Testing ICM -- some questions for discussion

[the shadow]

I've been rereading some threads on testing ICM, including:

eastbay's empirical equity study (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1617812&page=&view=&s b=5&o=),

eastbay's post in ICM: Does it match the results (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=singletable&Number=213 9238&Forum=All_Forums&Words=ICM&Searchpage=0&Limit =25&Main=2139049&Search=true&where=bodysub&Name=56 71&daterange=1&newerval=&newertype=w&olderval=&old ertype=&bodyprev=#Post2139238), and

schwza's computer programmers (x-post) (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2267540&page=&view=&s b=5&o=) .

For those who would like some background on the subject, check out Section 4.1 of my favorite threads (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=singletable&Number=191 8735&fpart=1&PHPSESSID=) list.

It seems to me that the simplest empirical test of a chip modeling hypothesis is whether the equity function -- the relationship between chip value (CEV) and equity ($EV) -- is linear in a heads-up freezeout. Sklansky seems to assert that the relationship is linear (TPFAP p. 151) ("[E]qual players in a symetrical situation must win exactly in proportion to the size of their stacks."). If true, the probability of hero winning equals the hero's chips as a percentage of total chips. If the hero holds 75% of the total chips, the hero's probability of winning is 75%, etc., etc. As I understand, ICM similarly assumes linearity, at least in freezeout format.

eastbay (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1632782&page=&view=&s b=5&o=&vc=1), zephyr (http://archiveserver.twoplustwo.com/showflat.php?Cat=&Number=1169224&page=&view=&sb=5& o=&fpart=all&vc=1), and others have questioned that hypothesis. JNash has proposed a S-Curve Hypothesis (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Number=1009515&page=&view=&s b=5&o=&vc=1) . He conjectures that the equity function is convex (http://www.algebra.com/algebra/lessons/graphing/convex.jpg) for small stacks and concave (http://www.algebra.com/algebra/lessons/graphing/concave.jpg) for big stacks, with the inflection point equaling average stack size. JNash speculates that the S-curve becomes more pronounced as blinds increase relative to the average stack. If true, the probability of hero winning would not necessarily equal the hero's chips as a percentage of the chips in play.

As far as I know, no one has tested whether the equity function is linear in a HU freezeout. While many other aspects of chip modeling are both interesting and await empirical analysis, such as incorporating a skill factor, this aspect strikes me as more tractable and interesting. Focusing on 2-person play would avoid many of the computational and data collection problems involved in 3- or more person play.

Even so, the result would still be interesting. As JNash suggests, under the S-curve hypothesis, it would be correct (+$EV) for the small stack to take a perfect coin flip (50/50 CEV), but incorrect (-$EV) for the big stack to accept the same coin flip. To put it differently, it could also be correct (+$EV) for the small stack to take some -CEV gambles.

So, this reasoning leads me to some questions for discussion. First, can anyone else suggest a better (i.e, simpler and still interesting) aspect of chip modeling to test?

Second, if we were to test this aspect empirically, what data should be collected? It seems to me that it would include the following fields:
Data source
Site (Party, Stars, etc.)
Buy-in ($5, $10, $20, $50, $100, etc.)
Tournament no.
Chips for player A
Chips for player B
Winner (1 for A, 0 for B, or some similar convention)
Are there any other fields that should be collected?

Third, what should be the size of the data set? Should n=500, 500,000, 5 million, etc.?

Fourth, related to the second question, what tests should be run to test the data for linearity or non-linearity? I recall several such tests from my econ days, but recall that they dealt with time series. I am not sure of their application to this problem.

Finally, aside from collecting data, are there other ways of testing the hypothesis of a linear equity function in a HU freezeout?

I have some thoughts of my own, but thought it might be best to pose some questions for discussion. I welcome any suggestions and comments.

The Shadow

P.S. If you find it hard to visualize JNash's S-curve hypothesis, take a look at M.C. Escher's print Concave and Convex (http://ccins.camosun.bc.ca/~jbritton/escher/convex_and_concave.jpg). A simple S-curve suddenly seem a lot easier. /images/graemlins/smile.gif

EDIT: Links edited out to fix legibility problem.

[shejk]

Please edit for readability.

[the shadow]

It was fine when I previewed it. I'm editing it now.

[Blarg]

OMG fix it. Not that I would probably understand it anyway.

[eastbay]

Shadow,

I think it's a good place to start. The difficulty I possibly see is that any deviations from linearity may be small, and since what we'll be looking for is small deltas, it will again be hard to find them (with any confidence) without huge sample sizes.

BUT, I think it's worth trying anyway, and if you do find the S-curve phenomenon, fame and riches await you. Well, not really, but it would be a very cool result nonetheless.

Before we get into questions of sufficient sample sizes, etc, do you have something in mind for how to do the data mining?

eastbay

[the shadow]

I have a few ideas.

First, the best place to start is with data from actual tournaments. The best place to get that data is from online sites. If the data to be collected were clearly defined, one might be willing to cooperate. Poker Room, for example, has posted the limit EV data that you and I have debated before. Since the data that would be collected would not reveal any site-specific or player-specific information, one or more sites might just be willing to help out.

Second, even if no online site were willing to assist, it might be possible to collect the data from a group of players, especially if the data set required for any meaningful analysis would not be insurmountable.

Third, if online sites were not cooperating and sufficient data were not available from players, I would give some thought to taking data from bot v. bot simulations. There are a number of increasingly sophisticated poker bots available on the market. While I would have concerns about the validity of applying bot data to human play, it might be illuminating in this context. For example, if the bot data were consistent with a null hypothesis of a linear equity function, my faith in that assumption would be different than if the data suggested a non-linear relationship.

Fourth, aside from collecting tourney or bot data, there are other ways to attack the problem. For example, online electronic markets (http://www.biz.uiowa.edu/iem/) have been used to predict presidential elections and economic indicators. One way of approaching the equity function would be to ask, how much money would a third-party be willing to pay for a stack size of x in a HU freezeout? There are ways, such as online markets, to determine that answer without having to collect and analyze a gazillion data points.

The Shadow (who appreciates your patience while I fixed the links)

[eastbay]

If you want an "empirical" study then, right, you have to collect the data, and online games is the only viable option. I am doubtful that you could get cooperation from the sites.

I think the best way to do it is with an observed game data mining program. This is a significant undertaking, but has obvious multiple utilities which may justify the "cost" of developing such a capability.

Collecting data from individual players has clear sample bias problems if you are looking for effects which occur in the mean of the pool of all players.

An alternative is with a game model and simulation. I have done this with some quite simple "push/fold" strategies. The results are remarkably linear, but the strategies may not include "real world" elements that generate the deviations from linearity. I have some pretty well-defined ideas about how to improve such simulations to search for the S-curve phenomenon. But any such study is not really "empirical" and there will always be doubts about the sufficiency of the strategy algorithms to capture all of the things that real players do which might generate the deviations from linearity.

Just a quick brain dump.

eastbay

[the shadow]

Thx for your thoughts.

The data mining idea sounds promising, but would be beyond my skills.

As a lawyer, I'm continually surprised by how much information you can get by simply asking. Assume that an online site were willing to assist. What and how much information would you be looking for?

[Nottom]

I actually think the bot vs. bot idea is a very good one, since having the same bot play each other would completely remove any bias in favor of the better player. This would give an excellent control set for any chip ratio you might desire.

I'm not sure how fast a computer could run a bot v bot heads up match, but in theory you could generate a substantial amount of data relatively fast with this method.

[gumpzilla]

[ QUOTE ]
As a lawyer, I'm continually surprised by how much information you can get by simply asking. Assume that an online site were willing to assist. What and how much information would you be looking for?

[/ QUOTE ]

The size of the blinds at the time the match gets to be HU seems like it would be a relevant factor.

One issue that interests me about the data collection is how many data points one should take from each tournament. Let's say I play 20 hands heads up against my opponent; should we use this as 20 data points? My gut feeling is no, we shouldn't, because any deviations from average behavior become magnified.

As an oversimplified, unrealistic example of what I'm talking about, say I get heads up with blinds of 100, my stack being 2000, villain's stack being 8000. Now let's say he's ridiculously tight, so tight, in fact, that he'll fold anything but aces. Now I should win this just about all of the time, and so I'm going to skew the data horrendously if you take multiple data points from this tournament that show my small stack overcoming a major disadvantage and winning.

[eastbay]

[ QUOTE ]
[ QUOTE ]
As a lawyer, I'm continually surprised by how much information you can get by simply asking. Assume that an online site were willing to assist. What and how much information would you be looking for?

[/ QUOTE ]

The size of the blinds at the time the match gets to be HU seems like it would be a relevant factor.

One issue that interests me about the data collection is how many data points one should take from each tournament. Let's say I play 20 hands heads up against my opponent; should we use this as 20 data points? My gut feeling is no, we shouldn't, because any deviations from average behavior become magnified.

As an oversimplified, unrealistic example of what I'm talking about, say I get heads up with blinds of 100, my stack being 2000, villain's stack being 8000. Now let's say he's ridiculously tight, so tight, in fact, that he'll fold anything but aces. Now I should win this just about all of the time, and so I'm going to skew the data horrendously if you take multiple data points from this tournament that show my small stack overcoming a major disadvantage and winning.

