Putting this one here because I think it will get pretty complicated. Question is playing SB and BB if the rest of the table has folded. This is different from HU because in SB youre now OOP.

I'm going to take a stab by modeling it with the [0,1] game (high real number wins at showdown).

Here's the model for which we'd like a solution (too complicated for me to get it exact, so rough is ok): SB=1, BB = 2. Stacks = 24. SB can fold, call or raise to 8. If SB raises BB can call or push. If SB calls, BB can raise to 6 or just call. If BB raises to 6, SB can then call or push.

I'm going to start with a simplified model, where BB has to be completely passive (can only call or fold). What are SB's ranges here?

SB clearly needs at least to call on [1/4,1] since it costs 1 to win 3.

A raise costs 7 to win 3. I'm pretty sure that means that SB should raise [.7,1].

Now BB is risking 6 to win 10. So, he needs to win 3/8 of the time when he calls. That would have him calling only with top (5/8)*.30 hands. Hence, bluffs would become profitable for SB.

A bluff is unprofitable for SB only if BB calls at least 30% of the time. So, that's what BB needs to do for bluff prevention (bluffs are neutral EV with that call %).

To do that, SB needs to bluff (3/8)*.3 = 9/80 = 1/9 more or less.

Hence, in this simplified model, SB bluffs [1/8,1/4], limps [1/4,.7] and value raises [.7,1].

BB calls a raise with [.7,1].

This isn't exact but close enough.

The reason I'm looking at this is that in practice, I'm trying to get away from just always raising or folding SB and making my % dependent on estimated fold % of BB. I think that's a bit too primitive. A limp should also be possible--including limp re-raise.

I'll get back to this later. The full model is going to be difficult.

I'll try one more simplification to get warmed up. Now, SB can't raise but just calls sometimes. Then BB has the option of raising, and SB can call or fold.

SB now clearly folds [0,1/4], and let's assume he completes the rest (he may need to play tighter OOP).

Now, in raising to 6, BB is risking 4 to win 4 on a fold. So, he needs to raise the top half of SB's range, namely [5/8,1].

Now SB has to call 4 to win 10, so he needs to win 3/7 of the time. That means BB needs to bluff 3 times for every 4 value bets, and (3/4)*(3/8) = 9/32.

Hence, BB would bluff [0,9/32], check [9/32,5/8], and value raise [5/8,1].

This clearly also drives SB's original limp range upward, because he's no long risking 1 to win 4. I just have trouble figuring out how much it drives it up.

Just guessing, I'd say BB's positional advantage in this extreme model should force SB to call only something like [3/8,1]. I think [1/2,1] is too conservative.

I think I've figured out an approach to an accurate solution on the passive SB model, which I think is easier to do in the lowball version (low real number wins).

Suppose SB calls on [0,x].

We know BB is value raising [0,x/2].

Now to call this raise, SB is risking 4 to win 8, so he needs to win 1/3 of the time. In order to make the call neutral at x/2, BB needs to be bluffing, then, once for every 2 value bets.

That means he's bluffing with a probability of x/4. So, the probability that BB will raise is x*(3/4).

So, here's what the tally now looks like for SB:

1) SB has [0,x/2]:
BB has [0,x/2]: wash, but SB put in 5 and won 7. So, SB wins 1/6 here.
BB has [x/2,x]: SB wins always for a profit of 3.
BB bluffs [x,x+x/4]: SB wins always for a profit of 7.
BB has [(5/4)*x,1]: SB wins 3.

EV for SB is thus: (x/12)+((3/2)*x)+((7/4)*x)+(3*(1-(5/4)*x)) = (x/12)+((18/12)*x)+((21/12)*x)+3-((45/12)*x) = 3 - (5/12)*x

2) SB has [x/2,x]
BB has [0,x/2]: SB folds for a loss of 1
BB has [x/2,x]: wash but SB put in 1 and won 3, so SB wins 1/2 on this interval
BB bluffs [x, (5/4)*x]: SB loses 1
BB has [(5/4)*x,1]: SB wins 3.

EV here is: (-x/2) +(x/4) - (x/4) + (3*(1-(5/4)*x)) = -(x/2) + 3 -(15/4)*x = 3 - (17/4)*x.

Both equations have a probability of x/2.

The sum of both is 6 - (5/12)*x - (51/12)*x = 6 - (56/12)*x = 6 - (14/3)*x.

To solve for a max, we can just multiply by x: 6*x - (14/3)*x^2, and that will have the same max as 18*x - 14*x^2.

Differentiating, we get 18 - 28*x = 0 for local max, so 18 = 28*x and x = 9/14, or pretty close to 2/3.

