Probability and combinations

[venmic]

Is there a Simpler way of doing this
Source Manhattan


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?


answer: [spoiler]17/33[/spoiler]

[Anurag@Gurome]

Is there a Simpler way of doing this
Source Manhattan


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?


answer: [spoiler]17/33[/spoiler]

Let p be the probability that Bill finds no pair of cards that have the same value. Then probability that Bill finds at least one pair of cards that have the same value = 1- p
Total no. of ways in which 4 cards can be drawn from 12 cards = 12C4 = 495

Number of ways in which Bill finds no pair of cards with the same value:
There are 6 cards within a suit and we want to select 4 cards. So, this can be done in 6C4 = 15 ways.
For 15 values each card can be of either suit.
1st card can be chosen from 2 suits, similarly 2nd, 3rd & 4th card can be chosen from 2 suits. We can do this 2^4 = 16 ways
So, number of ways to choose 4 cards so that no pairs exist = 15X16 = 240

So, p = 240/495 = 16/ 33
Hence, 1 - p = 1 - 16/33 = [spoiler]17/33[/spoiler]