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# Does anyone know this?

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From a standard deck of cards, the four Aces, four 2's, four 3's, and four 4's are drawn, forming a smaller deck of 16 cards. All 16 cards are dealt at random to four players, so that each player gets four cards. What is the probability that each player gets an Ace?

Feb 5, 2020
edited by AnimalMaster  Feb 5, 2020

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We can use a constructive approach, which isolates the unique probabilities for each card.  This gives us an answer of (1/16*c(16,1) + 1/15*c(16,2) + 1/14*c(16,3) + 1/13*c(16,4))/c(16,4) = 27/260.

Feb 5, 2020
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I'm not sure that this is correct, but:

There are 16 cards, 4 aces and 12 non-aces.

The first player must get 1 ace out of 4 and 3 non-aces out of 12:  4C· 12C3  =  880.

This is out of a total of choosing 4 cards out of 16 cards:  16C4  =  1820.

The probability that this occurs is:  880/1620.

There are now 12 cards, 3 aces and 9 non-aces.

The second player must get 1 ace out of 3 and 3 non-aces out of 9:  3C· 9C3  =  252

This is out of a total of choosing 4 cards out of 12 cards:  9C4  =  .495

The probability that this occurs is:  252/495.

There are now 8 cards, 2 aces and 6 non-aces.

The second player must get 1 ace out of 2 and 3 non-aces out of 6:  2C· 6C3  =  40

This is out of a total of choosing 4 cards out of 8 cards:  8C4  =  .70

The probability that this occurs is:  40/70.

There are now 4 cards, 1 ace and 3 non-aces.

The second player must get 1 ace out of 1 and 3 non-aces out of 3:  1C· 3C3  =  1

This is out of a total of choosing 4 cards out of 4 cards:  4C4  =  .1

The probability that this occurs is:  1/1

Now, we need to multiply these together:  (880/1620) x (252/495) x (40/70) x (1/1)  =  8 870 400 / 56 133 000.

Feb 6, 2020
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So 887000/56133000 is the answer? that's a long one

AnimalMaster  Feb 6, 2020
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From a standard deck of cards, the four Aces, four 2's, four 3's, and four 4's are drawn, forming a smaller deck of 16 cards. All 16 cards are dealt at random to four players so that each player gets four cards. What is the probability that each player gets an Ace?

I have not looked at Geno's answer, I just decided to think about it independently.

I certainly will not promise mine is correct.

There are   16C4*12C4*8C4*4C4 ways to choose the hands but I this will be double counting so I think I need to divide by

4!

=63063000/4!

=2627625   total selection of hands (maybe)

Take all the Aces out and give three cards to each

(12C3*9C3*6C3) /4!*4!  maybe = 369600

369600/2627625

369600/2627625 = 0.1406593406593407 = 64/455     (probably wrong)

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Tell us when you know the answer, preferably with working:)

Feb 10, 2020
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Nope! You're correct! Thanks Melody!

AnimalMaster  Feb 10, 2020
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Really!   I am astonished!

Melody  Feb 10, 2020