deck of cards common in **basic stats** courses. For example, the **probability** of choosing one card, and getting a certain number card (e.g. a 7) or one from a certain suit (e.g. a club).

*You might wonder why you’re learning about cards (what’s the point?). The answer is that finding probabilities (like the probability of contracting an illness) can be a tricky concept to grasp at first. So your instructor will try and simplify problems using cards, dice or Bingo numbers. Once you’ve grasped the basics, you’ll start to use “real life” data for probability (usually a bit later on in the class, for example in normal distributions).*

Here’s how to find the probability of picking something in a couple of simple steps.

## Probability of picking from a deck of cards: Steps

**Step 1:***figure out the total number of cards you might pull.*

Write down all the possible cards and mark the ones that you would pull out (in our case we’ve been asked the probability of a club**or**a seven so we’re going to mark all the clubs and all the sevens):- hearts: 2, 3, 4, 5, 6,
**7**, 8, 9, 10, j, q, k, A - clubs:
**2, 3, 4, 5, 6, 7, 8, 9, 10, j, q, k, A** - spades: 2, 3, 4, 5, 6,
**7**, 8, 9, 10, j, q, k, A - diamonds: 2, 3, 4, 5, 6,
**7**, 8, 9, 10, j, q, k, A

This totals

**16**cards. *- hearts: 2, 3, 4, 5, 6,
**Step 2:***Count the total number of cards in the deck(s)*. We have one deck, so the total =**52****Step 3:***Write the answer as a fraction*. Divide step 3 by Step 4:

**16 / 52**

That’s it!

**Tip**: It isn’t as easy as just adding the number of sevens (4) and the number of clubs (13). If you did this for this example, you’d get 17 cards, not the correct answer of 16. The reason for this is that one of the cards in our example is both a club AND a number 7.

## Probability of picking from a deck of cards: Using Excel

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It gets a **LOT **more complex if you’re playing a card game, you have a certain number of cards in your hand, and you want to know your odds of getting a certain card if you are drawing a certain number of cards. You have to use something called a **hypergeometric distribution** to figure out the odds. The formula is:

H (n) = C (X, n) * C (Y – X, Z – n) / C (Y, Z)

Where:

X is the number of a certain card in the deck

Y is the total number of cards in the deck

Z is the number of cards drawn

N is the number you are checking for

As you can see, the formula uses combinations and factorials — it can get a bit messy to do this by hand, so consider using technology like Excel. The command in Excel is: “=HYPGEOMDIST(N,Z,X,Y)”. For example, if you have a standard 52 card deck and draw 4 cards, what will be your chances of **not **drawing an ace?

X is 4

Y is 52

Z is 4

N is 0 (as you want zero aces!)

the formula would be:

=HYPGEOMDIST(0,4,4,52) you will get the chance for **not **drawing the card.

Like the explanation? Check out the Practically Cheating Statistics Handbook, which has hundreds more step-by-step solutions, just like this one!

**Next**:

this helps make perfect sense. When working problems, I would always come up with the wrong answer. I was not taking into account that one of the cards was 2 of the problems. Now, I understand that you must subtract an extra card in the equation.

I have the hardest time with this probability for some reason I get lost in the process and then I feel like I am doing it right and come to find out its all wrong every step. this really didn’t help me as this being an example.

1 is not a card in the deck

You are correct! I got carried away typing out all those numbers. Thanks for the correction :)

This happened to me and I have always wondered how to calculate the odds.

Take a 52 card deck and layout the game of Solitaire, then go through the remaining deck (each card vs every 3 cards) and not be able to make a single move of any kind. How does one calculate the odds for this scenario?

What is the prob of 1-12numbers cumming out 8times in a row on a roulette wheel ??

How to calculate say from a deck of 52 cards taking only 3 cards and the total points is 1 or 2-9 points assuming 10 is zero and J,Q and K is zero as well.

I found this very clear. I was wondering.

There’s an online game where, play starts from beint four cards from a deck.Your deck can have no fewer than eight cards, though iit can have more according to certain requirements my question involves a deck of eight.

The deck contains eight distinct cards. I’ll call them ABCDEFGH. I was wondering how you could figure the probability of getting say A, A and B, A and B and C, A and B and C and D, A or B, A or B or C, and A or B or C or D.Also the probability of the events not happening.

So I tried to figure out how many distinct combinations I could get. The cards may be played in any order, so I know I’m not concerned with permutations.( 8!=40,320 for the deck and 8!/ 4! = 1,680 for the premuations of possible four card hands. )

When I tried figuring out the combinations,I feel conused. 8!/ 4!/ 4!= 70. So I figure the probability of A being drawn and not drawn is 1/2 or 35/70 .Though how should I figure out the rest? Sorry, if this is too off-topic.

I meant A or B or C or D being drawn not A.

Hi, Hannah,

Thanks for visiting…perhaps one of our visitors could answer this?

Stephanie

Odds? I take a deck and start turning over cards, one at a time. Calling out the number..so the first card is an ace, second is 2 etc. jack is 11, queen 12, king 13 then back to ace.

I can’t get through whole deck without hitting the same card as the number. I.e. The fifth card is a 5. Or the 11th card is a jack.

What are the odds/chances of getting through the deck without hitting the same card?

I tried many times, but never made it.

The odds are practically nil

You have a 4/52 = 1/13 chance of losing on the first draw (an ace comes up).

On the second draw, you have a 1/13 chance of turning a two (assuming you didn’t pick a 2 in round 1). If you did pick a 2 the odds still aren’t good (3/52).

When you go down the list like this, you should be able to see there’s practically a zil chance of making it to the end.

If you want to get an algorithm, it might be easier to start with 12 cards to see the pattern more clearly (I don’t know of an algorithm for this!)

Find the probability that five cards chosen at random from a deck of 52 will contain:

a) no aces

b) atleast one ace

c) exactly 1 ace

Where do you get stuck?

Pick four cards from a regular deck without replacement. What is the probability that you get exactly two hearts at the first and fourth cards???????????????????

Probability of that first heart is 1/13.

As you don’t replace, probability of a heart on pick two is 12/51.

The probability of getting a heart on round three is 12/50.

The probability of getting a heart on round four is 12/49.

So, total probability = 1/13 * 12/49

In the example, where X=4, Y= 52, Z=4,& N=0, I understand everything except why does X=4?

X is the number of a certain card in the deck

So, there are 4 aces.