Cards Dealt Probability Problem

Lo all

Just a quick question which is mashing my brain a little ..

Lets say you have two pools of cards

Pool 1

A, K, Q, J, T

Pool 2

9, 8, 7, 6

What are the odds - in hold 'em - of being dealt 1 card from pool 1 and one from pool 2 (in any order)

I seem to have gotten myself in a fuddle working it out, my difficulties being the following :

If card 1 is from pool 1, you are left with 4 ranks and 4 suits (16 cards total) that card 2 MUST be

and

If card 1 is from pool 2 then your left with 5 ranks and 4 suits (20 cards total)that card 2 MUST be

Its the either/or situation that card 2 presents thats killing me

Can anyone help me out?? Oh .. and try and keep it simple for me, I ain't blessed with mathmatical prowess .. [img]/forums/images/icons/smile.gif[/img]

Thx in advance

T

Okay, there are 20 cards in pool 1, 16 in pool 2.
The probability of the first card being pool 1, and the second being pool 2 is (20/52)*(16/51). But since order is not important, we simply double this. ( There are 2 combinations: pool 1 then pool 2 or pool 2 then pool 1.)
The final answer is 0.2413273. No wonder I've been looking at K6o so much.

Bah! I had it like that 2 days ago but talked myself out of it!!!!! Doh!!!!

Thx!!!

T