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no

two cards, remember

i m getting the wrong answer

@Benjammin: I did the same thing but getting the wrong answer

are you supposed to get one of each card..one king and one queen?

@Benjammin: your calculations give the answer: 14/663

\[\frac{{4\choose 1}{4\choose1}}{{52\choose 2}}\]

\[\frac{8}{663}\]

yeah right.. How?

I gave the formula above

this is a classic hypergeometric distribution problem

but there shoulb be 8 cards right? 4 for kings and 4 for queens

yes..hence the two 4's in the solution

okey, means 8/52, now how did u get 4/51?

The way I see it, it's simply:
8/52 * 4/51

yes...same thing

pick from the 8...then don't pick a matching card...hence only 4 left to choose from

no i dont get it.. i was thinking it 7/51

see we've drawn 1 card, now 7 are left, so it would be 7/51

but by doing that you could get a matching pair...2 kings or 2 queens

that would work...that is what SmoothMath did

right. one thing more..

if we take 4 cards for 2nd turn, will it not be 49 in the denominator? 4/48

Oooh, I see. What I gave was to draw two cards, each of which would be a king or queen.

i m getting it now...

okey, now what is the probability that the cards are both jack?

(4*52)*(3*51)

the second will always be /51 if we draw 2 cards?

yes...if you are drawing without replacement

because you pick one of the 52 cards...there are only 51 cards remaining

alright, thanks.. i get it now!