This hand is from the Bellagio's 40-80 mix game.*

So there's no confusion, badacey is a split pot game. Half the pot goes to the best A-5 lowball hand (A2345). The other half goes to the best badugi hand (A234).

All the draw games have been good action games with many players playing a wide range of hands and going far with them. We're playing seven-handed, though only six players get dealt in for badacey.


A fairly tight player limps first-in. A laggy player raises on the button.

I'm in the small blind with A347Q

I'm not thrilled with this hand, but I don't think I can be folding it against LAG raises.

The big blind (also laggy) and the tight player call. Four players take the first draw with 8 small bets in the pot.

I draw two, keeping the A34.

Everybody else draws two as well.

I catch A34TK

I check and it gets checked to the LAG on the button. He bets.

I and the other three players call. Four players take the second draw with 6 big bets in the pot.

I draw two again. The Big Blind and Tight Player draw two as well. The LAG draws one.

I catch A2:345

So, I've got the nuts for the lowball half of the pot and no realistic chance of winning the badugi half of the pot in a showdown. I've yet to see a two-card badugi win in either the badacey or badeucey games.

I checked and so did the Big Blind and Tight Player.

The LAG bet. I called.

The Big Blind then raised. The Tight Player folded. The LAG called.

Then, I 3-bet and both players called.

Three players took the third draw with 15 big bets in the pot.

I stood pat and both opponents drew one.

I bet.

What do you think?





* In the last few days, the 40-80 mix game has been heavily loaded with draw games incluiding 2-7 Triple Draw, Badeucey, Badacey, and Badugi.

i've never played the game but get the concepts.

imo too loose to start with, but once you catch the nuts for half you played well. you are bluffing for the other half of the pot (probly wont work). you cant go wrong by betting. then again if you are convinced the middle player has the best badugi you should check to him, he will bet, have the LAG player call and CR. if its the reverse, LAG player with best badugi, bet, middle player calls, LAG raises, now you either 3 bet or flat depending on how strong u think MP guy is and whether he can fold a strong badugi.

i think predraw even tho the player is LAG some guy already limped pre so LAG guy probly isn't bluffing and is playing for value, which is a +1 to folding. if i ever play this game if i am drawing i would start out by only playing offsuit wheel cards predraw. you get 5 cards so its not hard to get 3 offsuit wheel cards.

I'd play it the same. I've only played this game a handful of times tho so take it fwiw.

Just curious, how bad is it here to stick with the A347?

I realize 7s are pretty mediocre in A5, its hard to compare baduecy and badacey hand strengths but.

I dont play badacey but I do play badeucy and when I do I typically value the 4 card hand more than the 5.

Am I thinking about this the wrong way?

Boy, if I thought I had even a little bit of fold equity against the limper I would threebet this pre intending to keep the 7. It sucks to have to discard it since you may very well want it later.

I think if they're poor handreaders, and won't figure out you're doing it to trap the field, donking third is definitely worth considering.

I love the bet on the end with your hand faceup and nothing they can do about it.

I agree with this concept in general but it is so obvious that you are check raising here that button might be hesitant to call 1 bet if he isn't closing the action but then again the pot is so huge that cr might be best.

I'd fold pre not close, I'd donk out after second draw and hope the LAG cooperates. As played river bet is certainly good.

So we should not be drawing to a37r in this spot? Is this too rough of an a-5 draw?

i reckon you have to ask "can I lock up half with this hand and make a cinch?"

in any split pot game the objective is to lock up half and then freeroll on the other.

my guess is A37 in badacey doesn't do it.

Change the 7 to a 6 and we keep it? Or we still always want to be drawing to the nuts?

One of the reasons I posted the hand was because, at the time, I saw merits in playing the A34, A47, and A347 as well as folding. Inexperience in the game made it a weird spot for me. So, right or wrong, I decided to play the smoothest draw to half the pot even though I only had two cards to the other half.


My bet on the river was called by one player who had a badugi and a 7-low. The other player folded. So, I got half the pot

The results of the hand seem to be an example why you don't play one-way hands in split-pot games. You're risking too many bets for too little chance of a payoff.


Responding to some comments above:

Lucius (or anyone else), do you know the exact odds of getting three offsuit cards Ace to Six? (also, Ace to Five)

Many of my opponents certainly seem to be playing looser than those standards. In one badacey hand, on the button, I folded 2,4,6,7 (two diamonds, two spades) with the Tc to a raise. When, I mucked, the cards accidentally flipped up and I got a "wow". I didn't inquire about what the "wow" meant, but I was under the impression he thought it was a very tight fold. Though, it's possible he didn't realize the 2,4,6,7 part of the hand only contained two suits.


