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

Questions about how to figure out the odds of getting certain cards drawn from a deck are 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 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!

* It isn’t as easy as just adding step 1 and step 2. 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.

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 was kind of confuse with the deck of cards because I have no idea how many cards were in a deck. The blog was able to break down the cards into suits and that was really helpful in solving the problems.

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 ??

Hi, Tickerdy Boo,

If you’d like to post this in our forums, one of our mods will be happy to help :)

Stephanie

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.

Sam,

Unfortunately, time constraints prevent me from answering stats related questions on the comments section. But please ask for help on our forums — one of our moderators will be glad to help!

http://www.statisticshowto.com/forums/

Stephanie

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