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!

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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.