tough probability ?

[finsisher]

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?

[Attack]

is the answer 1/11??

let the suites be Red and Blue numbered 1-6. (R1, R2...B1, B2, B3...)
suppose the first card picked up by Bill is R1, then the probablity of picking up B1 next is 1/11.
the 'at least' condition is fulfilled. hence 1/11
...
but i am not completely conviced with this . can anyone point out flaw if/any if this reasoning is wrong?

[stantheman]

I think the answer is 3/11.

He turns one card over, and now he turns 3 other cards over, hoping that one of the 3 is the pair of the first.
Card 2: (probability is the pair) 1/11 though if he doesn't get it,
Card 3: (2nd card was not a pair) 10/11 * 1/10 (3rd card is pair) if not,
Card 4: (2nd not pair) 10/11 * (3rd not pair) 9/10 * (4th pair) 1/9

Probability(Finds Pair from 4 Cards) = Probability(Card 2 is pair) + Probability(Card 3 is pair) + Probability(Card 4 is pair) = 1/11+1/11+1/11 = 3/11

[Fiver]

Not quite sure, but here's my take:
Find P(no pairs):
1] Choose 4 from 6 of one suit: 6C4 = 15
2] Choose 4 from 6 of the other one: 6C4 = 15
Total cases = 30
3] Total possible outcomes: 12C4= 11*5*9
Hence P (no pairs) = 2/33
Hence p (atleast one pair) = 1-2/33 = 31/33
What's the OA?

[parajulibharat]

P (at least one pair) = 1- P (no pair)
= 1 - 1*10/11*8/10*6/9
= 1- 16/33
= 17/33

[parajulibharat]

Fever's method is not working becasue you are ignoring the fact that you can pick one card from one deck and three from other deck, two cards from one deck and two form the other deck etc and still there is no gurantee that you will get the pair.

You can pick 4,0:3,1:2,2:1,3,4:0

(6C4)*(6C0) + (6C3)*(3C1) + (6C2)*(4C2) + (6C1)*(5C3) + (6C0)*(6C4) = 240 ways of not having pairs

Total ways of picking 4 cards 12C4 = 495

P (no pair) = 240/495 = 16/33
P (at least one pair) = 17/33

[Fiver]

Originally Posted by parajulibharat

Fever's method is not working becasue you are ignoring the fact that you can pick one card from one deck and three from other deck, two cards from one deck and two form the other deck etc and still there is no gurantee that you will get the pair.

You can pick 4,0:3,1:2,2:1,3,4:0

(6C4)*(6C0) + (6C3)*(3C1) + (6C2)*(4C2) + (6C1)*(5C3) + (6C0)*(6C4) = 240 ways of not having pairs

Total ways of picking 4 cards 12C4 = 495

P (no pair) = 240/495 = 16/33
P (at least one pair) = 17/33

Perfect . Thanks buddy.

[stantheman]

Originally Posted by stantheman

I think the answer is 3/11.

He turns one card over, and now he turns 3 other cards over, hoping that one of the 3 is the pair of the first.
Card 2: (probability is the pair) 1/11 though if he doesn't get it,
Card 3: (2nd card was not a pair) 10/11 * 1/10 (3rd card is pair) if not,
Card 4: (2nd not pair) 10/11 * (3rd not pair) 9/10 * (4th pair) 1/9

Probability(Finds Pair from 4 Cards) = Probability(Card 2 is pair) + Probability(Card 3 is pair) + Probability(Card 4 is pair) = 1/11+1/11+1/11 = 3/11

The above doesn't take into account that each time you pick one card up, you don't want to pick the pair up!!! answer is 17/33

[sh_vivek]

IMO 7/165.