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

Re: combinations/probability question please help!!

P(at least 1) = 1 - P(None). This is probably more useful here.

So there are $\displaystyle 6C4 $ possible ways to select 4 distinct cards. Then there are $\displaystyle 2^4 $ different suits for those cards. Finally there are $\displaystyle 12C6 $ different cards.

So you have

$\displaystyle P(at least one) = 1 - P(none) = 1 - \frac{6C4*2^4}{12C6} = 1 - \frac{16}{33} = \frac{17}{33} $

Re: combinations/probability question please help!!

Hello, warElephant!

Another approach . . .

Quote:

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?

. . $\displaystyle (a)\;\tfrac{8}{33} \qquad(b)\;\tfrac{62}{165} \qquad (c)\;\tfrac{17}{33}\qquad (d)\;\tfrac{103}{165} \qquad (e)\;\tfrac{25}{33}$

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): $\displaystyle \tfrac{12}{12}$

The second card can be any of the other 10 non-matching cards: $\displaystyle \tfrac{10}{11}$

The third card can be any of the other 8 non-matching cards: $\displaystyle \tfrac{8}{10}$

The fourth card can be any of the other 6 non-matching cards: $\displaystyle \ftrac{6}{9}$

Hence: .$\displaystyle P(\text{no match}) \;=\;\frac{12}{12}\cdot\frac{10}{11}\cdot\frac{8}{ 10}\cdot\frac{6}{9} \:=\:\frac{16}{33}$

Therefore: .$\displaystyle P(\text{at least one match}) \;=\;1-\frac{16}{33} \;=\;\frac{17}{33}$