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I have a collection of sixteen cards. Each card is rose, amber, emerald, or liberty (a shade of blue) . Also, each card is labelled A, B, C, or D. There is a card for each color-letter combination, e.g. there is an amber card that has a B on it.

These sixteen cards are dealt at random to four people, and each person gets four cards. What is the probability that each person gets a card with a letter A on it?

 Dec 7, 2019
 #1
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I  have seen this question posted several times in the last week or so and nobody seemed to have answered it.

Now, I am no good at answering probability questions, but I will give it a go and somebody should ckeck it one way or another!

 

You have 16 cards with 4 different colors and, in addition, each color is labled with 4 letters - A, B, C, D. These 16 cards are dealt to 4 different people randomly. So, the obvious question is: in how many way can these 16 cards be dealt to 4 people? Well, to my limited undestanding, there are: 16 C 4 =1,820 ways to do this. One of those 1,820 ways must have 4 "As" in it!!. So, the probability that each person gets a card with the letter "A" on it is simply: 1 / 1, 820 ? Somebody should check this !!.

 

Another way of looking at it is as follows:

The probability of the first person getting a card with the letter A on it is: 4/16

The probability of the second person getting a card with the letter A on it is: 3 / 15

The probability of the third person getting a card with the letter A on it is: 2 / 14

The probability of the fourth person getting a card with the letter A on it is: 1 / 13

So, the overall probability that each person gets a card with the letter A on it is:

[4 / 16]  x  [3 /15]  x [2 / 14]  x [ 1 / 13] =24 / 43,680 =1 / 1,820

 

Does this make sense? Somebody little more versed in this kind of probability should take a good look at it and point out the fallacy of my naive approach !!. Thanks.

 Dec 7, 2019
 #2
avatar+1852 
+1

Solution:

One method for visualizing a solution is to view the cards in a 4 X 4 matrix, where the columns represent the players and the rows represent the cards in the players' hands. Valid conditions for success occur when the players have single “A” cards located in any one of the four positions of their respective hands.

 

\(\left[ {\begin{array}{cccc} A & A & A & A \\ X & X & X & X \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {First arrangement of “A} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{cccc} A & A & A & X \\ X & X & X & A \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{cccc} A & A & X & A \\ X & X & A & X \\ X & X & X & X \\ X & X & X & X \\ \end{array} } \right] \text {Fifth arrangement} \\ \left[ {\begin{array}{cccc} X & X & X & X \\ X & X & X & X \\ X & X & X & X \\ A & A & A & A \\ \end{array} } \right] \text {Last arrangement} \)

 

 

From this, it’s easy to see there are (4^4) = 256 arrangements of success.

 

Dividing the number of successes by the total number of combinations

(44) / (nCr(16,4)), gives (64/445) ≈ 0.1406593.

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

Here are two other solution methods for this question:

 

https://web2.0calc.com/questions/probability-help_11#r5

Here, Melody presents the total arrangements without the target cards then adds the arrangements of the target cards.

 

https://web2.0calc.com/questions/probability-help_11#r13

Here, Rom presents the arrangements of each hand without the target cards then adds the arrangements of the target cards to each hand in sequence.

 

 

GA

 Dec 7, 2019
edited by GingerAle  Dec 7, 2019
 #3
avatar+108732 
0

Thanks Ginger and guest

 

I think I am still being put above a cat.  wink 

 Dec 9, 2019
edited by Melody  Dec 9, 2019
edited by Melody  Dec 9, 2019

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