Fold the King High flush in this situation?

[kingstalker]

Paradise 5+0
Omaha high
Blinds 50/100
I have 785 in chips after posting the blind
4 limpers, I am dealt /images/graemlins/club.gifK /images/graemlins/club.gif5 /images/graemlins/spade.gifA /images/graemlins/diamond.gif7 in the BB
Flop comes down
/images/graemlins/club.gifJ /images/graemlins/club.gif8 /images/graemlins/club.gif2
SB(1320) bets 200,I move all, all fold to the SB who has the /images/graemlins/club.gifA /images/graemlins/club.gif10.
If this was Holdem I woulnt have any regrets about losing to the ace high flush but I had to figure since there was 4 other players in the pot there was a good chance that the /images/graemlins/club.gifA was out,so should I fold here?

[Buzz]

[ QUOTE ]
so should I fold here?

[/ QUOTE ]

Yes.

Kingstalker - Rightly or wrongly, people tend to find excuses for playing suited aces in Omaha.

The chance of the ace of clubs, plus at least one more club, having been dealt to one of your opponents depends on how many opponents you originally had - rather than on how many limped.

How many players were originally dealt cards? Just five? Or ten? Makes a big difference in terms of the chances of encountering an opponent with the ace of clubs plus another club.

Assuming you had nine original opponents, after you have seen your hand and the flop, the probability one of them was dealt the ace of clubs is 36/45. Then we have to figure the probability of the person who might have been dealt the ace of clubs also having been dealt at least one more club.

What it all boils down to in a full, loose Omaha game is the 2nd nut flush figures to win about twice as often as it loses. In other words, if you go all-in three times with a flopped second nut flush, maybe you win a small amount two times and get knocked out of the tournament one time.

Just my opinion.

Buzz

[kingstalker]

It was ten handed.

[Buzz]

Kingstalker -

Step one. In a ten handed game, your nine opponents were dealt 36 cards. After the flop, when you see seven cards, none of which is the ace of clubs, the probability the ace of clubs was dealt to one of your opponents is 36/45. I don’t know if that’s obvious to you or not. If it’s not obvious, think this way: Any one of the missing 45 cards could be the ace of clubs. Another way of expressing that concept is: There’s one chance in 45 that any card you choose will be the ace of clubs. Your opponents have been dealt 36 of these cards. Hope that explains it, if it wasn’t immediately obvious.

Step two. We see five cards in the flush suit plus two non- flush suit cards. Leaves eight in the flush suit, including the ace, plus 37 non-flush-suit cards.
If someone was dealt the ace of clubs, the probability of that player being dealt no other clubs is approximately
37*36*35/45/44/43 = 0.548.
Therefore, the probability of the player who was dealt the ace of clubs having at least one club is approximately 1 - 0.548, or approximately 0.452.

Step three. Combnining probabilities, (36/45)*0.452 = 0.362 is the approximate probability your flopped king high flush might well lose to the flopped ace high flush in a game where someone would be almost sure to see the flop with the suited ace of clubs.

I think that’s the essence of it. Looks good for your king of clubs, doesn’t it? The odds are 638 to 362, or about 7 to 4 that none of your opponents has the suited ace of clubs. (I actually think you can make those odds work for you in a limit game if you can read your opponents well).

But who is going to want to play when the flop has three clubs and you go all-in? Seven times out of eleven, you’ll win the antes - but four times out of eleven you’ll encounter absolute disaster.

I made it 2 to 1 in the first post. It's actually closer to 7 to 4.

Buzz

[Big Dave D]

In a tournement its a clear pass, in a cash game, with a big stack, well you're kust playing poker.

gl

dd

[chaos]

I think it is a clear fold. Betting all in is a big mistake since you will probably only get called when you are beaten by the nut flush.