[/ QUOTE ]

Intuitively I agree that only one data point is permissible on independence of observations grounds. I'm not sure how to prove it, but I'm not sure that it needs proving.

eastbay

[tech]

If the linearity theory is indeed false, it would not be hard to disprove. You only need to do an analytic or simulated solution for three data points under one set of conditions. If you find a non-linearity in those (which I'm sure can be done given a correct choice of data points), then the equity function must be non-linear. In other words, you basically only need one counterexample.

On the other hand, I would think "proving" the s-curve hypothesis would be quite difficult.

[gumpzilla]

[ QUOTE ]
If the linearity theory is indeed false, it would not be hard to disprove. You only need to do an analytic or simulated solution for three data points under one set of conditions. If you find a non-linearity in those (which I'm sure can be done given a correct choice of data points), then the equity function must be non-linear. In other words, you basically only need one counterexample.

[/ QUOTE ]

I'm not clear on what you mean by this. The whole point of the exercise, as I understand it, is to see how well the analytical models match up against reality. Whether or not some specific model predicts linearity isn't really the question, but it seems to me that you're suggesting doing calculations to determine just that.

[tech]

[ QUOTE ]
The whole point of the exercise, as I understand it, is to see how well the analytical models match up against reality

[/ QUOTE ]

To me, data collection simply tests the assumption of symmetric players. If you assume symmetric players, the problem can be solved analytically (or approximated via simulation).

[gumpzilla]

[ QUOTE ]

To me, data collection simply tests the assumption of symmetric players. If you assume symmetric players, the problem can be solved analytically (or approximated via simulation).

[/ QUOTE ]

In what model? I still think you're missing the point.

[eastbay]

Seconded.

Don't confuse models with reality. The whole point of making measurements is to validate a model, whether it be a relatively simple ICM or a more involved full-scale simulation.

eastbay

[XChamp]

[ QUOTE ]
[ QUOTE ]

One issue that interests me about the data collection is how many data points one should take from each tournament. Let's say I play 20 hands heads up against my opponent; should we use this as 20 data points? My gut feeling is no, we shouldn't, because any deviations from average behavior become magnified.

[/ QUOTE ]

Intuitively I agree that only one data point is permissible on independence of observations grounds. I'm not sure how to prove it, but I'm not sure that it needs proving.

eastbay

[/ QUOTE ]

I don't see how this would create a bias in the data. You would be doing this for every tournament so the fact that you have 20 points of data from one tournament where one player was exceedingly tight is countered by the fact that you have hundreds of points of data from just a few dozen tournaments where players were reasonably loose.

[gumpzilla]

[ QUOTE ]

I don't see how this would create a bias in the data. You would be doing this for every tournament so the fact that you have 20 points of data from one tournament where one player was exceedingly tight is countered by the fact that you have hundreds of points of data from just a few dozen tournaments where players were reasonably loose.

[/ QUOTE ]

But consider that in loose contests, it's likely to end faster, so you will have proportionally fewer data points for the same number of SNGs. I think the simplest thing is just to take one data point per tournament.

[tech]

We agree to disagree.

[gumpzilla]

Can you please explain in the most simple English possible what precisely it is that you're going to do, perhaps by laying out in detail what calculations you think you can perform and what they'll demonstrate?

[tech]

Sure. My understanding is that the issue in question (the linearity of the equity function) is basically the same as saying that someone with x% of chips in a HU freezeout with someone of equal skill will win x% of the time. My point is that if that premise is false, to disprove it does not require extensive data collection. All you have to do is to construct a scenario where someone with x% of the chips wins something other than x%.

That says nothing about the s-curve or any other possible shape for the equity function. But if it can be done, it shows that the function is not linear in all cases.

[eastbay]

[ QUOTE ]
We agree to disagree.

[/ QUOTE ]

What I said is not a matter of opinion.

eastbay

[tech]

[ QUOTE ]
What I said is not a matter of opinion.

[/ QUOTE ]

No, but some of it is incorrect, specifically the part about the whole point of making measurements being to validate a model.

[gumpzilla]

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

[/ QUOTE ]

This is a very vague statement. We can construct scenarios where opponents have equal chips but one always loses, because the other one has AA. This isn't what you mean, but you haven't nailed down what you mean. What eastbay and I are trying to get at is that you can't nail down what you mean in any way that will answer the original question. Any calculation or simulation you make will have to assume certain strategies, certain values of the blinds, etc., and now you're not really talking about heads-up poker, you're talking about a model of heads-up poker.

[eastbay]

[ QUOTE ]
[ QUOTE ]
What I said is not a matter of opinion.

[/ QUOTE ]

No, but some of it is incorrect, specifically the part about the whole point of making measurements being to validate a model.

[/ QUOTE ]

Uh... whatever.

Do you know what ICM is? Do you understand what validating a model means?

eastbay

[tech]

OK. I see what you are saying. My understanding was that Sklansky's assertion was basically that over a large number of trials those things really didn't matter, so long as the players were of equal skill. If this is not what he is saying, then you are correct -- what I said can't be done.

[tech]

[ QUOTE ]
Do you understand what validating a model means?


[/ QUOTE ]

Yes. I have been doing academic research for the last 9 years, developing and validating models and publishing the results. There are lots of reasons to take measurements and lots of ways to validate models. They are not one and the same. Like I said, we agree to disagree.

[eastbay]

[ QUOTE ]
[ QUOTE ]
Do you understand what validating a model means?


[/ QUOTE ]

Yes. I have been doing academic research for the last 9 years, developing and validating models and publishing the results. There are lots of reasons to take measurements and lots of ways to validate models. They are not one and the same. Like I said, we agree to disagree.

[/ QUOTE ]

By definition there is only one way to validate a model: make measurements.

You may be confusing validation with verification or you may use the word in some kind of sloppy way which considers model to model comparison "validation." I don't, I expect anyone who is involved in academic research to know better.

eastbay

[gumpzilla]

For lack of a better thread to put it (this is enough of a triviality that I don't feel it merits its own complete post), here's an argument regarding heads-up linearity. It's not a very complicated argument at all, but it at least backs up the intuition that % stack should equal % win in a particular case. This is the kind of thing that seems very intuitive to me, but I'd never looked at any kind of argument for why it had to be.

The case that I'm considering is simple move-in poker - two players, with stacks x and y (assume without loss of generality that x > y), both go all-in every turn. Since they are pitting two random hands against each other, each has a 50% chance of winning each confrontation. We can now easily show that the equity formula E(x) = x / (x + y) is consistent. Half of the time stack x wins, half of the time stack x ends up in a new game with stacks x - y, 2y. So,

E(x) = .5 + .5 E(x - y) = . 5 (x + y / x + y) + .5 (x - y / x + y) = x / (x + y)

Notice that if we choose some kind of screwy function, like E(x) = 1 if x > y, 0 otherwise, for example, we don't necessarily get this consistency. So it's a strong argument that % stack = % win works for this particular situation. Also, it's important to note that this just assumes a relatively braindead strategy of pushing and calling every time; an actual game is more nuanced, which is where the argument comes in.

[tech]

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. /images/graemlins/smile.gif

[eastbay]

[ 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. /images/graemlins/smile.gif

[/ QUOTE ]

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

eastbay

[poincaraux]

[ 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.

[tech]

Basically yes, although I think you could just see if the actual percentage of wins matched the expected percentage under linearity instead of plotting.

[the shadow]

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

[schwza]

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.

[the shadow]

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. /images/graemlins/grin.gif)

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)

[gumpzilla]

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.

[eastbay]

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/unpublished/gamblersruin.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

[eastbay]

[ 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

[the shadow]

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

[schwza]

it's called auto mouse v1.3 (link (http://www.download.com/Auto-Mouse/3000-2347_4-4376190.html) ). you have to register it for $15 to make it run for more than 60 minutes, so it's not too functional if you don't do that, but you can mess around with it. if anyone does buy it, i'd be happy to share my algorithm for hand mining.

i'd be very interested in trading hand histories for observed STT's (i.e., not STTs that you played in).

(edited to add link, fix some wrong info)

[poincaraux]

[ 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.

[/ QUOTE ]

Well .. if this is ever going to be linear, my intuition tells me that it's going to be linear in the case where both players push/call any hand. So,

- if you have fewer than 1sb in chips, you're in linear world.

- if you're the bb and have fewer than 1bb, you're in linear world.

What if you're the sb and have a short stack? If you assume chip value is linear, you need to have more than 2.8 small blinds in your stack (it's possible that I made a mistake working this out, but it seemed pretty simple) before you can fold the worst possible hand (23o vs. a random hand is 32.3% vs. a random hand). So, it would have to be very nonlinear for there to be a difference with these stack sizes.


As far as the rest of your post: the whole claim is that it's linear heads-up in the absence of skill, right? That'll be hard to get at, since skill probably has something to do with the stack sizes when it gets heads-up. At the very least, it'll have to be datamining random tourneys, rather than looking at your own hand histories (unless I'm wrong, you're a little more skillful than the average opponent /images/graemlins/smile.gif).

This is interesting. I'm looking forward to having some free time to think about it soon.

[the shadow]

[ QUOTE ]
As far as the rest of your post: the whole claim is that it's linear heads-up in the absence of skill, right? That'll be hard to get at, since skill probably has something to do with the stack sizes when it gets heads-up.

