iPoker - $0.50 PL Hi/Lo - Omaha Hi/Lo - 9 players
Hand converted by PokerTracker 4

CO: $49.95 (VPIP: 0.00, PFR: 0.00, 3Bet Preflop: 0.00, Hands: 5)
BTN: $78.35 (VPIP: 50.00, PFR: 0.00, 3Bet Preflop: 0.00, Hands: 14)
SB: $56.29 (VPIP: 57.14, PFR: 7.14, 3Bet Preflop: 0.00, Hands: 14)
BB: $40.63 (VPIP: 50.00, PFR: 14.29, 3Bet Preflop: 0.00, Hands: 14)
Hero (UTG): $17.47
UTG+1: $29.88 (VPIP: 11.11, PFR: 0.00, 3Bet Preflop: 0.00, Hands: 9)
MP: $22.04 (VPIP: 28.57, PFR: 0.00, 3Bet Preflop: 0.00, Hands: 14)
MP+1: $15.00
MP+2: $27.38 (VPIP: 30.00, PFR: 10.00, 3Bet Preflop: 0.00, Hands: 10)

SB posts SB $0.25, BB posts BB $0.50, MP+1 posts penalty blind $0.50

Pre Flop: (pot: $1.25) Hero has Q A 6 2

Hero calls $0.50, fold, MP calls $0.50, MP+1 checks, fold, CO raises to $3.25, fold, SB calls $3.00, BB calls $2.75, Hero calls $2.75, MP calls $2.75, MP+1 calls $2.75

Flop: ($19.50, 6 players) 4 8 8
SB checks, BB checks, Hero bets $14.22,

I know usually bare nut low draw in PLO8 not great hand but I thought with this hand, other bare A2 hands may fold or AA with no low may fold. Also in my mind was the fact that I would probably have odds to call against an 8 as long as it is not A28.

Thanks as always.

Simplistically I think there are two key cards here,
the 8 and the 8.

We see seven cards after this flop, leaving 45 unseen cards including the two eights.
Your five opponents have between them 20 cards.
The probability an opponent has an eight is about 70%.

• Here's the math:
  P=1-C(43,20)/C(45,20)=
  P=1-25*24/45/44=0.6967

So it's about 7 to 3 that one of your opponents has an eight.

Since some opponents would usually would bet an eight from a blind while others usually wouldn't, we can't tell if either blind has a missing eight.

I like your bet for a variety of reasons. For one, you might steal the pot with your bet.

I don't think so, but you never know.

Kind of doubtful that any of your five opponents has A28*. It's doubtful enough that I wouldn't worry about it at this point in the hand.

I don't know. I think it depends on what cards your opponents are holding and there's no way to know that at this point.

Buzz

ex·pect·ed val·ue
noun MATHEMATICS
a predicted value of a variable, calculated as the sum of all possible values each multiplied by the probability of its occurrence.

Thanks, but I still don't know if Hero's play is +ev. We could run a simulation if we had an idea of Villain's hand

The way things stand, if an opponent has an eight, then I think the opponent is probably ahead of Hero.

ProPokerTools Omaha Hi/Lo Simulation

600,000 trials (Randomized)

board: 488

Hand	Pot equity	Scoops	Wins Hi	Ties Hi	Wins Lo	Ties Lo
QhAd6s2c	34.95%	6,129	6,569	202	404,385	13,819
8***	65.05%	180,193	593,229	202	19,998	13,819

But if an opponent has random cards, then I think the Hero is probably ahead.

ProPokerTools Omaha Hi/Lo Simulation

600,000 trials (Randomized)

board: 488

Hand	Pot equity	Scoops	Wins Hi	Ties Hi	Wins Lo	Ties Lo
QhAd6s2c	53.72%	168,410	194,228	12,494	380,959	21,025
****	46.28%	125,416	393,278	12,494	30,722	21,025

Am I doing something wrong or looking at this in the wrong way?

Buzz

I don't know man, don't think so though.

UTG, full ring, with no fd's I'm prob folding PRE 'cause it cant really flop anything too strong.

I think getting it in on the flop is the correct play given your stack size.

Of course there's rarely any way to know precisely what cards the opponents hold. I believe you're thinking about EV in terms of what a simulation would come up with if we were to know specifically which hands we're going to be gambling against. You can also think about the overall EV of the situation taking into account how a random sampling of opponents from the field would act given a sampling of hands they would tend to have given the action. Consider a range of opponents as well as a range of hands for each opponent.

