+0

# Plz help probability

+2
45
6
+138

You take the four Aces, four 2's, and four 3's from a standard deck of 52 cards, forming a set of 12 cards. You then deal all 12 cards at random to four players, so that each player gets three cards. What is the probability that each player gets an Ace, a 2, and a 3?

Jul 20, 2020

#3
+138
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plz help!

Jul 21, 2020
#5
0

The probability is (1/4*1/3*1/2)^4*4! = 1/13824.

Jul 26, 2020
#6
+110089
+1

Not sure but maybe ....  I admit that I really do not know what I am doing.

If you get the answer, with or without the logic, please share it here.

Take the Aces and hand them out 4! ways

Take the 2s and hand them out 4! ways

Take the 3s and hand them out 4! ways

4!*4!*4!

Or line all the cards up and hand them out in order  12! ways  I think I need to divide this by 3!3!3!3! = 6^4 = 1296

$$\frac{4!*4!*4!*1296}{12!}\\=\frac{4!*4!*1296}{5*6*7*8*9*10*11*12}\\ =\frac{24*24*1296}{5*6*7*8*9*10*11*12}\\ =\frac{3*2*1296}{5*6*7*9*10*11}\\ =\frac{*1296}{5*7*9*10*11}\\ =\frac{1296}{34650}\\ =\frac{72}{1925}\\ \approx 0.037$$

LaTex

\frac{4!*4!*4!*1296}{12!}\\=\frac{4!*4!*1296}{5*6*7*8*9*10*11*12}\\ =\frac{24*24*1296}{5*6*7*8*9*10*11*12}\\ =\frac{3*2*1296}{5*6*7*9*10*11}\\ =\frac{*1296}{5*7*9*10*11}\\ =\frac{1296}{34650}\\
=\frac{72}{1925}\\
\approx 0.037

Jul 27, 2020
edited by Melody  Jul 27, 2020
edited by Melody  Jul 27, 2020
edited by Melody  Jul 27, 2020