[/ QUOTE ]

I agree. I was trying to get at the same point in my suggestion to schwza (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2356175&page=0&view=c ollapsed&sb=5&o=31&vc=1) elsewhere in this thread.

As I think about it further, we may not be able to assume equal skill if we (that is, schwza, who seems to be the one starting the work /images/graemlins/smile.gif)take the results from the last two players in an SNG. After all, wouldn't the better, more skilled player tend, at least slightly, to have more chips once it became HU?

What do you think about taking the chip count at a preset period in a HU tourney, say once the blinds hit level 2 or 3? The skill effect may still be there, but it may be less pronounced.

The best way to do it would be to assign random starting chip amounts to players of equal skill. That way we could be sure that the difference in chips was not due to a difference in skills. That seems to argue for a bot v. bot simulation.

BTW, thanks, but I wouldn't make any assumptions about my skills relative to any player -- they'd be way to easy to disprove. I'm just trying to understand this game.

The Shadow

[schwza]

[ QUOTE ]
After all, wouldn't the better, more skilled player tend, at least slightly, to have more chips once it became HU?


[/ QUOTE ]

hard to say. it could be the case that bad players gamble too much when it's 3-handed and so you usually have a good short stack and a dumb big-stack LAG.

i've been thinking about including poker prophecy data as a measure of skill. i've never used the program, so i'm not sure exactly how it would work.

[gumpzilla]

[ QUOTE ]
i've been thinking about including poker prophecy data as a measure of skill. i've never used the program, so i'm not sure exactly how it would work.

[/ QUOTE ]

You should be aware that the Poker Prophecy data is both incomplete and inaccurate, according to the reports of just about everybody in STT land.

[schwza]

[ QUOTE ]
[ QUOTE ]
i've been thinking about including poker prophecy data as a measure of skill. i've never used the program, so i'm not sure exactly how it would work.

[/ QUOTE ]

You should be aware that the Poker Prophecy data is both incomplete and inaccurate, according to the reports of just about everybody in STT land.

[/ QUOTE ]

bummer. thanks for the heads up. i'm not really concerned about it being incomplete, but how the hell are they inaccurate?

[gumpzilla]

[ QUOTE ]

bummer. thanks for the heads up. i'm not really concerned about it being incomplete, but how the hell are they inaccurate?

[/ QUOTE ]

Why it's inaccurate is beyond me, but there are numerous reports of various posters having records where Prophecy has them playing buy-ins they've never touched, or more ITM finishes than they've actually had, etc. The incomplete is only relevant because it's still a young enough product that you'll find lots of good players with terrible Prophecy records and vice versa, or so I'm given to believe.

[Apathy]

[ QUOTE ]
I think the best way to do it is with an observed game data mining program. This is a significant undertaking, but has obvious multiple utilities which may justify the "cost" of developing such a capability.


[/ QUOTE ]

Why not talk to people that are already set up to do this? If you are really interested in doing this kind of study I'm sure you could talk to the people at Poker Prophecy and cut them in or give them some money. They are pretty greedy so I don't think they can turn down money, and they are already set up better then anyone I can think of to do what you're talking about.

[eastbay]

[ QUOTE ]
[ QUOTE ]
I think the best way to do it is with an observed game data mining program. This is a significant undertaking, but has obvious multiple utilities which may justify the "cost" of developing such a capability.


[/ QUOTE ]

Why not talk to people that are already set up to do this?


[/ QUOTE ]

Like who?

[ QUOTE ]

If you are really interested in doing this kind of study I'm sure you could talk to the people at Poker Prophecy and cut them in or give them some money.


[/ QUOTE ]

They aren't set up to do this. If they were, their results would be correct. They use the unreliable and insufficient method to just see how many players are at the table. We would need full observed HH information. Very different task.

eastbay

[Apathy]

That's too bad, I thought it might only be a small adjustment to get them to keep the tables open for the HU matches, but I suppose that would signifigantly take away from their actually business. They were my first thought since they have several dedicated machines that just sift through SNG data 24/7.

[eastbay]

[ QUOTE ]
That's too bad, I thought it might only be a small adjustment to get them to keep the tables open for the HU matches

[/ QUOTE ]

From the description of their method posted here, they don't open tables at all.

eastbay

[Apathy]

[ QUOTE ]
[ QUOTE ]
That's too bad, I thought it might only be a small adjustment to get them to keep the tables open for the HU matches

[/ QUOTE ]

From the description of their method posted here, they don't open tables at all.

eastbay

[/ QUOTE ]

Really? Do they determine who is at the table just by the little name popups on the right when the table is highlited? No wonder this is so inaccurate.

[schwza]

[ QUOTE ]
No wonder this is so inaccurate

[/ QUOTE ]

can you fill me in a little on how they're inaccurate?

[Apathy]

[ QUOTE ]
[ QUOTE ]
No wonder this is so inaccurate

[/ QUOTE ]

can you fill me in a little on how they're inaccurate?

[/ QUOTE ]

Well, since as you can see I don't even know specifically how they collect their data the only thing I can tell you is about my personal observation and the experience of myself and others. I don't know why they are so inaccurate but for whatever reason that have been known to report people playing in SNGs that they never did, reporting a 'loss' as a 'win' and most importantly many players that I talk to on a regualr basis say that their stats seem to be inflated.

Of course the fact that the only get about 25% of SNGs played isn't a big deal if their sample is still properly recorded and random, but all evidence seem to point to it not beng random samples that are recorded.

[the shadow]

The owner -- Adam Schultz, who posts here as AdamSchultz (http://forumserver.twoplustwo.com/dosearch.php?Cat=&Forum=All_Forums&Name=26798&Sear chpage=0&Limit=25&) -- has made several posts acknowledging some data collection problems and bugs, but asserting that those problems have been corrected and offering refunds to dissatisfied customers. According to Adam (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=singletable&Number=196 1114&Forum=,All_Forums,&Words=&Searchpage=0&Limit= 25&Main=1960747&Search=true&where=&Name=26798&date range=&newerval=&newertype=&olderval=&oldertype=&b odyprev=#Post1961114), at least one problem apparently concerns PP incorrectly listing 3-player games that actually have 4 players. Other posters complain about continuing incorrect information. I don't own, haven't used, and get no commission on Poker Prophecy, so I don't know from personal experience.

That said, if Adam or his colleagues are paying attention, I, for one, would welcome any assistance that they could offer on this subject.

The Shadow

[hansarnic]

[ QUOTE ]
We can now easily show that the equity formula E(x) = x / (x + y) is consistent. Half of the time stack x wins, half of the time stack x ends up in a new game with stacks x - y, 2y.


[/ QUOTE ]

With you so far.

[ QUOTE ]

So,

E(x) = .5 + .5 E(x - y) = . 5 (x + y / x + y) + .5 (x - y / x + y) = x / (x + y)


[/ QUOTE ]

You lost me here but it looks mightily impressive and, I'm sure, is correct. So the question I have is how much more complicated is this formula is to run for 3 stacks?

[gumpzilla]

[ QUOTE ]

You lost me here but it looks mightily impressive and, I'm sure, is correct. So the question I have is how much more complicated is this formula is to run for 3 stacks?

[/ QUOTE ]

I'll explain in English what it is that I did. We're assuming that your chances of winning with x% of the chips are x%. In other words, your equity (we'll assume winner takes all, which we can always reduce to by subtracting out second place) is x / (x+y), the fraction of the total chips that you possess. But, when we're playing move-in poker, we can easily determine the equity of the game another way. There are only two outcomes, and the equity of the game that we're playing must be related to the equity of those two outcomes. With two random hands going against each other, I will win contests 50% of the time (I'm neglecting splits but I don't think they really make too much of a difference here) and lose 50% of the time. I'll also assume that I have the bigger stack (x > y).

So when I win a particular hand, I've won the entire game. When I lose, I still have some chips left, x - y specifically. So, I can determine that the equity with x chips is just .5 + .5 * equity with x-y chips. If we now plug in our guess that tournament equity is just the total fraction of chips, we find that plugging this in on both sides of the equation leads to agreement. This suggests (but I don't think it proves) that this is probably the right form for the equity in this particular game.

As for 3 handed, this approach gets very hairy. First off, we need to decide how the game will be played. The simplest is just to say that everybody moves in on every hand, but this is a far worse assumption than in the heads-up case, because three handed play is going to look very different from this, whereas headsup play can frequently degenerate into a pure pushfest. However, even here you start encountering difficulties that make the problem substantially uglier. The one that I'm thinking of in particular is that you now have to start worrying about side pots. But, let's take a look: assume x >= y >= z, and that all three players push all-in each hand. As before, we ignore splits. Does E(x,y,z) = x / (x + y + z) lead to consistency?

1/3 time x wins main pot, and it's over.
1/3 time y wins main pot, which means that y must also win any side pot that exists: in this case the new E = E(x-y, 2y + z, 0)
1/3 time z wins main pot: this needs to be broken down into two cases:
x wins side pot: E(x + y - 2z, 0 , 3z)
y wins side pot: E(x - y, 2y - 2z, 3z)

It's ugly, but you can verify for yourself that if you weight everything properly and sum it up you do find that assuming E(x,y,z) = x / (x + y + z) for player x in this situation is a consistent solution.

