What if the question were a little different?

[bibhas.mondal]

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, for each value from 1 to 6, there are two cards in the deck with that 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 exactly two pair of cards that have the same value?

[gokul_nair1984]

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, for each value from 1 to 6, there are two cards in the deck with that 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 exactly two pair of cards that have the same value?

--What is the chance that Bill finds exactly two pair of cards that have the same value?--I understand that this is the part you changed.

Probability of choosing 4 cards out of 12= 12C4=495(This will be our denominator)

To find exactly 2 pairs:
Let the numbers be
1 2 3 4 5 6
1 2 3 4 5 6--------Let our case be a 3,3 and 5,5...

This can be done in the following manner,

Choosing 2 pairs out of 6 pairs: 6C2=30. Now divide this by 2! to avoid redundant pairs. ie;3,3,5,5 is the same as 5,5,3,3.
So 30/2=15
Therefore total probability of getting exactly 2 pairs:
15/495 =1/33

Makes Sense?

[bibhas.mondal]

Thanks Gokul. I reached the same answer but you reasoning "Now divide this by 2! to avoid redundant pairs. ie;3,3,5,5 is the same as 5,5,3,3.
So 30/2=15
" is not clear to me. I thought the answer would be 6C2/12C4 = 15/495=1/33.

[gokul_nair1984]

Thanks Gokul. I reached the same answer but you reasoning "Now divide this by 2! to avoid redundant pairs. ie;3,3,5,5 is the same as 5,5,3,3.
So 30/2=15
" is not clear to me. I thought the answer would be 6C2/12C4 = 15/495=1/33.

Sorry, there! I was thinking slot method but finally ended up with Combinations. Ya it should be 6C2/495. I was also analysing the slot method for selecting two pairs out of six. First can be chosen in 6 ways and the next in 5 ways.There you have to divide it by two 6*1*5*1/2!(due to repetition)

[rajivbhatia2007]

I did this question a different way and also got 1/33.

In the game, Bills picks four cards. For him to get two pairs:

First he picks up a card: Prob: 12/12 = 1

The second card must match the previous card or could be a new card, so imagine two branches of the probability tree with top and bottom branches:
top: 1/11 (matching previous card)
bottom 10/11 (another card)


For the third card:
top: prob: 10/10 = 1 (one pair made, bill picks up a new card)
bottom: prob: 2/10 (must match one of the two previous cards)

For the fourth card:
top: prob: 1/9 (must match the previous card)
bottom: prob: 1/9 (must match the one card that is not in the pair already).

Thus,
probability of top branch route = 1 x 1/11 x 1 x 1/9 = 1/99

probability of bottom branch route = 1 x 10/11 x 2/10 x 1/9 = 2/99

Adding both probability, we get 3/99 = 1/33

my website/gmat story:
http://gmathints.com/

[bibhas.mondal]

Thanks Rajiv. Great alternative explanation.

[tim]

looks good to me!