Problem Solving

Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?


8/33


62/165


17/33


103/165


25/33

Total number of ways to pick 4 cards from 12 unique cards: 12C4
= 12!/(8!*4!) = 495

there are 3 cases:
1. 2 pairs ex. 2,2,1,1
2. 1 pair, 2 individual ex. 4,4,5,3
3. 4 individual ex 1,4,5,6

For case one:
there are 6 pairs, so we will pick 2 of them: 6C2 = 6!/(4!x2!) = 30

We must divide this by 2! since there are two slots for the pairs and without doing this we are double counting solutions, (ex. 4,4,2,2 is the same as 2,2,4,4 so we divide by 2! to account for this).

30/2! = 15

For case two:
we know there are 6 ways to choose a pair, but we still have to pick 2 more cards. Regardless of which card is picked, it will have a matching pair. In order to avoid double counting errors, the fourth card can be picked 8 ways rather than 9.

10*8=80 and divide by 2! since, as in case one, alternate combinations are not considered alternate draws (ex. 2,2,4,1 is the same as 2,2,1,4).

=80/2!=40

so we are left with 6*40= 240 which is the number of card draws that contain exactly 1 pair.

We now have all the figures for each case of draws, so we compute the solution:

case 1 + case 2
240 + 15 = 255

divide the totals for cases 1 and 2 out of the total number of possible draws (12C4 = 495) , and you will have the answer:

255/495 = 17/33 which is answer C


alternatively as mentioned above by another guy whose name I don't have in front of me, 495 minus case 3 (all individual card draws) works too

here's a fast way to do case three:

(12 x 10 x 8 x 6)/4! = 240

I'm skipping numbers because I cannot pick a card's pair since I'm looking for the number of draws that have only individual cards. for example, if the first card drawn is 4, there are in fact 11 cards left, but one of them is another four which does not apply to the case I am trying to solve for, so remove that other four from my drawable cards, and am left with 10. this happens again with the second card, and so on.

I divide by 4! to reflect that for each draw, there are 4! ways to arrange the cards ( ex. 2,3,5,6 ->2,3,6,5 -> 2,5,3,6 -> etc). As illustrated here, regardless of the order, the draw is the same.

495 - 240 = 255 ; 255/495 = 17/33 which is answer C

Can anyone explain why this approach yields the wrong answer?

1st card is irrelevant. Let's say it's a 6.

2nd card has a 1/11 chance of being the other 6. Let's say it's a 5.

3rd card has a 2/10 chance of being either the other 6 or 5. Let's say it's a 4.

4th card has a 3/9 chance of being the other 6, 5 or 4.

Add together each draw probability - 1/11 + 1/5 + 1/3 = 15/165 + 33/165 + 55/165 = 103/165.

Ans D.

Can anyone explain why this approach yields the wrong answer?

1st card is irrelevant. Let's say it's a 6.

2nd card has a 1/11 chance of being the other 6. Let's say it's a 5.

3rd card has a 2/10 chance of being either the other 6 or 5. Let's say it's a 4.

4th card has a 3/9 chance of being the other 6, 5 or 4.

Add together each draw probability - 1/11 + 1/5 + 1/3 = 15/165 + 33/165 + 55/165 = 103/165.

Ans D.

is there a simple method to do this...

i m struggling...

raunekk wrote:is there a simple method to do this...

i m struggling...

Go with Brent Hanneson's way, that best explains the problem. It is not time consuming too.

Hi,
I got to the same answer as Brent but through a different method. Tell me whether this works for you guys.

It is easier to find out the chances of getting 4 card without any couple of them with the same value. This can be deducted from 1 to get the probability of at least one couple.

Card 1: can be any number at any suit. So we can chose 12 out of 12.
Card 2: can be any card from the remaining 11 besides the card, which we already hold its number. So it is 10 out of 11.
Card 3: can be any card from the remaining 10 besides the 2 cards, which we already hold the numbers. So it is 8 out of 10.
Card 4: the same idea, only now we have to deduct 3 invalid cards. So it is 6 out of 9.

Combining the option gives us:
Probability of no couples = 1*(10/11)*(8/10)*(6/9) = 16/33.
Now we can say that:
Probability of at least one couple = 1 - 16/33 = 17/33

1 2 3 4 5 6
1 2 3 4 5 6

So we can choose 12x10x8x6 ways but we need to divide it by 4!

= 240 ways ( NO PAIR )

C(12:4) =495

So : (495-240)/495

Both Shulapa and logitech are even faster at it. Good!

logitech wrote:1 2 3 4 5 6
1 2 3 4 5 6

So we can choose 12x10x8x6 ways but we need to divide it by 4!

= 240 ways ( NO PAIR )

C(12:4) =495

So : (495-240)/495

I like this method, however I am unsure of the 4! part. Why divide by 4!? Is it because there are 4 different cards you are choosing?

Why we have divided here by 4! ?

vertigo05 wrote:Why we have divided here by 4! ?

If we do not divide it by 4! here, then it would amount to the various arrangements (permutations) that we don't require. So, divide the permutations of 4 distinct objects by 4! to get the various selections (combinations), as required in the question.