However, I must stress again that this is about as far as you can get from real three-handed poker, so it's not a very meaningful result. One way to go about making it more realistic is to only allow for one player calling you; by this we could perhaps simulate the button always pushing and the BB always calling, which is still unrealistic but is closer to reality than the original model. However, it at least naively appears that position is important in such a setup. It might end up that it is not important, but that would be at least relatively non-trivial.

[poincaraux]

[ QUOTE ]
This suggests (but I don't think it proves) that this is probably the right form for the equity in this particular game.

[/ QUOTE ]

Satisfying that equation is necessary, but not sufficient. For example, I think E(x) = {x>y:1; x<y:0} also works (not saying it's realistic, just that there are other solutions to that equation).

[gumpzilla]

[ QUOTE ]
Satisfying that equation is necessary, but not sufficient. For example, I think E(x) = {x>y:1; x<y:0} also works (not saying it's realistic, just that there are other solutions to that equation).

[/ QUOTE ]

The example you cite is one that I specifically pointed out in an earlier post does not work:

Say x = .6. E(x) = .5 + .5*E(.2) => 1 = .5. No good.

I agree about the necessary but not sufficient, thus my comment that this doesn't prove anything. The point was just to make a reasonably compelling argument about why one might even think that % stack = % win, and I think it does nicely on that front.

[Pokerscott]

[ QUOTE ]

What do you think about taking the chip count at a preset period in a HU tourney, say once the blinds hit level 2 or 3? The skill effect may still be there, but it may be less pronounced.
The Shadow

[/ QUOTE ]

I think you are concerned with the skill effect relative to the probablity of winning given the current chip counts. It may be true that the more skilled players are likely to have more chips when it hits heads up, but the question is will that skill delta (over the second most skilled player on average) equate to a larger win probability, when compared to the chip share.

You are basically testing ICM from headsup to completion. I think that minimizes the impact of skill. The earlier you go into the tournament the more impact skill will have on your analysis. For example, everyone starts at 10% of the chips, but skill will clearly make certain players win more that 10% of the time.

Someone that enters heads up with 60% of the chips may indeed be better on average (debateable given the need to be tight early as a skilled player), but all the early skill has been captured in the 60% and thus doesn't impact the analysis.

Pokerscott

[Pokerscott]

[ QUOTE ]
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.


[/ QUOTE ]

There are certainly statistical techniques that can deal with data clustering (which is the problem you would have using multiple data points from the same SnG). However, all of those techiniques are going to require additional assumptions in order to properly correct for the clustering. I would strongly encourage you to at least perform the analysis using only the data assuming one data point per SnG (and the data point that corresponds to the first headsup situation seems like one that is as good as any).

It would be interesting to see if the cluster analysis corroborates the results though,

Pokerscott

[gumpzilla]

I have a better argument for the linearity of heads-up move-in poker now. It's a little less user-friendly to the 2+2 masses than the first one, but it's more mathematically convincing. The only assumption I'll make is that the equity function f(x) is an analytic function of x.

For e < .5, we know f(1 - e) = .5 + .5 f(1 - 2e). We only need to worry about e < .5 because we can clearly see that for e > .5, f(1 - e) = 1 - f(e) = 1 - f(1 - e') for some e' < .5. Taylor expanding both sides yields

1 - f'(1) e + (1/2) f''(1) (e^2) + . . . = 1 - f'(1) e + f''(1) (e^2) + . . .

Equating powers of e, we see that all derivatives above the first must be equal to zero if this is to be equal. Since we know the boundary conditions that f(0) = 0, f(1) = 1, this means that f(x) = x / (x+y).

Yes, I'm still making an assumption that f(x) is analytic, but I think this is a highly reasonable assumption to make.

[the shadow]

I raised the drunken walk issue simply because I was wondering whether the equity function was linear once the small stack's chips were less than the small blind. My intuition suggested that if a player was down to 1 chip, his or her equity might be < E(SB) * (1/SB) where E(SB) is the equity of the small blind and SB is the amount of the small blind. My intuition was wrong.

As you and eastbay suggest, the one-dimensional drunken walk I posited is a linear relationship. Elsewhere in this thread, poincaraux (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2359999&page=0&view=c ollapsed&sb=5&o=31&vc=1) made the same point where the small stack is less than the small blind (and maybe even when it's less than 1.5xBB). You guys are way faster at math than I. /images/graemlins/cool.gif

For anyone else interested in the subject, google one-dimensional random walks with absorbing boundaries. Also take a look at the math dealing with gambler's ruin (which more players should pay attention to, given all the posts we have about whether their bankroll is enough). Both of those forumulas apply here and both are linear.

Not only does the math show that the equity function is linear where the small stack < SB, but so does common sense. Assume that the small blind is 200 and the small stack is 100. Since the small stack is automatically all-in, the probability of winning is a perfect coin flip. That means that it's equity is half of a stack equal to the SB, which is a linear relationship.

Assume that there're 1000 chips in play. This doesn't prove that E(200) = 200/1000 = 20%, but it does show that E(100) = E(200)/2.

The Shadow

[schwza]

[ QUOTE ]
f(x) is analytic

[/ QUOTE ]

what does that mean /images/graemlins/confused.gif

[gumpzilla]

It means that I am allowed to expand it in a Taylor series, which basically says that it's a nice, differentiable, smooth function. This is a bad assumption if you think that the equity function is going to have all kinds of jumps and discontinuities in it. However, for the kind of poker I described, I think it's not a bad assumption. My chances of winning if I have .5001 of the chips should be just slightly better than if I have .5, since I'm still alive the half of the time I lose the first hand but am going to have virtually no chance of winning since I have to double up a bunch of times just to get back to even and will lose if I miss out on even one of those.

[Pokerscott]

[ QUOTE ]
It means that I am allowed to expand it in a Taylor series, which basically says that it's a nice, differentiable, smooth function. This is a bad assumption if you think that the equity function is going to have all kinds of jumps and discontinuities in it.

[/ QUOTE ]

Nothing other than intuition, but it seems like there may be a jump in the function when the little stack gets above 3xBB and is able to effectively steal blinds again from the big stack.

Hmm there may not be an asymetry if both players can no longer steal when one stack is <3xBB. hmmmm.

Pokerscott

[gumpzilla]

[ QUOTE ]

Nothing other than intuition, but it seems like there may be a jump in the function when the little stack gets above 3xBB and is able to effectively steal blinds again from the big stack.

[/ QUOTE ]

My argument applies specifically for the case where both players push all-in every hand. The closest actual situation that this could model is something like "average stack = 5 BBs"; in other words, a situation where it is commonly accepted that you are now playing crapshoot poker. If you make the strategy at all more complicated, the basic formula f(1 - e) = .5 + .5*f(1 - 2e) becomes highly, highly nontrivial to produce, I would assume. In other words, it becomes harder to put useful delimiters on what the equity function has to look like because there are more outcomes. I'll try and put together a quick example of what I'm talking about.

[poincaraux]

[ QUOTE ]
The example you cite is one that I specifically pointed out in an earlier post does not work

[/ QUOTE ]

Yeah, well, at the very least, you should have listed that example right after you gave your equation so that I couldn't miss it. Oh, wait, you did. Oops. /images/graemlins/blush.gif

I'll refrain from posting any more on this thread until I actually have time to *read* the other posts.

[marv]

I've been mulling over this question for ages (if two people play an optimal strategy for HU freezeout tourney, is the equity function linear).

One observation: If each game in the match were fair (each player's expected chip gain/loss before they see their hole cards is zero) then the linear equity function is true, because the bankrolls will follow an unbiased random walk (this is provable).

But the games aren't fair in two ways:
1) the small blind has a positional advantage.
2) the big blinder may be forced all-in while the small blind will still have the option to fold.

Bias 1 favours the big stack, since the advantage arises from playing second on later rounds. With the small stack you are more likely to be all-in before then and thus lose this advantage.

Bias 2 may actually be zero in holdem with sane blinds since the small blind should probably *always* call a big blinder who is all-in on their blind. Otherwise, it favours the big stack since he's the one who is less at risk of being blinded all-in in subsequent hands.

So the small stack is playing a -ev game which gets progressively worse as he approaches the point of being blinded all-in. I think this implies the S-shaped equity function.

Also, if the linear hypothesis were true, then the optimal strategy for each player would be the same whether they are the small stack or the large one - which goes against the wisdom that the big stack should try to 'bully' the small stack - force him into a passive game and let the -ev eat him up.

Marv

[marv]

[About the idea that the small stack is playing a -ev game and this leads to the non-linear equity function]

[ QUOTE ]

But the games aren't fair in two ways:
1) the small blind has a positional advantage.

[/ QUOTE ]

Arrh. I've just realised this effect is independent of stack sizes since the big stack's extra chips play no part in the game one both players have posted blinds.

So the only way I can see the game not being fair is when the big blind is blinded all-in (and even that effect will be zero if the small blind should always call).

Marv

[eastbay]

[ QUOTE ]
I've been mulling over this question for ages (if two people play an optimal strategy for HU freezeout tourney, is the equity function linear).

One observation: If each game in the match were fair (each player's expected chip gain/loss before they see their hole cards is zero) then the linear equity function is true, because the bankrolls will follow an unbiased random walk (this is provable).


[/ QUOTE ]

What's provable? That unbiased random walk leads to a linear function? Or that "if the each game in the match is fair" (you mean hand in the tournament?) it is equivalent to unbiased random walk?

