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View Full Version : K-xs Close Enough to be Nut Flush?

Festus22
07-01-2003, 01:35 PM
I like to play K-xs as a flush draw-only hand (flop 2 of the suit or fold - period!). Is this probability correct:

Assume a 3 suit board. K-xs can only lose to A-xs (forget boats or greater for this discussion). So that hand must have the suit A (18/45 or 40%) and the other card must be the suit (13 - my 2 - board 3 - suit A = 7 so 7/44 = 15.9%). Combining, there's only a 6.4% chance that my K-xs will be beat be an A-xs given a 3 flush board. Of course, when the board shows a 4 flush, now I'm only a 60% favorite. Is this correct? I've had a lot of success with these hands yet I don't see this play mentioned a lot. Unless the miracle K-K-x flops, the key is bailing even if a lone K is flopped.

punkass
07-01-2003, 01:38 PM
Maybe I'm missing something, as Im no expert in probability. Where's the 18/45 coming from? particularly the 18.

Festus22
07-01-2003, 02:20 PM
Assuming a 10 handed HE game, you have two cards, the board has 5 and the other players have 18 of the remaining 45 unseen cards. So there's an 18/45 chance one of them holds one specific card. In my question, that card is the A of the suit I hold.

TJSWAN
07-01-2003, 02:37 PM
Where I play you will beat a lot of smaller flush draws with KXs then the rare time the AXs is out there.

Copernicus
07-01-2003, 04:05 PM
Your approach isnt quite correct but the result is close enough. For a particular one of the other 9 at the table to have you beat he must have started with the A (1/45) and other (7/44) of the suit, times 2 because order doesnt matter. That is only a .707% probability. To find the probability for any one of the 9, subtract that from 1 (giving the probability he doesnt have it) raise it to the 9th power (giving the probability none of the 9 have it), and subtract that from 1, or 6.2%.