Chase the low draw?

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Good enough to peel one.

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I disagree. I discount the backdoor outs down to near zero. Normally, each backdoor flush/straight draw is worth 1 out each. I'd discount this to about .1 outs with the paired board and high hand. What kind of hands are staying in on this flop? Even if you're not drawing dead for high, any hand that contains a Ten in it would be staying in and taint your flush/straight outs. I think the discussion really should be limited to the backdoor low draw because of this. I could be wrong, but when you're looking for backdoor draws at 1/2 the pot which might not even be good if it hits, it just seems like a recipe for disaster.

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I second this. Flop fold seems trivially obvious to me.

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It seemed obvious to me too until I ran my excel spreadsheet counting specific cards with specific results. depending on what assumptions you want to use its relatively close.

-g

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OK I'll bite. What assumptions? runner runner highs are going to win x% or something? Low's won't get quartered?

It'd be cool if you could share the spreadsheet. I think Gooper did that once somehow.

[gergery]

Don't know how to do that. PM me i'll email it

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PM me i'll email it

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Yeah like I'm going to give you my email address so you can track me down and sodomize me? Nice try you sick fk.

[jedi]

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It seemed obvious to me too until I ran my excel spreadsheet counting specific cards with specific results. depending on what assumptions you want to use its relatively close.

-g

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Gergery,

What kind of assumptions are you making about other hands that are staying in? Any Ten will stay in the hand, and if it's not an ATxx hand, he's still got as many as 9 outs to have you drawing dead for hi. There may even be a hand with 2 diamonds in it, sticking around for some other draw and have you drawing dead to the diamonds.

I just don't see that the backdoor hi possibilities help in any way with this paired/2 flush flop. I think the analysis should really only be confined to the backdoor low draw.

[gergery]

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It seemed obvious to me too until I ran my excel spreadsheet counting specific cards with specific results. depending on what assumptions you want to use its relatively close.

-g

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Gergery,

What kind of assumptions are you making about other hands that are staying in? Any Ten will stay in the hand, and if it's not an ATxx hand, he's still got as many as 9 outs to have you drawing dead for hi. There may even be a hand with 2 diamonds in it, sticking around for some other draw and have you drawing dead to the diamonds.

I just don't see that the backdoor hi possibilities help in any way with this paired/2 flush flop. I think the analysis should really only be confined to the backdoor low draw.

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I was assuming no value for high, since i think that will be good <2% of the time. and you see no flop/turn raise and get quartered ~20% of time and its ~EV neutral. give or take.

-g

[jedi]

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It seemed obvious to me too until I ran my excel spreadsheet counting specific cards with specific results. depending on what assumptions you want to use its relatively close.

-g

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Gergery,

What kind of assumptions are you making about other hands that are staying in? Any Ten will stay in the hand, and if it's not an ATxx hand, he's still got as many as 9 outs to have you drawing dead for hi. There may even be a hand with 2 diamonds in it, sticking around for some other draw and have you drawing dead to the diamonds.

I just don't see that the backdoor hi possibilities help in any way with this paired/2 flush flop. I think the analysis should really only be confined to the backdoor low draw.

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I was assuming no value for high, since i think that will be good <2% of the time. and you see no flop/turn raise and get quartered ~20% of time and its ~EV neutral. give or take.

-g

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I see. The discussion we were having with TxRedMan was really centered on the possibilities of the backdoor high adding value to the hand. I didn't think it did, but even if it doesn't, you're still saying that taking one off on the flop is neutral EV? Something just seems wrong there. I'll have to reread everything, but taking one off here for a backdoor low, no high possibility, for a potential 1/2 of the pot just seems to be contrary to everything I've been taught about Omaha.

[gergery]

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It seemed obvious to me too until I ran my excel spreadsheet counting specific cards with specific results. depending on what assumptions you want to use its relatively close.

-g

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Gergery,

What kind of assumptions are you making about other hands that are staying in? Any Ten will stay in the hand, and if it's not an ATxx hand, he's still got as many as 9 outs to have you drawing dead for hi. There may even be a hand with 2 diamonds in it, sticking around for some other draw and have you drawing dead to the diamonds.

I just don't see that the backdoor hi possibilities help in any way with this paired/2 flush flop. I think the analysis should really only be confined to the backdoor low draw.

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I was assuming no value for high, since i think that will be good <2% of the time. and you see no flop/turn raise and get quartered ~20% of time and its ~EV neutral. give or take.

-g

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I see. The discussion we were having with TxRedMan was really centered on the possibilities of the backdoor high adding value to the hand. I didn't think it did, but even if it doesn't, you're still saying that taking one off on the flop is neutral EV? Something just seems wrong there. I'll have to reread everything, but taking one off here for a backdoor low, no high possibility, for a potential 1/2 of the pot just seems to be contrary to everything I've been taught about Omaha.

