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  #31  
Old 05-09-2005, 01:40 AM
eastbay eastbay is offline
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Default Re: Testing ICM -- some questions for discussion

[ QUOTE ]
This thread is in full hijack mode largely because of me, so this will be my last post on the subject.

Yes, I am all too familiar with verification and validation. I understand everything you have said, and I disagree with the points I noted. You think I am completely wrong, and I am fine with that. Life would be boring if we all agreed on everything. All the best. [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]

Then I have no idea what you could disagree with but I don't care enough to find out either.

eastbay
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  #32  
Old 05-09-2005, 01:47 AM
poincaraux poincaraux is offline
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Default Re: Testing ICM -- some questions for discussion

[ QUOTE ]
All you have to do is to construct a scenario where someone with x% of the chips wins something other than x%.

[/ QUOTE ]

Hi,

This sentence is one of those things that *sounds* clear on its face, but is actually pretty murky. I'm going to rephrase some of what you said in words that I understand .. can you tell me if this is what you mean?

The claim: Given two opponents with identical strategies, chip value is linear in a heads-up freezout.

Oops .. I'm going to interrupt here .. we have to mean something specific by "identical strategies" .. if our strategy is to wait for a huge hand and fold a lot with a small stack, but push a lot with a big stack, we're dead in the water, right? Ok, let's assume we can ignore that .. does making the opponents play a good, pure, jam-or-fold strategy work? Intiutively, I think it might. Jamming/calling every hand has to be linear, but it seems a bit too trivial.

Right, back to the question at hand. If chip value is linear, we can put things on a graph where the X axis is the number of starting chips and the Y axis is the percentage of tournaments won if two opponents sit down and play all possible tournaments at this structure (i.e. every possible run of cards).

Two points make a line, so if we can find three non-colinear points, we know the claim is false. Heck, we don't necessarily need to do more than one calculation: we already know that (0,0) and (100,100) are on the line.

So, what you want to do is this:

1) pick a strategy that isn't obviously broken (e.g. the pure jam-or-fold strategy I mentioned earlier).

2) pick a structure that you think might lie off the 45-degree line (a very small stack is probably a good bet, but it must be larger than the blind).

3) run a *ton* of simulations looking for a monte-carlo-type conclusion about y(x).

if you get a value that's off of the 45-degree line in some statistically significant way, and you can convince yourself that your strategy isn't broken, you win.

right? is that what you meant? i'm tired, so i could easily be misinterpreting.
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  #33  
Old 05-09-2005, 02:07 AM
tech tech is offline
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Default Re: Testing ICM -- some questions for discussion

Basically yes, although I think you could just see if the actual percentage of wins matched the expected percentage under linearity instead of plotting.
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  #34  
Old 05-09-2005, 12:58 PM
the shadow the shadow is offline
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Default Re: Testing ICM -- some questions for discussion

1. To poincaraux:

[ QUOTE ]
(a very small stack is probably a good bet, but it must be larger than the blind)

[/ QUOTE ]

Why do you impose the requirement that the small stack must be greater than the big blind in your hypothetical simulation? I have some thoughts myself that suggest why the equity function is non-linear once the small stack is less than the big blind, but would like to hear your reasoning.

2. To everyone:

Assume that we collect a gazillion data points from actual SNGs. Assume that each one is from a separate SNG to avoid any arguments about the independence of the data. Assume that we get the following information:

We have n number of observations where player A's chip stack (Ca) is x fraction of the total chips in play (Ct). At that stack size, player A wins a mean of y fraction of the time with a standard deviation of s.

Assume that x > y. In other words, on average, player A apparently wins less than than what a linear equity formula would suggest.

How large does n need to be for us to say with confidence that y lies off of a linear equity function?

Of course, we have data for not only x=1, but also x=2, x=3, x=4, . . . where 0 < x < 1. What should we do with that data?

The Shadow
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  #35  
Old 05-09-2005, 01:14 PM
schwza schwza is offline
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Default Re: Testing ICM -- some questions for discussion

hey all,

just noticed this thread.

here's an update on my progress (see link in original post)that touches on some things in this thread. i found a shareware program that i'm using to automate datamining. it does 4/hour now, but i can increase it by adding more than one skin and reducing the time between opening and closing stt's.

a 2+2'er has agreed to write a text parser to turn hand histories into data. and i'm in touch with an econ professor at u. md. who is going to work on the data analysis. he said that it is possible to use multiple data points from the same stt using some clever statistical technique. he also said that 250 stt's would be enough to start looking at preliminary models.

we're going to look at heads up first. i'll keep you posted.
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  #36  
Old 05-09-2005, 01:40 PM
the shadow the shadow is offline
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Default Re: Testing ICM -- some questions for discussion

Assume that a drunk is standing 1' to the right of a cliff. It's dark, so he doesn't realize it, but he's walking parallel to a cliff, which runs in an absolutely straight line. If he takes a single step to the left, he falls to his death.