The first is obvious and the second seems to be a big assumption.

There's a difference between the tournament having zero value (modulo button initialization) and _each hand_ having zero value. I don't have any reason to believe that chip disparity can't lead to non-zero value from that point forward. That big stack can threaten the tournament for little stack but not vice-versa seems like one mechanism by which this could be the case.

eastbay

[marv]

[ QUOTE ]
[ QUOTE ]
I've been mulling over this question for ages (if two people play an optimal strategy for HU freezeout tourney, is the equity function linear).

One observation: If each game in the match were fair (each player's expected chip gain/loss before they see their hole cards is zero) then the linear equity function is true, because the bankrolls will follow an unbiased random walk (this is provable).


[/ QUOTE ]

What's provable? That unbiased random walk leads to a linear function? Or that "if the each game in the match is fair" (you mean hand in the tournament?) it is equivalent to unbiased random walk?

The first is obvious and the second seems to be a big assumption.

There's a difference between the tournament having zero value (modulo button initialization) and _each hand_ having zero value. I don't have any reason to believe that chip disparity can't lead to non-zero value from that point forward. That big stack can threaten the tournament for little stack but not vice-versa seems like one mechanism by which this could be the case.

eastbay

[/ QUOTE ]

I meant the first (the obvious one) was provable. (You're correct that I meant each hand in the tournament, rather than the tournament as a whole.)

If each hand in the tournament is fair, then player 1's bankroll will follow an "unbiased random walk" in the sense that the change in his number of chips over each hand may have a complex distribution dependent on both player's strategies, but it will have mean zero, so we have our random walk.

I accept that alternating blinds and the button advantage may make a hand unfair - this complicates things. But if we had the players flip a coin for the button at the start of each hand, then unless the big stack's extra chips give him an an +EV edge on some particular hand, we must have the linear equity function. If we just played with antes and no blinds for example, I don't see that he'd ever have an edge?

Marv

[eastbay]

[ QUOTE ]
unless the big stack's extra chips give him an an +EV edge on some particular hand, we must have the linear equity function. If we just played with antes and no blinds for example, I don't see that he'd ever have an edge?

Marv

[/ QUOTE ]

Right, but isn't that the whole question (does big stack have an advantage)? I think it's plausible that a big stack may have an edge for (at least) the reason I stated - he can take risks the small stack cannot without busting.

We're of course talking about a freezeout game and not a cash game.

eastbay

[the shadow]

For the moment, let's assume that the big stack has an advantage that is based upon the fact that he or she has more chips, but is greater than his or her fraction of the chips in play. When would that effect be the least and when would it be the greatest?

It seems intuitive that the effect would decrease as the chips approached 50/50 and would virtually (but perhaps not entirely) disappear at 51/49. After all, if the big stack tried unsuccessly to bully the small stack, he or she would be virtually busted if he or she lost the push.

The effect would vanish once the small stack was less than the small blind. In light of poincaraux's observation (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2359999&page=1&view=c ollapsed&sb=5&o=31&vc=1), the effect would likely fade away once the stack stack fell below about 1.5 big blinds.

It seems to me that the effect would be greatest around 75/25 and perhaps above. At that point, if the big blind pushed and lost, at worst he or she would be back to even; however, if the small stack lost, say good-bye.

I'm aware of the work suggesting that the optimal push-call percentages change as the ratio of stack size to big blind decreases. Is there any analysis that might suggest that the big stack should change his or her strategy as the ratio of the two stacks varies? If so, that might support this speculative effect.

The Shadow

[marv]

[ QUOTE ]
[ QUOTE ]
unless the big stack's extra chips give him an an +EV edge on some particular hand, we must have the linear equity function. If we just played with antes and no blinds for example, I don't see that he'd ever have an edge?

Marv

[/ QUOTE ]

Right, but isn't that the whole question (does big stack have an advantage)? I think it's plausible that a big stack may have an edge for (at least) the reason I stated - he can take risks the small stack cannot without busting.

We're of course talking about a freezeout game and not a cash game.

eastbay

[/ QUOTE ]

Understood.

I take issue with the idea that the big stack's ability to 'force the small stack to play for the tournament' actually offers any advantage at all:

In my example of a freezeout tournament with antes but no blinds and where the players flip a coin for the button at the start of each hand, let's imagine we're the big stack at the start of a hand (before the coin flip). Here we will still be able to force oppo to 'play for the tournament', but our extra chips have no bearing on the up-coming hand as we'll never get to use them. Thus the hand we're about to play is a symmetrical game and we have no edge(*) in it. (Hence our stack size during the course of the tournament does an unbiased random walk and we get a linear equity curve.)


Marv

(*) By 'edge' I mean E(C1)-C0, where C1 is the number of chips we have at the end of the hand and C0 is the number at the start, before the coin flip.

[eastbay]

[ QUOTE ]

In my example of a freezeout tournament with antes but no blinds and where the players flip a coin for the button at the start of each hand, let's imagine we're the big stack at the start of a hand (before the coin flip). Here we will still be able to force oppo to 'play for the tournament', but our extra chips have no bearing on the up-coming hand as we'll never get to use them. Thus the hand we're about to play is a symmetrical game and we have no edge(*) in it.

[/ QUOTE ]

We do get to "use" them, in the sense that they prevent us from getting frozen out if we lose. This is the key point. Our N chips at risk in the hand have less value than small stack's N chips. At least, this seems plausible to me.

eastbay

[the shadow]

[ QUOTE ]
For e < .5, we know f(1 - e) = .5 + .5 f(1 - 2e).

[/ QUOTE ]

gumpzilla, bear with me, but does we really know this?

As I understand the formula, it says that the equity of the big stack -- f(1 - e) -- equals half of the equity in the freezeout -- 0.5 -- plus half of the equity left after subtracting the small stack and the big stack's chips equal to the small stack -- .5 f(1 - 2e).

e = the equity in the small stack
f(1 - e) = the equity in the big stack
0.5 = half of the equity in the freezeout
f(1 - 2e) = the equity left after substracting the small stack twice

Am I correctly following you so far?

If so, then do I understand correctly that you are attributing to the big stack half of the equity in the amount of chips by which the big stack has the small stack covered?

If my understanding of your formula is correct, then it seems to me that your formula assumes the conclusion that you reach. I'm inclined to agree that the equity function is linear if the big stack gets only "half credit" for his or her extra chips.

But what if the big stack gets more than "half credit" or maybe even "extra credit" for the extra chips? What if the "extra credit" that the big stack gets increases as the amount of the extra chips grows?

That's what I was saying (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2375078&page=0&view=c ollapsed&sb=5&o=31&vc=1). It also seems to be a logical inference from what eastbay was suggesting (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2378340&page=0&view=c ollapsed&sb=5&o=31&vc=1).

To put if differently, isn't it the case for a non-linear S-curved equity function f(1 - e) would not necessarily equal .5 + .5 f(1 - 2e)?

I'm interested in your thoughts.

The Shadow

[gumpzilla]

[ QUOTE ]
[ QUOTE ]
For e < .5, we know f(1 - e) = .5 + .5 f(1 - 2e).

[/ QUOTE ]

gumpzilla, bear with me, but does we really know this?

[/ QUOTE ]

It's important to note that I am talking about an extremely limited form of HU poker, where both contestants agree to push all-in every hand. Given this, and ignoring splits, you're flipping a coin to decide who wins the hand, essentially. Where my formula comes from, then, is noticing that when the big stack (1 - e) wins a hand, he wins the tournament because he has all of the chips. When he loses, he now has (1 - 2e) chips and whatever equity goes along with that. Thus, f(1 - e) = .5 f(1) + .5 f(1 - 2e).

It's crucial that they are just pushing in every hand, as this leads to the coinflip between winning and losing. This is also a symmetric way to play, or fair to use the term that marv uses. Because of this, we don't have to worry about who is in what seat. If we did, then things would be more complicated, obviously, and this formula wouldn't be so simple.

So the result I posted is of little direct relevance; as I think I mentioned somewhere, you can maybe use it to argue that with super-high blinds you should have linearity, because pushing every hand is going to be pretty close to correct play in those circumstances, but that's about it for real applications.

One thing that I think would be sort of interesting to look at are the ways in which linearity is broken when you introduce "skill" asymmetries between the players. To use a very unrealistic example, let's say that both players will push and call with any two, but player A pushes from the SB/ calls from the BB every time, and player B flips a coin to decide whether to push/call or fold. Player A's strategy is obviously superior to player B, because in this example player B is "tightening up" without any resulting gain in equity in the situations where they do clash, so the occasional folding can only help player A. Consequently, player A should always be able to do better than the move-in game I described earlier. Graphically, you can see that what this means is that for short stacks he must grow faster than linear and for large stacks slower than linear. Here's a crappy drawing of what I mean:

http://img26.echo.cx/img26/5686/superlinear9jh.png

The line is the unbiased game, the curve is an idea for what I think player A's equity function would look like in the biased game. Also interesting is that since f(A) + f(B) = 1, this means that B grows more slowly than linear for small stacks and faster than linear for big stacks (I think, correct me if I'm wrong here.)

The reason that I think this kind of thing is interesting is that it's probably much more representative of low-level SNGs (where there are likely to be substantial skill differences) than trying to figure out how the equity function behaves against two opponents playing optimally.