I think the OP's title is a shorthand way of asking "is this ever +EV vs a certain field of opponents as a holisitic average" and not "when the cards are turned over are we ever in a +ev situation?" Even if we knew the specific cards of our opponents in this hand and we could say +EV or -EV, stamped and done, we still could prove nothing about the situation in general! Because we just answered a completely different question! So the next time we ran into this exact scenario we would only know a tiny bit more information, namely this is how we perform when we run into this one exact slice of our opponents' ranges. This is akin to running into the top or bottom of an opponent's range and placing too much emphasis on the result.

But if we analyze the situation as a whole, even though we probably are going to have to make some course estimations, we will know more about the EV of future similar situations.

The initial leg work of the analysis from you looks good except for this. We actually do gain information from checks if we consider that some opponents would bet out with an eight some of the time. If we consider our opponents this way then some basic bayesian statistics would tell us that it it is somewhat less likely these first two opponents hold an eight. To argue otherwise would be to argue that position in poker confers no advantage.

Bayes rule:
P(A|B) = P(B|A) * P(A) / P(B)

A=holds an eight
B= check
| = given that

In this situation reads as "The probability that an opponent holds an eight given than they check equals the probability that they check given that they are holding an eight ...

We can definitely get down and dirty and make better estimations than this. And of course we have to deal with all the contingencies of running into multiple opponents, which will happen quite often here. It is rather complicated though, and I gots to go!

when in doubt, pot.
here you're calling off the rest of your stack anyway. Pot has $19.50 and hero has $14.22 behind. Hero needs 30% equity if one player pots with an 8, and the rest fold, which he has. That is to say if CO pots when checked to, bets, and it's folded to hero he has to get it in anyway expecting 35% equity or so.
If you're going to stack off by calling, then clearly shoving and maximizing whatever FE we have got is the correct play.

Generally if there's pot or less in the stack, and you have to call anyway if villain pots it, it is better to just pot it yourself.

Answering the topic. Yes, this shove is +EV

If there was some confusion, by + EV, I mean on when we shove have this $14 into the pot in spots like this at this level, do we expect to get at least $15 back on average?

Also maybe I am over waiting this factor but can't we reduce the probability that someone holds an eight here. My understanding is that buzzes math is based on random hands ( I need to learn how to do thst at the table by the way) but the thing is some hands are more/less likely to be played than others. To illustrate my point, if the flop were AA4 in a five way raised pot, we would say it is almost a certainty that someone holds a ace, right?

So I was thinking the chance someone holds an eight is less than many other cards because an eight is unlikely to be the reason a starting hand is played. I think one or two of these players were really loose so maybe discount that for them but again, I would only call a raise with an eight in my hand if I had two or three other good cards to go with it meaning most of my eights would now be in the muck. am I over emphasizing that factor?

Yes. I think so.

There's $19.50 in the pot, so if everyone folds, we win the $19.50. If you can guess the fraction everyone will fold, multiply the $19.50 by that fraction. For example if we expect everyone to fold 30% of the time,
$19.50*0.30=$5.85. That's the amount we win when everyone folds.

We started with $17.47 and we expended $3.25 on the first betting round. Hero has $14.22 left, all of which he bets on the second betting round.

If one opponent with a random eight calls, then the pot becomes
$19.50+$14.22+$14.22=$47.94, of which we have 34.95% pot equity. I think that means we should expect to average winning 0.3495*$47.94=$16.76.

So 70% of the time we win $16.76.
$16.76*0.70=$11.73 is how much we win when an opponent with an eight contests the pot.

$5.85+$16.76=$22.61 And our cost is $14.22. So if my numbers are right, it looks to me like we should average a profit of roughly $8 per hand by betting $14.22 here.

Somebody please correct me if I'm thinking incorrectly with my understanding of pot equity or some other concept.

My computation is obviously too simplistic, but I don't know any other simple way to think about this.

Maybe, but on the other hand, five opponents are seeing the flop. Some of them could have all kinds of crap.

Good point.

(Under game conditions, I'd actually use an approximation method that would get me to an answer of about P=75%, which for my purposes is close enough to the more rigorously calculated value of about P=70%. And then I'd cut that in half because I'd guess the probability would be lower because eights are not favored cards).

Good point.

Mine too. If you're playing at a table where you can see your opponents, sometimes you can pick up a fairly reliable tell from someone who mucked a hand with an eight.

I don't know. For the loose game you're describing where five opponents saw this flop for $3.25 each, maybe you are. In a tighter game it would seem more of a factor.

Please, anyone, feel free to correct me if I have something wrong or if I'm not looking at the problem from the proper perspective.

Buzz

Thanks Buzz,

That helps.

I guess the real danger is that one of the 5 opponents has an 8 and another one is willing to overcall with the same nut low A2 draw.