Tapirboy, I think we can assume the tight player limped in with a three-card badugi. So, he isn't likely to fold it even if it's a bit weak. There's not much limp-folding in this game. The limp itself was unusual.

TheGusPair, if I've got the 6 instead of the 7, it's got to be playable as either A46 or A346. It's got to be too tight to discard a three-card 6 badug draw that has an Ace.

What do people think about drawing to 43xx ?

Why 43 instead of 4A?

I should be able to figure that for you.

First let's not immediately worry about suits. There are 10+6+3+1=20 possible three rank groupings that include exactly three ranks chosen from the six ranks A, 2, 3, 4, 5, and 6.
(If we exclude sixes, there are 6+3+1=10 possible three rank groupings that include exactly three ranks chosen from the five ranks A, 2, 3, 4, 5).

Then for each of these three rank groupings, there are ten possibilities, with five cards total, as follows for A23.

AAA23
A2223
A2333
AA223
AA233
A2233
AA23X, where X is anything other than A, 2, or 3.
A223X
A233X
A23XY where Y is anything other than A, 2, or 3.

Then there are a different number of ways each of first six these can be dealt.

hand	method	# ways
AAA23	4*4*4	64
A2223	^^	64
A2333	^^	64
AA223	6*6*4	144
AA233	^^	144
A2233	^^	144
total	.	624

That's going to be the same for each three rank grouping chosen.

The next part is kind of a bitch. The problem with X and Y is they can be cards belonging to ranks 4, 5, and 6. And that causes duplications.

The problem is greatly simplified if we take your "three offsuit cards Ace to Six" to mean there cannot be four or five cards ace to six.

In other words, if we choose X and Y to be anything other than A, 2, 3, 4, 5, or 6, the solution is much easier.

hand	method	# ways
AAA23	4*4*4	64
A2223	^^	64
A2333	^^	64
AA223	6*6*4	144
AA233	^^	144
A2233	^^	144
AA23X	6*4*4*28	2,688
A223X	^^	2,688
A233X	^^	2,688
A23XY	4*4*4*28*27/2	24,192
total	.	32,880

That's out of 2,598,960 possible hands. But remember there were 20 different groupings consisting of exactly three low ranks.
So the probability, if we don't worry about having all three low ranks the same suit, is 20*32,880/2,598,960=0.25302429, about one time in 4 (or odds of 3 to 1 against). That does not include four low rank or five low rank groupings. We'd have to calculate them separately. (We could do that, but you didn't ask for it).

But that's also without consideration of all three low ranks being the same suit. To correct for same suitedness, we only have to be concerned about the A23XY term (because where there's a pair, there can't possibly be such a grouping that we can't choose more than one suit).

So in other words, we have to modify the last term, the A23XY, to account for A, 2, and 3 all being the same suit. That's actually very easy.

There are 4*4*4=64 ways to chose an ace, deuce, and trey. Of these, only 4 have all three cards the same suit. Thus there are 60 ways to choose three specified low ranks without having all three cards the same suit.

Thus

A23XY

4*4*4*28*27/2

24,192

subtracting concerns about the ace, deuce and trey being the same suit, becomes

A23XY

60*28*27/2

22,680

And the completed table, corrected for suitedness becomes

hand	method	# ways
AAA23	4*4*4	64
A2223	^^	64
A2333	^^	64
AA223	6*6*4	144
AA233	^^	144
A2233	^^	144
AA23X	6*4*4*28	2,688
A223X	^^	2,688
A233X	^^	2,688
A23XY	60*28*27/2	22,680
total	.	31,368

The probability is 20*31,368/2,598,960=0.24138886, or still about one in four, or still odds of about 3 to 1 against. (Consideration of all three low ranked cards being of the same suit doesn't amount to much).

That's what you asked for, Dynasty. But maybe what you really want is how often anyone will be dealt a hand with three, four, or five different low ranks, not all of the same suit. If that's really what you want, Dynasty, I'll make a separate tabulation for four different ranks and another for five different ranks. (I think that's the approach to use).

(I remember doing that years ago for various lows in Omaha-8).

Buzz

Great reply + service.

1 in 4 hands seems to be the right amount of hands to play too. So not like you are a nit for going that way.

Um, Buzz, "offsuit" in this case means "of three distinct suits." It's Badugi.

Thanks.

Rats.

I did it for a five card deal too. Basically I figured it for a three card low in five card lowball, ace to five.

Three distinct suits is more complicated but shouldn't be too hard. Just much more tedious. Let me think about it. (I'll go do something else and maybe my brain will work on it).

