Model for short-stack all-ins

[Aisthesis]

This is another [0,1] game, and one I have been unable to solve in all but the very simplest case. If we can solve it, it would no doubt also be worth discussing the limitations of the model itself, but the game is this (the intended application should be fairly obvious):

n players, all with stacks of $10. Player A0 is to the left of the button and posts a blind bet of $1. A1, etc. then can move in as first to act, call an all-in or fold.

Unless I'm missing something here, a general solution for n players is going to be hopelessly complicated, and I have my doubts as to how far we can go with n and still get any solution at all.

Just to give you an idea: For n=2, I have a solution. Already on n=3, I'm stumped.

I'll post my solution for n=2 as well as one possible simplified version of the 3-player problem. Hopefully someone can come up with some simplification(s) that will make this thing doable. I do think it would be a really interesting model for study if anyone here can help me solve it!

[Aisthesis]

Ok, if A is the button and B is the blind, the solution isn't that hard (I hope!). Anyhow, here's what I get:

A moves in on [101/121,1]
B calls on [10/11,1]

The value of the game for A is 10/121.

This actually makes sense. A is moving on on roughly the top 20% of holdings. B is calling on roughly top 10%, and A picks up about 1/12 of the BB every hand with this strategy.

[Aisthesis]

Now I'll try to just explain where I'm at on the 3 player game. A is UTG, B is button, and C is BB.

Ok, if A folds, the game simply reduces to the 2-player scenario, so that's already solved.

Now, if we set a as A's threshold for moving in, and b as B's threshold for calling, we now have 2 (presumably different) thresholds for C:

1) If A moves in and B folds, then C will call on [c,1]

2) If A moves in and B calls, then C will call on [d,1]

Clearly d > c, as C is going to have to be very tight about calling all-ins from both opponents.

So, I tried this by indifference equations and indeed get 4 equations with 4 variables, but they're just horribly messy since they're non-linear.

One thing I haven't yet explored is trying the same problem with stack-sizes that might make it easier (like $4 or $3), but I did try this simplification:

Let's just assume that A always moves in (call it a new game where A moves in blind).

The solution for c here is easy, but I won't even bother doing that. The real question is how it sets up the relationships between b and d.

In this case, for d, I get a value of the square root of 19/31 = 0.7958

and a value for b of 0.5238 (I won't even try to write this out combining fractions and square roots).

So, if A is always moving in, then:

B calls on [.5238,1], and
C calls B's all-in on [.7958,1]

Also haven't figured out the value of this game for B and C (presumably B will get the biggest share). But the main thing of interest is just the relative proportions for the call on the part of B and C.

Given a value of a, I don't think the equation is going to be linear, unfortunately, but this should at least allow some kind of linear approximation.

In any case, A is clearly going to be much tighter here in moving in than he was in the 2-player game, and B will call the all-in roughly in the top half of A's all-in range, and C will call both players somewhere in the top 40% of B's calling range. (if B folds, C's calling criterion will also be roughly in the top half of A's range, since he's risking $9 to win $11).

[Aisthesis]

Ok, I eventually just gave up on an exact solution in the 3-player game (any suggestions on how to do this are much appreciated), but I did come up with a way of getting an approximation that should be pretty accurate--and applicable to the larger problem as well.

Basically, I just assumed (inaccurately, but fairly close) that B would call A's all-in on the top half of A's raising range. Similarly for C if B folds. But I also assumed that C would call BOTH all-ins on the top half of B's raising range (which also seemed close enough given the results when A moves in on any hand at all).

Here's what I come up with:

A moves in on [.9535,1]
B calls on [.9767,1]
C calls if B folds on [.9767,1]
C calls if both players are all-in on [.9884,1]

The value for A (UTG) here is $.0325
The value for B (button) is $.0849

It's worth noting that this is a very slight improvement for the button over the 2-player game, which had a value of $.0826

It's also worth noting that A is already extremely tight in the 3-player game in terms of moving in, and he'll just have to get that much tighter with more players involved.

In terms of applications to a poker context, I think one problem is the stack-size assumption. While 10xBB seems rather plausible as a dividing line for moving in, this is generally happening in practice (for me, anyway) when the blinds are big and antes have been introduced.