With that range, SB is winning 6 - (14/3)*(9/14) = 6 - 3 = 3, and multiply that by 9/14. So, it's about 2.

Anyhow, that means that even with a huge positional disadvantage (SB is forced to complete passivity here), SB still has to call top 2/3 roughly.

You lost me on line 3

My math in poker stops with the very simple basics on flushes and straights.

I wish I could do this though. GL

Ok, another simplified game that hopefully gets us closer.

Now SB can only raise to 8 or fold. To the raise, BB can fold, call or push.

Again the lowball version seems easier, although the equations may get a bit uglier on this one.

Let's again suppose SB raises on [0,x].

Now it costs BB 6 to call for a pot of 10 on a win. So, BB needs to win 3/8 of the time to justify a call. So, BB at least calls on [0, (5/8)*x].

When does BB value shove? Here it costs BB 22 to win 10. So, BB needs to win a bit more than 2/3 of the time. Let's just call it 3/4 for simplicity's sake. That means a value shove is [0,x/4].

On the shove, SB has to pay 16 to win 32. So, SB has to win 1/3 of the time to call.

That means BB has to bluff once for every 2 value shoves.

So, we have the following values for BB:

BB value shoves [0,x/4], calls [x/4, (5/8)*x], bluff shoves [(5/8)*x,(3/4)*x].

SB calls the shove on [0,x/4].

Now let's solve for x by calculating SB's EV in terms of x and then maximizing:

1) SB has [0,x/4]
BB has [0,x/4]. Wash, but SB loses only 23 and wins 25. So, SB makes a profit of 1/24 here.
BB has [x/4,(5/8)*x]. SB wins 9.
BB bluffs [(5/8)*x,(3/4)*x]. SB wins 25.
BB folds [(3/4)*x,1]. SB wins 3.

EV here is x/96 + (27/8)*x + (25/8)*x + 3*(1-((3/4)*x). The x/96 is hardly going to matter. So, I'm going to leave that part out. We get (52/8)*x + 3 - (9/4)*x = 3 + (52/8)*x - (18/8)*x = 3 + (70/8)*x = 3 + (35/4)*x.

2) SB has [x/4, (5/8)*x].
BB has [0,x/4]. SB loses 7.
BB has [x/4,(5/8)*x]. Wash, but SB loses 7 and wins 9 for a 1/8 gain.
BB bluffs [(5/8)*x,(3/4)*x]. SB loses 7.
BB folds [(3/4)*x,1]. SB wins 3.

EV is -(7/4)*x + (3/64)*x - (7/8)*x + 3*(1-(3/4)*x) = 3 - (112/64)*x + (3/64)*x - (56/64)*x - (144/64)*x = 3 - (299/64)*x. Let's just call this 3 - (37/8)*x.

3) SB has [(5/8)*x,x]
BB has [0,x/4]. SB loses 7.
BB has [x/4,(5/8)*x]. SB loses 7.
BB bluffs [(5/8)*x,(3/4)*x]. SB loses 7.
BB folds [(3/4)*x,1]. SB wins 3.

EV is -(7/4)*x - (21/8)*x - (7/8)*x + 3*(1-(3/4)*x) = 3 - (14/8)*x - (21/8)*x - (7/8)*x - (18/8)*x = 3 - (15/2)*x.

Now for the total EV:

In equation 1 we multiply by x/4. In 2 by (3/8)*x, and in 3 by (3/8)*x. For max, we can just multiply all of that by 8.

So, I get:

1) 2*x*(3 + (35/4)*x) = 6*x + (35/2)*x^2
2) 3*x*(3 - (37/8)*x) = 9*x - (111/8)*x^2
3) 3*x*(3 - (15/2)*x) = 9*x - (45/2)*x^2.

So, we want to maximize this on the interval [0,1]:

24*x - 5*x^2 - (111/8)*x^2 = 24*x - (151/8)*x^2.

Setting the derivative at 0:

0 = 24 - (151/4)*x
(151/4)*x = 24
x = 96/151.

Wow! That's a LOT of raising! There's so much addition and subtraction in here that I could easily have messed something up, but that would mean that raising almost top 2/3 is correct for SB here.

Even if we round way down to 1/2, it would look like this:

SB raises top 1/2 (tighter than optimal). BB value shoves roughly top 10%. BB calls on other top 30%. BB bluff shoves 30-35%.

Haven't repeated the math, but i'm curious how sensitive this is to stack sizes. I'm guessing very. Why did you use 24 btw?

PS: BK, can you stop trolling all his posts. Or at least make a half decent strategy post yourself to establish a little credibility before you do.