Buzz

Before you get going again, it is a five-card deal. We're looking for a three-card low in A-5 that is also a three-card badugi.

Yeah so we are looking for something like 654 and better 3-card Badugi hands in a 5 card deal.

I reckon the 3 card 6s will be situational but I could be wrong, that's why I specified originally 3 offsuit unpaired cards 5 or below out of the 5 dealt, thinking that would be a good range of draw-2 hands to play.

OK. Let's do it.

There are 10+6+3+1=20 possible three rank groupings that include exactly three ranks chosen from the six ranks A, 2, 3, 4, 5, and 6.
(If we exclude sixes, there are 6+3+1=10 possible three rank groupings that include exactly three ranks chosen from the five ranks A, 2, 3, 4, 5).

Then for each of these three rank groupings, there are ten possibilities, with five cards total, as shown in a previous post for A23.

For AAA23, A2223, and A2333 we'll choose AAA23 as the prototype.
There are 4 ways to exclude an ace and four ways to include a deuce. Thus there are 16 ways to come up with an ace and a deuce. Of these, 4 are of the same suit (because there are four suits).
Thus there are 16-4=12 ways to come up with an ace and deuce of different suits.
Then there are two ways to choose a trey of a different suit.
(4*4-4)*2=24 OK. That was easy enough.

For AA223, AA233, and A2233, we'll use AA223 as the prototype.
6 ways to choose 2 aces.
6 ways to choose 2 deuces.
Can't have the ace and deuce of the same suit. Only 4 ways could have the ace and deuce of the same suit. So 32 instead of 36 ways to do this.
Then whatever two suits are chosen for the ace and deuce, there are 2 ways to choose the trey.
Thus 32*2=64 ways to do this (rather than 144).
OK that wasn't too bad either.

Now AA23X, A223X, and A233X. We'll use AA23X as the prototype.
6 ways to choose the pair of aces.
4 ways to choose a deuce.
24 ways in all.
The ace and deuce chosen cannot be of the same suit.
4 of the 24 ways would have the ace and deuce of the same suit.
24-4=20 ways to choose the ace and deuce.
Then the trey has to be of a different suit. 2 ways to do that.
20*2=40 ways to select the ace, deuce, and trey. And the suit of the high card is immaterial. Thus 40*28=1120 ways to do this. That was a snap.

Now A23XX.
4 ways to choose the ace.
Then 3 ways to choose the deuce.
Then 2 ways to choose the trey.
4*3*2=24 ways to choose ace, deuce, and trey of different suits. Then the suits of the other two cards are immaterial.
Thus 24*28*27/2=9072 ways to do it. Also a snap.

Here's my table:

hand	method	# ways
AAA23	4*3*2	24
A2223	^^	24
A2333	^^	24
AA223	(6*6-4)*2	64
AA233	^^	64
A2233	^^	64
AA23X	(6*4-4)*28	1120
A223X	^^	1120
A233X	^^	1120
A23XY	4*3*2*28*27/2	9072
total	.	12696

That's out of 2,598,960 possible hands. But remember there were 20 different groupings consisting of exactly three low ranks.
So the probability, if we don't worry about having three low ranks in different same suits, is 20*12696/2,598,960=0.09770062, about one time in 10 (or odds of 9 to 1 against). That does not include four low rank or five low rank groupings. We'd have to calculate them separately.

I think that's right. Never a guarantee about my math, but I try to do it correctly and will edit if I think I've made a mistake.

Buzz

I can't get my head around all that lol

Just to be clear: 9 to 1 against to get dealt 543 offsuit or below in 5 cards?

No. 9 to 1 against to get dealt 654 offsuit or below in 5 cards.

It would be 10*12696/2,598,960=0.04885031 or roughly 19 to 1 against getting dealt 543 or lower of different suits.

That doesn't include hands that would have four or five ranks below 654 or 543. We'd have to do separate tabulations for those. I don't think it would be difficult to do those; just would take a while to crank out the math.

I actually did that, but my math, my charts, and my reasoning needs to be double checked. I'll tell you my tentative bottom line result, for what it's worth. I think you'll get dealt a three or four card badugi with ranks six or lower about one time in ten. (My bottom line result, 0.103, is very close to one in ten).

Buzz

Great!

How would you go about adding that 19 to 1 probability to the probability of getting 6543 (4 card badugi) or below in the 5 cards dealt?

Trying to gauge how tight the game should be played.

Using the numbers I still haven't checked, I get 0.0544 as that probability, about one hand in 18. (Odds 17 to 1 against).

Buzz

Sorry, I meant 4Ao. Why not draw to 4Ao? If I played this thats what I would draw.