So, let's say you're UTG with a stack-size of "x" with blinds at 100/200 and antes at, say 10. The pot before betting at a 9-player table is thus 390, or call it 400 as another approximation. The means that if your stack is 2000, the all-in is gaining you 400 rather than just 200.

As a model for this, I'm really thinking it would probably be better to set stack-sizes actually at just $5 here and keep the single blind structure (otherwise, it's going to get hopeless complicated). This will give BB much better odds to call than is the case in practice.

So, better yet, why not set up the game so that there's simply $1 dead money in the pot (BB gets it if everyone folds) and BB only puts up $1 blind. Stack-sizes now back to $10, so that you're risking $10 to gain $2.

That seems to me like at least a reasonable approximation of the kind of tournament situation we're talking about here. This should definitely loosen up all players in terms of taking a stab at the pot.

Any thoughts?

Anyhow, since this is pretty complicated even with my linear approximation method, I'd like to have the problem set up in such a way that it will hopefully shed some light on the plays to make in certain practical contexts.

[Aisthesis]

Ok, after my suggestions on the 3-player game, I think here's the way to set up the game:

No blind at all (this will simplify the math and maybe even make it doable) but $1 dead money in the pot, which simply goes to the button if everyone folds.

Stacks are all at $5 and the same options apply. I'll try working on this one and see what I come up with. I'm still a bit pessimistic about being able to solve it exactly, but I think this one should definitely be doable with some approximations.

[Aisthesis]

In the game with $1 dead money and $5 stacks, I made the following simplifying assumption:

If A (UTG) moves in at a, then anyone at the table will call the all-in with the top half of A's raising range. Due to the other players remaining to act, this isn't quite right, but it's pretty close. Also, the odds one is getting are slightly better than 2:1, which also changes things a bit, but again, it's still pretty close to accurate.

Now at the point a (where A is indifferent), A will ALWAYS lose if anyone calls according to these calling criteria. Hence, in order for A to be indifferent, the probability of everyone folding has to be 5/6. If anyone calls, A loses $5, but if everyone folds, then A has successfully stolen the blinds and wins $1.

So, for n players at the table, we simply have:

5/6 = (1/2 + a/2)^(n-1)

This gives the following solutions for up to 10 players:

2: 66.67%
3: 82.57%
4: 88.21%
5: 91.09%
6: 92.84%
7: 94.01%
8: 94.86%
9: 95.49%
10: 95.99%

The EV for all players is extremely complicated, so I'm not even going to try figuring that. The completely accurate solution for 2 players is 69.44%, which is for my taste close enough to the 66.67% given by the approximation. Anyhow, all of these solutions are going to be a little on the loose side, I'm assuming.

Finally, to translate into poker hands, we need a hand ranking system. But the top ones should stay pretty much the same. The 95% should translate roughly to 99-AA and AQ/AK--unless one wants to quibble over whether 99 is better than AJs. That's pretty re-assuring to me, as those are exactly the hands I'd move in with as shortstack even UTG.

As one gets to around 90% (5-player game), one can start thinking about hands like 55/66, A9s, ATo and KQs.

[parappa]

Just thought I'd mention that I'm reading all these and finding them very entertaining, but I'm not posting because it's way over my head.

Good stuff!

[Aisthesis]

Cool! Thanks for the encouragement! I know when I got off on one of my math binges, it can get a bit difficult to follow (particularly since I'm often struggling myself).

I do hope the "curve" is clear in terms of increasing quality of hands necessary for moving in in the earlier vs. the later positions.

I think that's really the important result here--which also corresponds nicely with similar results obtainable through other methods.

Also, if you or anyone else browsing through the thread can let me know of specific parts or steps in need of clarification, I'd be glad to explain (with the linear simplifications I had to do, the math here actually isn't all that bad).

[rachelwxm]

Parappa,
actually I mentioned this problem to Aisthesis as a direct application to SNG games. We are on the same page. Thanks Aisthesis for great work! [img]/images/graemlins/smile.gif[/img]

[well]

Hey Aisthesis,

here are the values for the tresholds as I found them:

A: .912296
B: .955165 (A all-in) and 101/121
C: .974216 (both A and B all-in), .951763 (only A all-in) and 10/11 (only B all-in).

For the non-rationals here, only approximations were found.

Regards.