For every step that the drunk takes forward, he lurches 1' either to his left or right. As best as we can tell, he veers left or right absolutely randomly.

The drunk takes 1 step forward every few seconds or so. He never sits down, stands still, or moves backward.

Luckily for him, 10' to the right of the cliff is a shallow ditch filled with soft clover. If the drunk reaches it, he falls down and gets to sleep it off. (When he wakes up, he can start multitabling SNGs all over again. [img]/images/graemlins/grin.gif[/img])

To put it visually, the drunk is walking north towards the top of the screen:

Cliff Safety
C--------S
C--------S
C--------S
C--------S
C--------S
C--------S
CD-------S

What are the chances that the drunk will fall to his death?

What are the chances that the drunk will wake up in a bed of clover?

What does this have to do with whether the equity function for a freezeout SNG is linear or not? (I'm not sure but have some thoughts.)

The Shadow (who is glad that no drunks were actually harmed in this hypothetical)
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  #37  
Old 05-09-2005, 01:51 PM
gumpzilla gumpzilla is offline
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Default Re: Testing ICM -- some questions for discussion

I'm not sure I see how this random walk problem meshes well with HU poker. Perhaps you are suggesting that each step to the right is a double up, and that our intrepid hero will need to double up 10 times to win the tournament?

Let D(x) be the probability of dying starting at 1', 2', 3', . . . 9'. If he can make it to 10', he lives. We can come up with a system of equations for determining the equity of each step:

D(1) = .5 + .5*D(2)
D(2) = .5*D(1) + .5*D(3)
.
.
.
D(9) = .5*D(8)

It looks to me, spending a minute working on this, that D_1 = .9; that is, starting where he does, he makes it to the clover 10% of the time. If we go with the assumption that you intend steps to be doublings, then he wins 10% of the time although he's going to have something on the order of 1 / 2^9 of all chips, so there's a nonlinearity here.

Interesting. I'll have to think about this some more.

EDIT: Okay, here's the flaw in this argument: until such point as the hero has acquired half of the chips, he's always effectively 1' away from the cliffside, so this isn't really an analogous situation; after doubling up, he's either ahead, in which case he's 1' away from the clover, or behind, in which case he's 1' away from death. The intermediate steps can't exist.
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  #38  
Old 05-09-2005, 02:07 PM
eastbay eastbay is offline
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Default Re: Testing ICM -- some questions for discussion

I didn't read carefully but this sounds like a random walk. A random walk in the continuous limit is governed by a diffusion equation. Diffusion analysis of tournament equity has been studied quite a bit.

The solution to the diffusion equation is indeed exactly linear in 1D (HU game).

Here's an exact solution for the 3-player problem:

http://www.math.ucla.edu/~tom/papers...mblersruin.pdf

Bozeman compared ICM to the diffusion solution in a great post quite awhile back. They are surprisingly close. You'll have to search for it if you're interested.

eastbay
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  #39  
Old 05-09-2005, 03:10 PM
eastbay eastbay is offline
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Default Re: Testing ICM -- some questions for discussion

[ QUOTE ]
hey all,

just noticed this thread.

here's an update on my progress (see link in original post)that touches on some things in this thread. i found a shareware program that i'm using to automate datamining.

[/ QUOTE ]

Care to share which one? Keep us posted.

eastbay
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  #40  
Old 05-09-2005, 10:57 PM
the shadow the shadow is offline
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Default Re: Testing ICM -- some questions for discussion

schwza,

Thx for the update.

It occurs to me that when you gather and analyze the data, you need to record seat position in addition to stack sizes. It would not be enough to simply call the short stack "Player A" and the big stack "Player B." If you did that, I would expect to find that Pa (probability of A winning) < Ca/Ct (A's chips/total chips), especially towards the extremes of the range (A down to 25% or less of the chips to pick an arbitrary number). After all, someone has to win the freezeout. If the chips have flowed from A to B, maybe that reflects a skill difference between the two players. If so, maybe B can do more with the big stack than a linear equity function would suggest.

To avoid the problem, assign "Player A" to the lower seat number and "Player B" to the higher seat number. That way, we should still be able to preserve the assumption of equal skills on average between the two positions.

The Shadow
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