This is exactly what happened in this hand and I got quartered.

Play full stack and fold it pre utg 9 handed. With your stack size I don't know why you limp - I would raise or fold. Limping seems bad.Why are you not playing full stack?

If I limped I'd probably want to cap it pre after CO's raise but the VPIP's look lol so I guess a stop and go is the only feasible option with your stack.

Every time I open a thread in this forum....



Oh, and Go Gators!

This is a good 1st pass. Now that it seems profitable to get it in with any A2 (our a2 isn't especially strong on this flop and it seems very profitable at a first pass), apply that same perspective to our opponents and assume they are going to gamble A2** as well. Then take a 2nd pass at the problem and do some rough and dirty math to see how likely it is to get in 3 ways. But yeah, as far as a really simple way to handle the problem, afraid there isn't one.

I would look at the count function ProPokerTools has with:
board: 4d 8c 8s
hand 1: qh ad 6s 2c
hand 2: (8, 44, a2) : 50% = 16641

Then:

board: 4d 8c 8s
hand 1: qh ad 6s 2c
hand 2: 50% = 71265

So we are looking at each opponent seeing the flop with a 50% range (adjust as you see fit) and then seeing how likely it is they hold a2, 44, or an 8.
16641/71265 = 23.4%

So the estimation for all 5 opponents folding: (1 - 0.234)^5 = 26.4%

Then it's pretty likely we get called it 2 or more spots since this chance is rather small. I don't know of a great way to do this, but to get a rough idea let's dead either A2 or an 8 and assuming that one opponent has already called, look at the chance that the other 4 fold:

board: 4d 8c 8s
dead: ac 2h
hand 1: qh ad 6s 2c
hand 2: (8, 44, a2) : 50% = 10436

board: 4d 8c 8s
dead: ac 2h
hand 1: qh ad 6s 2c
hand 2: 50% = 52659

So, a 10436/52659 chance that each of the 4 remaining opponents call:

(1 - 10436/52659)^4 = 41.3% chance that when one opponent has already called, the other 4 all fold.

Then there's also a pretty decent chance to get it in 4 ways+ but let's not do that and just put it all together from what we have so far:

Case 1: Everyone folds outright to our bet.
P = 0.264
Equity = $19.50 * 0.264 = +$5.15

Case 2: Only 1 caller
P = (1- 0.264) * 0.413 = 0.304

ProPokerTools Omaha Hi/Lo Simulation

600,000 trials (Randomized)

board: 4 8 8

Hand	Pot equity	Scoops	Wins Hi	Ties Hi	Wins Lo	Ties Lo
qh ad 6s 2c	39.15%	40,717	99,523	31,153	244,585	148,121
(8, 44, a2) : 50%	60.85%	156,316	469,324	31,153	34,602	148,121

0.3915 * (19.5+14.22) - 0.6085 * 14.22 = 4.55 * 0.304 = +$1.38

Case 3: Two callers or more
P = 1 - 0.304 - 0.264 = 0.432

ProPokerTools Omaha Hi/Lo Simulation

600,000 trials (Randomized)

board: 4 8 8

Hand	Pot equity	Scoops	Wins Hi	Ties Hi	Wins Lo	Ties Lo
qh ad 6s 2c	22.20%	6,783	17,855	7,776	119,530	251,653
(8, 44, a2) : 50%	38.94%	90,611	278,244	22,809	24,017	151,449
(8, 44, a2) : 50%	38.85%	90,161	277,888	22,740	23,942	151,508

0.222 * (19.5+14.22+14.22) - 0.778 * 14.22 = -0.42 * 0.432 = -$0.18

5.15 + 1.38 - 0.18 = +$6.35 equity on our shove (when called by 2 or less players)

Our equity gets rather bad 4 ways but since this will happen less frequently than everyone folding outright, we can still conclude that this is a very + EV spot:
P = ?
.1548*(19.5+14.22*3) - (1-.1548) * 14.22 = -$2.4 * P
P is less than or equal to 1, so our equity is >= 6.35 - 2.4

Also if opponents decided to gamble with A34* or A3TT type hands, we are doing even better so I don't think there are any surprises that can swing us into the red here.

Hmmm actually this calculation is only positive because of the good chunk of equity we get when everyone folds right away. The assumption we are making when doing this sort of calculation:
(1 - 0.234)^5 = 26.4%

is that each hand our opponents' hold is independent of each other, as if the cards are being replaced and re-dealt after we check to see if each player folds or not. It's a fine approximation when there are a large number of un-dealt cards but in problems like these, it may no longer hold up. Taken to the extreme, if all cards were dealt this assumption would completely collapse.

