Omaha probability question

I was playing Omaha Hi-Low and held A-3-4-8 with a 3-6-8-8 board, giving me eights full of threes. I knew I couldn’t be beat by quad eights, only eights full of sixes. I did a quick calculation and thought that, given the cards exposed, it was better than 40:1 that one of my opponents held 8-6. Later in the session I ran into the same situation with K-6-x-x and a K-6-T-K board. Yes, I was beat both times :-b

Afterwards I tried to figure out the exact odds and came up with

Probability of holding the case 8 = 1 / 44
Probability of holding a 6 = 3 / 43
Number of opponents = 3

1/44 * 3/43 * 3 = 9 / 1892 ~ 0.48% ~ 209:1 against

Is this the correct or do I have to include 6 (number of 2 card combinations in 4 hole cards) somewhere? Thanks.

Aren't there 5 board cards in Omaha? You can still get beat with a river card.

Oh sure. There were plenty of river cards to cause me trouble. If a 3 came out, I was beat by someone with 33xx. Or if, for example, a Jack hit the river, I was beat by someone with J8xx or JJxx. The 6d and 8d were on the board, so a 5d, 7d, 9d on the river might have given someone a long shot straight flush.

My question is: when I held the second nut high full house on the turn, what was the probability that I was already beat by someone holding 86xx and what is the correct formula for figuring the probability?

Havent done these in a while so here goes
Probability of any ONE guy holding 86 = 4/44 * (1-40/43*39/43*38/43) = 2.3%

Also the correct probability should involve some P(given) calculation, since your opponents cards are not completely random. Given that he stayed in the pot on a 3-6-8 flop, his hand most likely consists of 4 low cards, so the actual odds are higher that he connected. For the king ten I would actually adjust LOWER unless he got a free look from the blinds, since not that many king ten hands are playable in Hi/Low.

I think you are overlooking 2 important factors in this:how many people were in the hand at the beginning and the fact that someone else was in against you at all changes the probabilities.

If for example you are playing holdem and have pocket AA, 4 players go to the flop and it comes J 5 5, you bet and one person calls, you have to take into account you were playing against 4 players with 8 cards at the beginning. The fact that one person called you knowing you raised preflop greatly increases the probability he has a 5 beyond what simple mathematics would determine. That's the thing about poker, there are many more variables and skills to perfect than determining probability in a vacuum.

Oh I just saw it was hi-lo you were playing. Yeah that was pretty unlucky!

Any answer to this question based on the random probability of a particular card combination being dealt isn't relevant. The hands out against you in such a situation would be only those that had called preflop in the first place, and more importantly, were still in the pot at that point. Therefore, those hands are nonrandom. (For instance, you are unlikely to be up against QQQ2, but that is a "possible" hand.)

So you should look at it this way: the board is 3688. Now, unless you are playing with morons, no one will put any action in there unless they have the nut low (A2 something something), or they have an 8. Furthermore, no one would get particularly excited with that 8, because the three low cards on the board, PARTICULARLY three that were not an A or a 2, would make it very likely that someone had a made low hand. The only way a reasonable player would feel even somewhat comfortable with ANY high hand in a situation like that would be if it was the "nut" full house. Similarly, a player holding the nut low would be check-calling, because he would only be eligible for half the pot in any case, and could very easily get quartered. The nut full, at least, can feel reasonably good about his chances to get half (not that he wouldn't be subject to suckouts).

So my answer to the question, how likely is it that my second-nut full house is beaten, would be, "pretty damn likely". Anyone taking the lead in the betting on the turn would be MORE LIKELY THAN NOT to have the nut full house, simply because no other made hand would have a particularly good expectation.

So the answer to your question should be, "about 75%, if there's any significant action, and there are three or more players remaining". Heads up, a bettor could be trying to chase away a weak high or a weak low. Three or more players--no hope of that, so a bettor at that stage would have to have a legitimate hand, and as I noted, the only legitimate hand (with a realistic profit expectation) would be the NUT full house.

I should add that I learned how worthless the "second nut" full house is in Omaha only through bitter experience: in Omaha, there are only two hands that count: the nuts, and a draw to the nuts. And in Hi-Lo Omaha, there is only one hand that counts: a scooper.