[marv]

[ QUOTE ]
[ QUOTE ]

In my example of a freezeout tournament with antes but no blinds and where the players flip a coin for the button at the start of each hand, let's imagine we're the big stack at the start of a hand (before the coin flip). Here we will still be able to force oppo to 'play for the tournament', but our extra chips have no bearing on the up-coming hand as we'll never get to use them. Thus the hand we're about to play is a symmetrical game and we have no edge(*) in it.

[/ QUOTE ]

We do get to "use" them, in the sense that they prevent us from getting frozen out if we lose. This is the key point. Our N chips at risk in the hand have less value than small stack's N chips. At least, this seems plausible to me.

eastbay

[/ QUOTE ]

But we can't use the extra chips on that hand, so that hand (in isolation, forgetting about the subsequent hands in the tournament) is a symmetrical 'game'.

Next hand (if there is a next hand) will be another 'game', probably a different one since the depth is likely to be different, but that too will be symmetrical.

This is all we need for linearity, that on each hand (considered in isolation) neither player has an edge.

Marv

[the shadow]

[ QUOTE ]
It's important to note that I am talking about an extremely limited form of HU poker, where both contestants agree to push all-in every hand.

[/ QUOTE ]

I see now. Thanks. I agree that if the hand is decided by a perfect coin flip, then the equity function will be linear. That's true where stack stack < small blind, where small stack < abt 1.5 big blind, and where both players push every hand. I'm now inclined to believe that the same is true of a real-life HU freezeout poker tourney, as long as the skill of the two players is identical, for example, where two players have a life-time HU win-loss record against one another of 31415 wins v. 31415 losses (or something statistically equivalent).

I'm think you're heading in the right direction when you suggest that the shape of the equity function depends on the relative skill of the two players. This harkens back to what eastbay was working on -- building a skill factor into ICM (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=singletable&Number=183 8524&Forum=All_Forums&Words=skill&Searchpage=0&Lim it=25&Main=1838524&Search=true&where=sub&Name=5671 &daterange=1&newerval=&newertype=w&olderval=&older type=&bodyprev=#Post1838524).

I'll give some more thought to your points. In the meantime, you may have seen the spat between Negreanu and Sklansky. Negreanu challenged Sklansky to a HU freezeout and offered 11-10 odds. Sklansky countered with 5-4 odds (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=tv&Number=2332808&Foru m=All_Forums&Words=&Searchpage=0&Limit=25&Main=229 8516&Search=true&where=sub&Name=5&daterange=1&newe rval=&newertype=w&olderval=&oldertype=&bodyprev=#P ost2332808). Assuming that those odds (or something in between) accurately reflect their relative skills, is that info sufficient to sketch an equity function for such a match? How would we do so -- adding chips to Negreanu's stack a la eastbay, assuming that p -- the probability of Negreanu winning each hand -- is 0.50 + x, or some other way? What would the equity function look like?

The Shadow

[the shadow]

I'm not sure I agree. Isn't that asking us to ignore the man behind the curtain?

If your argument were correct, why wouldn't it equally apply in a SNG? At the bubble, the big stack has 5,000 chips, the two medium stacks have 2,000 chips, and the small stack has 1,000 chips. We don't ignore the big stack's extra chips then. Why should we HU?

The Shadow

[marv]

[ QUOTE ]
I'm not sure I agree. Isn't that asking us to ignore the man behind the curtain?

If your argument were correct, why wouldn't it equally apply in a SNG? At the bubble, the big stack has 5,000 chips, the two medium stacks have 2,000 chips, and the small stack has 1,000 chips. We don't ignore the big stack's extra chips then. Why should we HU?

The Shadow

[/ QUOTE ]

With 3+ players you won't always have the symmetry I used to guarantee each hand is 'fair'.

In your example, with stacks of 1, 2, 2 and 5, if we randomize seats and button position at the start of each hand (to avoid the complications of position - which is a much bigger change in 3+ players than with 2 anyway) then the next hand is effectively played with stacks of 1,2,2,2 .

This isn't symmetrical, so the small stack may be at a disadvantage (in terms of EV in chips) on this hand. And given the small stack may be at a disadvantage, the middle stacks must worry about becoming the small stack on later hands - when they suffer.

But with only 2 players, on every hand (with randomized button again), it's always symmetrical so each player can play a strategy which gets at least 0 EV in chips over that hand, and hence gets them at least the equity that the linear model suggests - so no S-shaped equity function.

There's another complication with 3+ players. The variance of our opponents effects our equity. In a 2 player game, the variance of each 'step' in the random walk our stack follows doesn't effect the linear equity function (provided every step has mean 0), but with 3 players, insanely loose agressive opponents increase our equity (for any sensible payout structure)

Marv

[eastbay]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

In my example of a freezeout tournament with antes but no blinds and where the players flip a coin for the button at the start of each hand, let's imagine we're the big stack at the start of a hand (before the coin flip). Here we will still be able to force oppo to 'play for the tournament', but our extra chips have no bearing on the up-coming hand as we'll never get to use them. Thus the hand we're about to play is a symmetrical game and we have no edge(*) in it.

[/ QUOTE ]

We do get to "use" them, in the sense that they prevent us from getting frozen out if we lose. This is the key point. Our N chips at risk in the hand have less value than small stack's N chips. At least, this seems plausible to me.

eastbay

[/ QUOTE ]

But we can't use the extra chips on that hand, so that hand (in isolation, forgetting about the subsequent hands in the tournament)


[/ QUOTE ]

Well, that's the whole point. I don't think you can forget about the rest of the tournament and arrive at a legitimate conclusion.

The hands in a tournament are just turns in the game, they are not independent games themselves.

eastbay

[marv]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

In my example of a freezeout tournament with antes but no blinds and where the players flip a coin for the button at the start of each hand, let's imagine we're the big stack at the start of a hand (before the coin flip). Here we will still be able to force oppo to 'play for the tournament', but our extra chips have no bearing on the up-coming hand as we'll never get to use them. Thus the hand we're about to play is a symmetrical game and we have no edge(*) in it.

[/ QUOTE ]

We do get to "use" them, in the sense that they prevent us from getting frozen out if we lose. This is the key point. Our N chips at risk in the hand have less value than small stack's N chips. At least, this seems plausible to me.

eastbay

[/ QUOTE ]

But we can't use the extra chips on that hand, so that hand (in isolation, forgetting about the subsequent hands in the tournament)


[/ QUOTE ]

Well, that's the whole point. I don't think you can forget about the rest of the tournament and arrive at a legitimate conclusion.

The hands in a tournament are just turns in the game, they are not independent games themselves.

eastbay

[/ QUOTE ]

OK, you're still not convinced so I'll try a slightly different tack:

First, note even if both players play an identical strategy (so they're certainly equally skilled), we may *not* get a linear equity function:

Example: suppose each player looks at his chips at the start of the hand, if it's <25% of the total, he just folds at his first action. If its 25%-75% he calls at every opportunity and if its >75% he makes a min raise at his first opportunity.

For these players the equity function looks like _/- it starts at 0, is horizontal up to 25%, then rises linearly to reach 1 at 75% and is then horizontal again until 100%.

So for linear equity we need some assumption about the players' common skill level.

Now suppose both players play the entire tournament optimally (in the game theory sense, so they each know each other's playing tendencies perfectly, in all situations and so play unexploitably).

Each player can guarantee an EV of at least 0 over each hand, as each hand (in isolation) is a symmetric game (I'm assuming the button is randomized). Thus each player can guarantee themselves least the equity given by the linear equity function, by just adopting the optimal strategy over each hand for all the remaining hands in the tournament.

But if player 1 can guarantee himself an equity of x% with x% of the chips, then the most player 2 can hope to get is (100-x)%. But he can also guarantee at least this by adopting an optimal strategy over each hand too, hence he should do this, and with (100-x)% of the chips, his equity is exactly (100-x)%, and we get the linear equity function.

Marv

[the shadow]

I've been thinking a bit more about your graph.

It may be trivial, but it seems to me that in the case of a HU freezeout, if the skills of the two players differ, the equity of player A cannot equal that of player B when the chips are equal. Do you agree and what can we infer from this?

The Shadow

[holdem2000]

I think bot data would be far better than human data for this - you're looking for situations in which player A and player B have identical strategies (which can of course vary with stack size, but the whole strategy for various chip sizes must be the same).

[eastbay]

[ QUOTE ]

OK, you're still not convinced so I'll try a slightly different tack:

First, note even if both players play an identical strategy (so they're certainly equally skilled), we may *not* get a linear equity function:

Example: suppose each player looks at his chips at the start of the hand, if it's <25% of the total, he just folds at his first action. If its 25%-75% he calls at every opportunity and if its >75% he makes a min raise at his first opportunity.

For these players the equity function looks like _/- it starts at 0, is horizontal up to 25%, then rises linearly to reach 1 at 75% and is then horizontal again until 100%.


[/ QUOTE ]

Yes, I've pointed variations of this out as a counterexample to DS's "proof" many times.

[ QUOTE ]

So for linear equity we need some assumption about the players' common skill level.

Now suppose both players play the entire tournament optimally (in the game theory sense, so they each know each other's playing tendencies perfectly, in all situations and so play unexploitably).


