• Nov 28th 2012, 06:08 PM
warElephant
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!!!:)
• Nov 28th 2012, 09:49 PM
Scopur
P(at least 1) = 1 - P(None). This is probably more useful here.
So there are $6C4$ possible ways to select 4 distinct cards. Then there are $2^4$ different suits for those cards. Finally there are $12C6$ different cards.
So you have
$P(at least one) = 1 - P(none) = 1 - \frac{6C4*2^4}{12C6} = 1 - \frac{16}{33} = \frac{17}{33}$
• Nov 28th 2012, 10:15 PM
Soroban
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?

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

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

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

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

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

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

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