Probability of dealing 4 cards from pack of 52 cards

In summary: Your 12, 11, 10 would be for picking the same suit, because there are 12, 11, 10 left.For the same value, there are 3, 2, 1 left.
  • #1
something42
2
0

Homework Statement


The question is:
Four cards are dealt from a well-shuffled pack of 52 cards.
What is the probability that all four cards bear the same face value?

Homework Equations


n/a


The Attempt at a Solution


My guess is that first of all, I pick a card out of random, 1/52. Then after that, I have 12 chances of picking the same face value out of 51 cards, so that is 12/52. Then I continue this for the other cards, so I get:

1/52 + 12/51 + 11/50 + 10/49 = 0.68

The thing is, I can't help but think that's a bit high. Have I missed something glaringly obvious, or have I done it right?
 
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  • #2
Welcome to PF!

Hi something42! Welcome to PF! :smile:
something42 said:
The thing is, I can't help but think that's a bit high. Have I missed something glaringly obvious, or have I done it right?

You want the joint probability (of independent events), so you need to multiply, not add. :wink:
 
  • #3
It depends on whether you specify the value of the card before you start pulling them from the pack.

If you don't specify the card then the odds are 1 * 3/51 *2/50 * 1/49 (the 1 at the start is because you must pull a card from the pack)

If you do specify the value the odds become 4/52 * 3/51 *2/50 *1/49
 
  • #4


Hey guys! Thanks for taking the time to answer my question!

tiny-tim said:
Hi something42! Welcome to PF! :smile:
Yay! Thanks! :blushing:

tiny-tim said:
You want the joint probability (of independent events), so you need to multiply, not add. :wink:
You know, after I posted this, I knew it was the addition that gave me problems. However, when I saw the results when I multiplied them afterwards, it gave me a shock as to how low the value was, so I didn't think that was appropriate. But wow! That low? :eek:

Thanks for the tip, though! In my moment of hair pulling, it helps to get some confirmation. :D

Jobrag said:
It depends on whether you specify the value of the card before you start pulling them from the pack.

If you don't specify the card then the odds are 1 * 3/51 *2/50 * 1/49 (the 1 at the start is because you must pull a card from the pack)

If you do specify the value the odds become 4/52 * 3/51 *2/50 *1/49

Thanks for taking the time! If you don't mind, I want to "challenge" your post.

First of all, the 1 at the start completely makes sense. I don't know why I didn't think of this sooner. So I guess that answers your question of the card being specified or not (from what I've seen in the question, it doesn't matter).

However, I don't get why you have 3/51, 2/50, 1/49. I always thought the definition of

probability of E = number of sample points in E/total number of sample posts

and considering that, if the card pack gets less and less for a suit (because that's what I'm reading it as), surely we take into account the number of available suits, and not the amount of cards currently needing to be picked?

Sorry about the "challenge", but I really want to get this stats stuff right. It seems so simple, yet if I got it wrong, I'd be constantly slapping my head for the rest of my life :(

But thanks everyone who has posted :D
 
  • #5
Hi something42! :smile:
something42 said:
First of all, the 1 at the start completely makes sense. I don't know why I didn't think of this sooner. So I guess that answers your question of the card being specified or not (from what I've seen in the question, it doesn't matter).

However, I don't get why you have 3/51, 2/50, 1/49. I always thought the definition of

probability of E = number of sample points in E/total number of sample posts

and considering that, if the card pack gets less and less for a suit (because that's what I'm reading it as), surely we take into account the number of available suits, and not the amount of cards currently needing to be picked?

Sorry about the "challenge", but I really want to get this stats stuff right. It seems so simple, yet if I got it wrong, I'd be constantly slapping my head for the rest of my life :(

But thanks everyone who has posted :D

Jobrag :smile: is right …

your 12, 11, 10 would be for picking the same suit, because there are 12, 11, 10 left.

For the same value, there are 3, 2, 1 left. :wink:
 

1. What is the probability of getting a specific card from a deck of 52 cards?

The probability of getting a specific card from a deck of 52 cards is 1 in 52, or about 1.92%. This is because there are 52 cards in a deck and each card has an equal chance of being chosen.

2. What is the probability of getting 4 aces in a row when dealing 4 cards from a deck of 52 cards?

The probability of getting 4 aces in a row when dealing 4 cards from a deck of 52 cards is (4/52) * (3/51) * (2/50) * (1/49) = 1/270725, or about 0.00037%. This is because the probability of getting an ace on the first draw is 4/52, on the second draw is 3/51, on the third draw is 2/50, and on the fourth draw is 1/49. These probabilities are multiplied together because each draw is independent.

3. What is the probability of getting at least one face card when dealing 4 cards from a deck of 52 cards?

The probability of getting at least one face card when dealing 4 cards from a deck of 52 cards is 1 - (40/52) * (39/51) * (38/50) * (37/49) = 0.597, or about 59.7%. This is because the probability of not getting a face card on the first draw is 40/52, on the second draw is 39/51, on the third draw is 38/50, and on the fourth draw is 37/49. To get the probability of getting at least one face card, we subtract this probability from 1.

4. What is the probability of getting a pair when dealing 4 cards from a deck of 52 cards?

The probability of getting a pair when dealing 4 cards from a deck of 52 cards is (13 * C(4,2)) * (48/52) * (44/51) * (40/50) = 0.422, or about 42.2%. This is because there are 13 possible pairs in a deck of 52 cards (2 of each rank), and there are C(4,2) ways to choose 2 of the 4 cards to make the pair. The remaining 3 cards can be any of the 48 cards that are not part of the pair, and each subsequent draw has a decreasing number of available cards to choose from.

5. What is the probability of getting a flush when dealing 4 cards from a deck of 52 cards?

The probability of getting a flush when dealing 4 cards from a deck of 52 cards is (4 * C(13,4)) * (48/52) = 0.002, or about 0.2%. This is because there are 4 possible suits for the first card, and C(13,4) ways to choose 4 cards of the same suit. The remaining 3 cards must also be of the same suit, and there are 48 remaining cards in the deck to choose from that suit.

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