Jmx wrote:QUESTION: 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?

ANSWER: 17/33 using (1-x) technic.

BUT calculating directly P I get 18/33: close but different! Where I am going wrong??

P = Total Desired Outcomes / Total Possible Outcomes

Total Possible Outcomes = Choice of 4 cards out of 12 = C(12,4) = 11x5x9 = 495

Total Desired Outcomes = Choice of 2 similar cards x Choice of 2 complementary cards (similar or not) = 6 x C(10,2) = 6x5x9 = 270

So, P = 6x5x9 / 11x5x9 = 6/11 = 18/33

Please help: where am I going wrong???

Many thanks!!!

The (1-x) technique sounds good (P = 1 - (1 x 10/11 x 8/10 x 6/9) = 17/33

For the other technique, the problem is in blue. You first determine the guaranteed pair and then choose two other cards. This allows for us to count the same combinations twice.

For example, our 2 similar cards can be 5 and 5, and then the 2 complementary cards can be 4 and 4. This would count as one outcome.
For example, our 2 similar cards can be 4 and 4, and then the 2 complementary cards can be 5 and 5. This would count as another outcome, even though it is the same as the first outcome.

So, the number 6x5x9 ia too large. We need to subtract all of the outcomes that were counted twice. These are all of the outcomes where we have two pairs of values (e.g. 4, 4, 5, 5)
In how many ways can we select two pairs of values?
We have 6 different numbers and we need to select 2 of them to be paired up.
This can be accomplished in C(6,2) ways (or 15 ways).

So, the answer is now [6x5x9 - 15]/11x5x9 = 17/33