[/ QUOTE ]

That's not what game theory optimal is. No knowledge of the opponent's tendencies is required for game theory optimal play. In fact, that's kind of the whole point of game theory optimal - that you don't need to know anything about your opponent's tendencies to not lose to him. That's just an aside here.

[ QUOTE ]

Each player can guarantee an EV of at least 0 over each hand, as each hand (in isolation) is a symmetric game (I'm assuming the button is randomized).

[/ QUOTE ]

Grr.

There is no reason to believe that optimal play for a single hand (to achieve 0 chip EV) is optimal play for the tournament. This is something you keep assuming and it's still unsubstantiated.

Giving up chip EV on one hand may allow the finding of a bigger edge later, producing better results for the tournament, even if it gives up something on some particular hand. The tournament is the game. The hands are not.

Let me repeat:

You have given no evidence that optimal play of each hand produces optimal play for the tournament.

Therefore, no conclusions about tournament equity functions for optimal play for the tournament can be derived from optimal play for each hand.

eastbay

[marv]

[ QUOTE ]

[ QUOTE ]

Each player can guarantee an EV of at least 0 over each hand, as each hand (in isolation) is a symmetric game (I'm assuming the button is randomized).

[/ QUOTE ]

Grr.

There is no reason to believe that optimal play for a single hand (to achieve 0 chip EV) is optimal play for the tournament. This is something you keep assuming and it's still unsubstantiated.


[/ QUOTE ]


OK eastbay, here's a proof.

Suppose A and B are playing a HU tournament starting with 50 chips each. They play each hand with a randomized button, and at some point the tournament, at the start of a hand (before they flip a coin for the button) A has x chips, and B has 100-x.

Claim 1: A can play the remainder of the tournament in such a way that

P(A wins the tourney | A starts with x chips) >= x/100

no matter how B plays. A suitable strategy for A is to play each hand using an optimal strategy for the game G which represents the 'hand in isolation'. This is the game which starts with a chance node representing the selection of the button, ends at the end of the hand, and whose payout for each player is simply the number of chips they have at the end of the hand. Different hands will have a different game G, so A's strategy will change from hand to hand.

Claim 2: The equity function for a player who plays optimally for game which represents the remainder of tournament as a whole (i.e. the game which ends when one player has all the chips) lies on or above the linear equity function, and hence the equity function for a pair of players each using an optimal strategy for the remainder of the tournament as a whole is the linear equity function.


Proof of 1
----------

Let ca denote the number of chips A currently has (so ca=x) and cb the number of chips B has (so cb=100-x). Recall that G denotes the 2-player game which amounts to stopping the tourney at the end of the next hand and paying each player in proportion to their chips at that point.

If ca>cb, the extra chips A has play no part in the hand, except to increase the value of the game G by (ca-cb) since the extra chips never end up in B's stack. Similarly if ca<cb, B's extra chips just increase the value of the game in his favour by cb-ca since they'll always be his at the end of the hand.

Thus G is a symmetric game (other than one player getting a constant bonus payoff), so if A plays an optimal strategy for G, we have E(CA) >= ca, however B plays, where CA denotes the (random) number of chips A has at the end of the hand.

Now consider the number of chips A has at the start of each successive hand in the remainder of the tourney if he uses the method just given to dicate his play on each hand. Denote this sequence

CA_0, CA_1, CA_2, ...

Where CA_0 = x since A currently has x chips.

Since E(CA_{n+1} - C_n | all information up to the start of hand n) >= 0, we can write the process CA_n as

CA_n = CA_0 + M_n + Z_n

where M_n is an 'unbiased random walk' (a martingale) with M_0=0, and Z_n is a random, non-decreasing process with Z_0=0.

A standard result for martingales is that

M_0 = E(M_T)

for any suitable 'stopping time' T. An example of a suitable stopping time here is T={least n for which CA_n=0 or CA_n=100}, i.e. T is the moment after the last hand of the tournament.

Thus

E(CA_T) = CA_0 + 0 + E(Z_T)

and note that the LHS is

0*P(A loses the tourney|A starts with x chips) + 100*P(A wins|A starts with x chips)

since at time T we'll either have CA_T=0 if A lost or CA_T=100 if he won. Also note that the RHS is >= x since CA_0 = x, Z_n is non-decreasing and Z_0=0.

Thus we have

100*P(A wins | A starts with x chips) >= x

This compeletes the proof.


Proof of 2
----------

Any player with x chips can guarantee themselves an equity of at least x% no matter how oppo plays (from the claim), so a player playing optimally for the tournament as a whole must get at least this. In other words if A plays using a strategy which is 'tournament optimal' then

P(A wins|A has x chips) >= x/100.

This proves the first part of claim 2.

But if B uses the strategy indicated in the proof of the claim, then B's equity is at least

P(B wins|B has 100-x chips) >= (100-x)/100.

Hence as

P(A wins|A has x chips) = 1-P(B wins|B has 100-x chips)

we have

P(A wins|A has x chips) <= x/100,

and so

P(A wins|A has x chips) = x/100.

which proves the second part.

Marv

[eastbay]

There's a lot here for me to figure out and/or be convinced of. So I may not respond for awhile, but that doesn't mean I'm ignoring your demonstration. It looks very interesting and may settle some key points that are always in dispute when discussing this sort of thing.

Thanks,
eastbay

[eastbay]

[ QUOTE ]
There's a lot here for me to figure out and/or be convinced of. So I may not respond for awhile, but that doesn't mean I'm ignoring your demonstration. It looks very interesting and may settle some key points that are always in dispute when discussing this sort of thing.

Thanks,
eastbay

[/ QUOTE ]

Marv,

I still haven't quite figured out your proof but I do believe it is correct. I'll post why shortly in a new thread.

eastbay

[John Paul]

Hello,
I have been following this thread over the past few days. If I understand everything, Marv has showen that if both players push every turn, then their % equity =% chips. Shadow has pointed out that a the meaning of being a "better" player than your oppenent is that your %equity > %chips.

Taken together - It might be worth thinking about how betterness occurs. If the short stack is small enough - their blind will put them all in. At this point, for both players %equity = %chips. So neither can play better here.

Even if the blinds do not put the short stack all in, it sems to be the 2+2 dogma that you can have a stack so short that you have to push any 2 cards, and the big stack should always call. In this case neither player can play better, just worse than pushing. Can this stack size be determined analytically in terms of stack sizes and blind sizes?

So, if you are going to be better than someone HU, that is have your $Equity>Chip Equity, you must have to do it when the stacks are more equal in size. That is, there has to be some fixed chip level where pushing/calling pushes is not the best possible play. But I wonder if you could devise an optimum strategy based on the equity of individual hands vs. random hands. I don't think that you could necessarily come up with an analytical solution for every situation, but I suspect that you could narrow down the possiblities.

What do folks think?

John Paul

[John Paul]

Sorry, It was gumpzilla who made the little drawing of the $ equity vs. chip equity for a player who is better than their opponent heads up. Just want to give credit where it is due.
JP

[the shadow]

I've given more thought to gumpzilla's chart (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2378768&page=2&view=c ollapsed&sb=5&o=31&vc=1) of an equity function where the hero is more skilled (or playing a better strategy) than the villian. Even though the artistry leaves a little to be desired /images/graemlins/smile.gif, I think gumpzilla's chart is a correct way of looking at it.

I approached the problem by using the gambler's ruin formula (http://www.quantnotes.com/edutainment/betting/gamblersruin.htm). Here's the hypothetical -- hero and villian start with a stack of 500 chips apiece for a total of 1,000 in play. For each play, hero and villian bet $1 on the flip of a coin. If it comes up heads, hero wins $2. If it's tails, villian wins $2. Using that scenario, if we know hero's stack size, can we figure out his equity?

Now, if the coin is perfectly balanced, then the equity function is linear. The hero's equity equals his or her percentage of the chips in play.

But what if the coin just slightly favors the hero? I assumed that the coin was weighted so that it falls heads 50.1% of the time. Using the gambler's ruin formula, I then calculated the hero's equity for the following stack sizes:

Hero Equity
0 0.00
10 0.04
50 0.18
100 0.34
200 0.56
300 0.71
400 0.81
500 0.88
600 0.93
700 0.96
800 0.98
900 0.99
1000 1.00

Here's a chart of the equity function using these assumptions:

http://i6.photobucket.com/albums/y221/arthowe/Equityfunctionchart.gif

As you can tell, the two charts, using different assumptions, are both concave.

That means that gumpzilla's statement

[ QUOTE ]
Also interesting is that since f(A) + f(B) = 1, this means that B grows more slowly than linear for small stacks and faster than linear for big stacks (I think, correct me if I'm wrong here.)

[/ QUOTE ]

is correct. If the hero starts with a small stack, his equity grows fastest when he's short-stacked. For example, as the hero's stack grows from 100 to 200 chips, his equity grows from about 1/3 to over 1/2.

(Another interesting aspect of this chart is that a minor skill disparity of only 50.1% can have major effects when multiplied over 1000s of coin flips. That may illustrate why a push-or-fold strategy at high blinds may take away much of an opponent's skill advantage. If an unskilled player can force the tournament to come down to a single flip of a coin, he's more likely to win then if the tournament is decided by repeated coin flips at which a minor disadvantage is multiplied time and time again.)

