Thread: combinations/probability question please help!!

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 at least one pair of cards that have the same value?
8/33
62/165
17/33
103/165
25/33

this is my solution but it is wrong:

an Arbitrary deck of hands with at least one pair of similar cards = A, B, C, A
the number of patterns that A,B,C,A hand appears = 4C2
Therefore, number of ways A is chosen= 12
when the first A is chosen, it locks in the value for the second A
number of ways B is chosen = 10(because by choosing a value for A, two cards are taken off the choice list)
number of ways C is chosen = 9
P(at least two cards have the same value) = (12 x 10 x 9 x 4C2)/(12 x 11 x 10 x 9) = 18/33

I hope that your explaination you will elaborate:
1. why my method is wrong
2. the actual solution.
Thank you very much for even considering my post!!!
Adhil

P(at least 1) = 1 - P(None). This is probably more useful here.
So there are possible ways to select 4 distinct cards. Then there are different suits for those cards. Finally there are different cards.
So you have

Hello, warElephant!

Another approach . . .

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.
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 probability that Bill finds at least one pair of cards that have the same value?

. .

We will find the probability that there are no matching pairs among the four cards.


The first card can be any card (it doesn't matter):

The second card can be any of the other 10 non-matching cards:

The third card can be any of the other 8 non-matching cards:

The fourth card can be any of the other 6 non-matching cards:

Hence: .


Therefore: .