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that also assumes all the players who stayed in this hand actually stay in, ie. 4 on turn, 2 on river. I'm not saying its good, i'm just saying its close. Personally, I'd have folded on the flop. my point is only that its closer than it appears. and you can make some assumptions (ie. tight agro table) where its EV-

-g

[Buzz]

Hi Ben - Keep in mind that I'm not a mathematician - just someone like you trying to make sense of it all.

Conclusion: Fold after a flop with only one low card when you have a back-door low draw but only a poor chance to win for high.
I'll explain this below.

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it seems to me that these backdoor shots are a little bit diferent from a situation where you draw twice at the same thing,like an inside straight draw

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Quite different.

I generally use two-card combinations in these situations, rather than two-card permutations. I'll do the math using two-card combinations.

After you have seen the four cards in your own hand and the three cards of the flop, there are 45 unseen cards for the turn and thus 990 unseen two-card combinations for the turn and river. If the 990 is unfamiliar to you, there are at least two ways to calculate the 990:
C(45,2) = 990. 45!/(2!*43!) = 990. Or simply (45*44)/(1*2) = 990.

At any rate, if you do not care about which card comes first, from the vantage point of a player who has seen one four-card hand plus the three-card flop, 990 is the total number of possibilities for the turn plus the river in Omaha-8. (Unless there are exposed cards, that’s our vantage point when we’re playing Omaha-8). Each of these 990 two-card possibilities is called a two-card combination.

Note that the result of dividing by 990 is not much different than dividing by 1000. And to divide by 1000, all you do is move the decimal point three places.

In truth, suits are very important in Omaha-8, much more important than in Texas hold ‘em. In a typically pre-flop-loose game, high is won by a flush somewhere in the neighborhood of one hand out of every four that goes to the showdown. (Many times flushes are beaten by higher flushes, full houses, quads or straight flushes). But at any rate, suits figure prominently in the play of hands in Omaha-8.

However, for the purpose of purely figuring your chances of making low, suits don’t matter. Holding 2346 after a flop of AAT, you will end up with:
• a wheel 36/990 (the nut low plus a chance to win high)
• the nut low 183/990 (but no realistic chance at high)
• the 2nd nut low 27/990, and
• no low 744/990.
The above is tabulated and calculated, not simulated.

I think the above break-down is correct. The numbers 24, 195, 27, and 744 are subtotals from a 90 line chart I made listing every possible two card combination given this hand and this flop. My method was tedious and there were lots of places to go wrong - but I've checked it over this time and feel certain my numbers are correct.

I’ll save the chart for a while, and if you or anyone else I believe wants to see it, I’ll replace each tab with five dots and post it. No big deal - just seems pedantic for me to post it without a request to do so.

I read a number incorrectly and didn't check my math the first time through. My bad. You don't really need a complete complete chart and I didn't make one for the first post or the correction. But I did make a full chart this time.

The exact numbers from the chart aren't something to remember. What you have to remember is the conclusion shown in the second paragraph of this post.

At any rate, whether to continue or not after the flop always depends on various factors which vary from game situation to game situation.

One of the factors, the possibility of making the nut low hand, purely involves the cards you, yourself, hold and the particular cards on the flop. This does not vary from game to game.

Although the particular combination of opponents I face varies from table to table and varies from the start of a session to the end at any particular table, I don’t think I ever face a combination of opponents such that I have favorable odds to show a profit by playing what is almost purely a back-door low draw after the flop. Thus the conclusion shown in bold type in the second paragraph of this post.

It’s not the starting hand, 2346s, that is poor here. It’s the combination of the starting hand with the flop, 2346s/ATTn.

There is a slight chance this 2346s hand will win high after a flop of ATTn seen by five opponents, but that chance seems too slight to give much consideration. Hero’s best chance (by far) to win high involves making a wheel, but with three (out of five) opponents staying after this flop, even if we give the three non-folding opponents random cards, Hero only scoops about 2%, and that is almost exclusively with a wheel when no opponent makes a full house or backs into a winning flush.

Of the 2.7% Hero ends up making the 2nd nut low, I think Hero will not lose to the nut low about half and will lose to the nut low about half.

Thus I think Hero with 2346/ATT probably will:
• scoop ~2%,
• win half the pot for low ~13%,
• win a quarter or a sixth of the pot ~9%, and
• lose ~76%.
Since it’s very difficult to know what kind of hands Hero’s three opponents would be playing after this flop, I’m just making semi-educated guesses, based mainly on simulations and my own experience.

