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Face cards in a regular set of playing cards consist of Jack, Queen, King, and Ace. With these 4 face cards occurring in all 4 suits (hearts, diamonds, clubs, and spades), that gives 16 total face cards.

If we draw 4 cards without putting them back, the probabilities are calculated on each draw as (number of face cards remaining)/(total number of cards remaining)

1st draw: 16/52 = 0.307692308

2nd draw: 15/51 = 0.294117647

3rd draw: 14/50 = .28

4th draw: 13/49 = 0.265306122

These are the individual probabilities of each step independent of each other. To calculate the probability when depending on the success of the previous step, we simply multiply each step's probabilities together.

Probability of drawing N face cards in a row:

1: (16/52) = 0.307692308

2: (16/52)*(15/51) = 0.0904977376

3: (16/52)*(15/51)*(14/50) = 0.0253393665

4: (16/52)*(15/51)*(14/50)*(13/49) = 0.00672268908

This translates to approximately a 7 in 1000 chance of drawing 4 face cards in a row out of a deck of 52 cards!

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14y ago
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13y ago

First, determine the probability of three sevens and two non-sevens, then worry about the permutations. The probability is the number of sevens, sevens minus one, sevens minus two, non-sevens, and non-sevens minus one; all over the number of cards left in the deck with each draw. The permutations of this configuration are 10 (you can verify this by arranging five cards as desired).

P(exactly three sevens) = 10*(4/52)(3/51)(2/50)(48/49)(47/48)

= (47)/(13*17*25*49) = 47/270,725

Those are pretty small odds, haha.

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13y ago

ORIGINAL POSTER: Okay, I kinda of see what you mean, but you answers reflect the probability of two events... I want to know the "overall probability..."

So if I weightedly average your two numbers, (weighting in the chances that each would occur) I get...

(12/102) x (20/26) + (11/102) x (6/26) = 3/26

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The 102 comes from the 2 in 1/2 multiplied by the 51 in 12/51 or 11/51.

(1/2) x (12/51) is (12/102)

(1/2) x (11/51) is (11/102)

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ORIGINAL POSTER: Sorry, I'm not quite sure where the 102 is from. I see the (12/51) and the (11/51), but for those I see that if each is multiplied by the probability of that event happening, I get my answer. Essentially:

(1/2)X( (12/51)X(20/26) + (11/51)X(6/26) ) = (3/26)

--------------------------

I left your original answer in, because I have not completely reviewed it.

The probability of drawing a black card is 1 in 2.

The probability of subsequently drawing a face card is 12 in 51, if the first card was not a face card, or 11 in 51, if the first card was a face card.

Simply multiply these two probabilities together, since they are sequentially interdependent, and you get 12 in 102 (0.1176) or 11 in 102 (0.1078).

-----------------------------

This post is by the original asker.

I know the probability of initially choosing a black card is (1/2).

I know the probability of initially choosing a black face card is (6/52).

I know the probability of initially choosing a black non-face card is (20/52).

---

So I figure it can be said that given a black card has been removed, I have (on "average")...

Black face cards: (20/26)X(6) + (6/26)X(5) = (150/26) = 5.7692 cards

Red face cards: 6 cards

Face cards: 6 + (150/26) = (153/13)

So, from there, I can see that my probability, on "average" of choosing a face card from a deck of 51 cards, with one black card removed is (153/13)/51.

This simplifies to... (3/13)

Now, of course, I have to multiply by (1/2) because of the initial probability of picking the black card in the first place. So...

(3/13)X(1/2) = (3/26)

Therefore, I've determined that the probability of drawing a face card immediately after drawing a black card is (3/26), or approximately 11.54%.

HOWEVER:

The probability of choosing a black face card from a deck of 52 cards is (6/52) or (3/26), the same as my answer. Am I going in circles, or is this just a coincidence? I have tried answering this question using several methods, but this one makes the most logical sense. Your answers or input are greatly appreciated.

--

TheBrink

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12y ago

In a standard 52 card deck, there are 13 clubs and 12 face cards, three of which are clubs, so there are 22 (13 + 12 - 3) different ways of drawing a club or a face card. This is a probability of 22 in 52, or 11 in 26, or about 0.4231.

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Q: Four cards are drawn at random from a pack of 52 playing cards Find the probability of getting all face cards?
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