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?

Guest Dec 7, 2019

#1**+1 **

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.

Guest Dec 7, 2019

#2**+2 **

**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

**(4 ^{4}) / (nCr(16,4))**, gives

**-------------------------------------------------**

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

GingerAle Dec 7, 2019