In real life, I think after both a turn and a river both favorable to Hero, at least one of Hero’s three opponents very well may fold, and if that doesn't happen, then I think Hero is probably getting quartered or sixthed. When Hero ends up with a scooper, somebody who flopped trips but didn’t improve may pay him off, but there’s also a distinct possibility Hero may end up not collecting from anybody when he makes a wheel that is a scooper.

When I wrote my blue, red, white colored chips example for Tex, I was bending over backwards to give Tex's point of view the benefit of the doubt. I thought the chip stacking demonstration might be able to clearly and graphically demonstrate that Hero did not have 8 to 1 implied pot odds, nor even 4 to 1 implied pot odds. Maybe the directions for the colored chips example were not entirely clear. I could possibly make them clearer if someone I believe thinks that might be helpful to him/her.

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In this case,before the flop Boolean has 27 cards out of 45 to pick up a low draw,3 to 2 in his favor to do this.If he misses on obtaining the pick-up,he throws his hand away.When he "hits",he will receive about 4 to 1 from the pot to draw to either 20 or 21 cards out of 44,about 6 to 5 against.

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That approach is going to involve permutations rather than combinations. I don't think permutations is the easiest way to look at this situation. (See above).

However, I guess you can use permutations to get the solution.

If you do, I think Hero folds when the turn is A, 9, T, J, Q, or K, a total of 21 (not 18) cards, and you end up with:
• 21/45 Hero loses one bet and folds,
• 8/45*24/44 Hero loses one small bet plus one big bet and folds,
• 8/45*20/44 Hero makes nut low,
• 3/45*23/44 Hero loses one small bet plus one big bet and folds,
• 3/45*21/44 Hero makes nut low,
• 4/45*24/44 Hero loses one small bet plus one big bet and folds,
• 4/45*9/44 Hero makes a wheel
• 4/45*11/44 Hero makes nut low,
• 9/45*23/44 Hero loses one small bet plus one big bet and folds,
• 9/45*4/44 Hero makes a wheel,
• 9/45*11/44 Hero makes nut low.
• 9/45*6/44 Hero makes 2nd nut low.

Phew! I think that's all of it. I haven't computed, and collected terms, but I think if you leave all the denominators alone, you probably end up with twice the totals I did using 990 above.

The denominators for this second method are all going to be 1980 because this second method uses permutations rather than combinations. Notice that 1980 is twice as big as 990.

Lot's of places to go wrong in these (rather tedious) computations. At any rate, I think you can approach the solution in either of the two ways outlined above (basically by using either combinations or permutations).

I feel a bit sheepish trying to explain this, since I'm not a mathematician.

Buzz

P.S. What the heck, let's see if it works:

• 21/45*44/44 = 924/1980.... -1
• 8/45*24/44 = 192/1980.... -3
• 8/45*20/44 = 160/1980.... nut low,
• 3/45*23/44 = 69/1980.... -3
• 3/45*21/44 = 63/1980.... nut low,
• 4/45*24/44 = 96/1980.... -3
• 4/45*9/44 = 36/1980.... wheel
• 4/45*11/44 = 44/1980.... nut low,
• 9/45*23/44 = 207/1980.... -3
• 9/45*4/44 = 36/1980.... wheel,
• 9/45*11/44 = 99/1980.... nut low.
• 9/45*6/44 = 54/1980.... 2nd nut low.

Collecting terms:
• 924/1980.... -1
• (192+69+96+207)/1980.... -3
• (160+63+44+99)/1980.... nut low
• (36+36)/1980.... wheel
• (54)/1980.... 2nd nut low

Combining terms:
• 924/1980.... lose one small bet
• 564/1980.... lose three small bets
• 366/1980.... nut low
• 72/1980.... wheel
• 54/1980.... 2nd nut low

Looks like the numerators add up to 1980. (That's my first check).

And the wheel, nut low, and 2nd nut low permutation totals are twice as much as for combinations shown above. (That's my second check).

Now we just have to decide what the nut low is worth and what the wheel is worth.

I don't think they're worth enough to me to make up my losses by continuing after the flop when drawing mainly for runner-runner low in my typical games.

O.K., I think I'm done.

Again, let me emphasize I'm not a mathematician, just an Omaha-8 player and enthusiast.

Buzz

[chaos]

Nicely done!

[Boolean]

Wow, just wow. My little post has really encourage some really great discussion. I for one want to thank you all, most of all Buzz for his incredible analysis, for your insight. I will be taking Buzz's conclusion to heart as it makes complete sense to me. Also, thanks for the insight on the preflop call. It's a marginal hand, I know. I like playing marginal hands from time to time, leftover from my Hold'em days. I'll quit that.

Again, thanks all.