Probably need a better way to estimate fold equity in problems with large numbers of players to the flop. I imagine the equity is still positive, but not massively so.

You're welcome.

That was the idea.

Too bad. Rightly or wrongly, I probably would fold with the same nut low A2 draw if faced with an all-in raise after this 884 flop.

Depending on how your opponents would react, perhaps there's nothing you could do. AhQd6s2c is not what I'd consider a premium starting hand, but I think it's playable, roughly better as a starting hand than about 85% of the starting hands you could be dealt. I think not playing AhQd6s2c is too tight.

There are those who evidently think you'd be more successful if you raised pre-flop with this hand, and perhaps that is true. But perhaps if an opponent with a hand identical to yours would stay with you after you bet $14.22 on this flop, the same opponent with the same identical hand would also have stayed with you if you had raised before the flop. And maybe a different opponent would have won high with a different high hand if you had raised before the flop.

When you're a two to one favorite, you should expect to lose one time in three. (I don't know exactly how much of a favorite you were here).

My bottom line: I think getting it all in on the flop is your best chance of success with this hand/flop. I think you played perfectly.

Buzz

Two things:
(1) Rightly or wrongly, I wouldn't gamble with a hand identical to Hero's hand when Hero pushed after this flop. And I think most of my strong opponents also would probably not continue with a hand identical to Hero's hand when Hero pushed after this flop. Note that Hero's opponent with the identical nut low draw also got quartered.
(2) Granted that a large chunk of the profit comes from Hero's fold equity (which I estimated perhaps too highly at 30%), but my own more simplistic calculation shows Hero with a small profit against one opponent.

Good point.

That would be nice, but it's still going to be very highly specific opponent dependent. What I mean is the distribution of cards could be exactly the same on two adjacent tables, but corresponding players on the two tables might estimate the value/playability of their (identical) cards differently.

After I gave my equity estimate, I thought it was probably too high.

Thanks.
Buzz

Yeah I jumped the gun to say that since an A2 shove is positive, then our opponents are calling with all A2s. However, clearly that is going to happen as the OP found out, and I imagine it's going to happen a lot at middle to low stakes, so good to look at it from this angle as well. Interesting to note that this play becomes better the stronger your opposition is and the more they respect the game of chicken.

As Buzz calculated, vs one opponent this is always quite a positive shove. Even if they do call with all A2s, I messed with the numbers a bit and it's going to be a shove even if we only have something like 15% outright fold equity and run into 2 opponents quite often. The times that we do take it down or get HU are so profitable that you have to stomach the slightly negative EV of getting it in 3 ways or more.

This is a definite shove

Fwiw I agree with this.

(Under game conditions, I'd actually use an approximation method that would get me to an answer of about P=75%, which for my purposes is close enough to the more rigorously calculated value of about P=70%. And then I'd cut that in half because I'd guess the probability would be lower because eights are not favored cards).

Buzz, would you mind to remind me of how to do this quickly in your head?
There are 2 more 8s and 45 unseen cards. So there is roughly a 1/22 chance that anyone random card turned over is an 8. Therefore I think it is 4 times as likley that one random hand holds an 8, or 4/22 (i.e 18%), so basically an 80% chance agains any one random hand including an 8. So then .8 * .8 *.8 * .8* .8 = 32% five random hands and none include an 8.

That would be my rough way of doing it but is there a better way? I even used a calculator for last part. Thx

IloveCLocks,

You have sold me on pro poker tools, I didnt know it could do so much. As a holdem player all these years, there was never any math I couldnt figure out. Looks like there will be in the future with this game.

Thx

I'm thinking "P=1-C(43,20)/C(45,20), which reduces to
P=1-24*23/45/44 ...I can see that.
Then I'm thinking 24/45 and 23/44 are both roughly 1/2 (a little more than 1/2).
Then I'm thinking (1/2)*(1/2)=1/4 (a little more than 1/4).
Then I'm thinking P=~1-a little more than 1/4.
So P is a little less than 75%. (It's actually about 70%, rounded to two digits).

When I finished college, I had never even seen a computer or even a hand calculator. We used slide rules. (I still have my old K&E). We did most of our math to three digits. But the slide rule doesn't set the decimal place. To do that we needed to approximate.

1-.32=.68
So then it's about 68% at least one hand includes at least one eight. That's even closer than my approximation... about the same, but even closer.

I don't know if the way I did it (shown above) is better or not. Another time I might have approximated differently. I wouldn't generally multiply the probability one hand has something by the number of hands (because that method doesn't work unless the probability one hand has something is very small).

Buzz