Given that gumpzilla and marv (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2389072&page=2&view=c ollapsed&sb=5&o=31&vc=1) may have put to bed any debate about the shape of an equity function for equally matched players each using an optimal strategy, I'm inclined to agree with gumpzilla's suggestion that developing an equity function for skill disparities may be more interesting.

The Shadow

[eastbay]

[ QUOTE ]

Given that gumpzilla and marv (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2389072&page=2&view=c ollapsed&sb=5&o=31&vc=1) may have put to bed any debate about the shape of an equity function for equally matched players

[/ QUOTE ]

Let's not get confused, now. We know how to construct examples for equally matched players who have strongly nonlinear equity functions.

Marv's argument is based on optimal play only.

eastbay

[the shadow]

Good questions. Thanks for joining the discussion.

[ QUOTE ]
So, if you are going to be better than someone HU, that is have your $Equity>Chip Equity, you must have to do it when the stacks are more equal in size.

[/ QUOTE ]

John Paul, I'm either not understanding what you're saying or I think you may be mistaken.

If the hero is more skilled than the villian, then the equity of the hero's chips is greater than the hero's chips as a percentage of all chips in play for all levels of the hero's stack between 0 and 100%. In other words, the hero's skill advantage will manifest itself at any stack size and may be more powerful at smaller stack sizes.


[ QUOTE ]
Even if the blinds do not put the short stack all in, it sems to be the 2+2 dogma that you can have a stack so short that you have to push any 2 cards, and the big stack should always call. In this case neither player can play better, just worse than pushing. Can this stack size be determined analytically in terms of stack sizes and blind sizes?

[/ QUOTE ]

If you haven't done so already, take a look at the heads up and hand ranking threads in the favorite threads (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=singletable&Number=191 8735&fpart=1&PHPSESSID=) list. A lot of work already has been done to calculate push top x% and call with top y% of hands at different blind levels. poincaraux also has posted some similar analysis (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=singletable&Number=220 3844&PHPSESSID=&fpart=) on the poker stove website.

The Shadow

[the shadow]

Good clarification. I agree.

[pergesu]

[ QUOTE ]
[ QUOTE ]
As a lawyer, I'm continually surprised by how much information you can get by simply asking. Assume that an online site were willing to assist. What and how much information would you be looking for?

[/ QUOTE ]

The size of the blinds at the time the match gets to be HU seems like it would be a relevant factor.

One issue that interests me about the data collection is how many data points one should take from each tournament. Let's say I play 20 hands heads up against my opponent; should we use this as 20 data points? My gut feeling is no, we shouldn't, because any deviations from average behavior become magnified.

As an oversimplified, unrealistic example of what I'm talking about, say I get heads up with blinds of 100, my stack being 2000, villain's stack being 8000. Now let's say he's ridiculously tight, so tight, in fact, that he'll fold anything but aces. Now I should win this just about all of the time, and so I'm going to skew the data horrendously if you take multiple data points from this tournament that show my small stack overcoming a major disadvantage and winning.

[/ QUOTE ]

I haven't read all the posts yet, I'll do that when I have more time, so I don't know if you guys came to a conclusion on how many data points to use.

It seems to me that you could just use the final hand of each tourney. There are only 4k possible starting stacks in an 8k chip tourney, and 5k possibilities in a 10k chip tourney. So if you managed to get 5 million final hands, that ought to give you sufficient data. Then you can examine the deviation between stack% and win%. I don't know how useful that would be in coming up with a new chip modelling theory, but it could be used to validate the linear theory that Sklansky suggests.

Again, this may have been brought up already, and perhaps even shot down thoroughly. Just a thought I had while reading through the first page.

[John Paul]

[ QUOTE ]
Good questions. Thanks for joining the discussion.

[ QUOTE ]
So, if you are going to be better than someone HU, that is have your $Equity>Chip Equity, you must have to do it when the stacks are more equal in size.

[/ QUOTE ]

John Paul, I'm either not understanding what you're saying or I think you may be mistaken.

If the hero is more skilled than the villian, then the equity of the hero's chips is greater than the hero's chips as a percentage of all chips in play for all levels of the hero's stack between 0 and 100%. In other words, the hero's skill advantage will manifest itself at any stack size and may be more powerful at smaller stack sizes.


[/ QUOTE ]

I don't think you can be a better player when one stack is really small HU, although in a way it is a trivial result. If I only have 50 chips, and the blinds are 100/200 then both me and my opponent are going all in next hand (and the one after that if I win) and there is no possibility to show any skill. If an always push strategy results in %equity=%chips, then for someone to show skill, it has to be at a time when the blinds don't force one player all in HU. Like I said, this is pretty trivial, but if I am following the debate, that would imply that equity relationship is linear at the extremes. This would not depend on any characteristic of the 2 players, as the blinds dictate their strategies. However, I may mis-understand the debate here.

Thanks for the links. I have been playing limit ring games for a few months, but I am pretty new to SnG's so I am still catching up to the rest of the class. I hope folks keep pursuing these things both analytically and empirically. Perhaps there will be some results that help folks play, and it is interesting in its own right anyway.

John Paul

[marv]

[ QUOTE ]
[ QUOTE ]

If the hero is more skilled than the villian, then the equity of the hero's chips is greater than the hero's chips as a percentage of all chips in play for all levels of the hero's stack between 0 and 100%. In other words, the hero's skill advantage will manifest itself at any stack size and may be more powerful at smaller stack sizes.


[/ QUOTE ]

I don't think you can be a better player when one stack is really small HU, although in a way it is a trivial result. If I only have 50 chips, and the blinds are 100/200 then both me and my opponent are going all in next hand (and the one after that if I win) and there is no possibility to show any skill. If an always push strategy results in %equity=%chips, then for someone to show skill, it has to be at a time when the blinds don't force one player all in HU. Like I said, this is pretty trivial, but if I am following the debate, that would imply that equity relationship is linear at the extremes. This would not depend on any characteristic of the 2 players, as the blinds dictate their strategies. However, I may mis-understand the debate here.

Thanks for the links. I have been playing limit ring games for a few months, but I am pretty new to SnG's so I am still catching up to the rest of the class. I hope folks keep pursuing these things both analytically and empirically. Perhaps there will be some results that help folks play, and it is interesting in its own right anyway.

John Paul

[/ QUOTE ]

Here's a thought:

For small stack sizes, certainly we'd expect our players' difference in skill level to be smaller as there will be less preflop play. At the extreme, as people have noted, once one of you is blinded all-in, then skill has left the building, you just call.

But if the short stack doubles up a few times, he may now still have less equity than the linear equity function would suggest if he tends plays a short stack poorly, so his equity vs chips graph is linear in the left hand corner but its angle is much less that 45%. This might show up in the data - that short stack players play worse than they should.

One thing I learned from my tussle with eastbay is that if you can play each hand in a +cev way, you're certain to get at least the linear equity. In fact your additional equity is exactly the expectation of the sum of the 'cev edge' you have over your opponent over the remaining hands.

I'd like to know if shortstacked players (when there are 3+ players) do (or should) take -cev actions? That can't be too hard to test?


Marv

[the shadow]

Marv, thanks for offering the proposed proof of a linear equity function for a HU match where each player uses an optimal strategy. I'm still thinking about it and am looking forward to eastbay's comments.

This thread started with a suggestion for some empirical work, but due to the contributions of you, gumpzilla, eastbay and others, has made more progress on the analysis side. To that end, you write:

[ QUOTE ]
I'd like to know if shortstacked players (when there are 3+ players) do (or should) take -cev actions? That can't be too hard to test?


[/ QUOTE ]

Well, if you're counting me, shortstacked players certainly take -CEV actions, but not because it's the right thing to do. /images/graemlins/crazy.gif Seriously, it's a good question. I'll give it some thought.

In the meantime, it seems to me that there're still some productive questions to explore about heads up play. In addition to incorporating skill differences, here're two perhaps easier ones to start with:

Assume that we're down to the last two players of a SNG. The payout is 0.5 to 1st place and 0.3 to 2d place. Each player uses an optimal strategy. Is the equity function still linear? What does it look like? Should one player make -CEV plays?

Assume that one player gives another player odds (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2394286&page=0&view=c ollapsed&sb=5&o=31&fpart=3#Post2403618) in a heads up match. Again, each player uses an optimal strategy. Is the equity function still linear? What does it look like? Should one player make -CEV plays?

The Shadow

[Degen]

let there be flame


Andre

[marv]

[ QUOTE ]

Assume that we're down to the last two players of a SNG. The payout is 0.5 to 1st place and 0.3 to 2d place. Each player uses an optimal strategy. Is the equity function still linear? What does it look like? Should one player make -CEV plays?


[/ QUOTE ]

(The following assumes the proof I gave is OK and we play with a randomized button.)

Yes, it's linear: at the moment the 3rd place player is decided, if one of the remaining two players has 100x% of the chips and both are playing optimally, his equity is 0.3 + 0.2x .

If the players are using optimal strategies they'll never make -CEV plays, even if their opponent were suddenly to deviate. If one of them deviates to the point where he does make a -CEV play worth -100x% chips, his oppo gains 0.2x equity.

[ QUOTE ]

Assume that one player gives another player odds (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2394286&page=0&view=c ollapsed&sb=5&o=31&fpart=3#Post2403618) in a heads up match. Again, each player uses an optimal strategy. Is the equity function still linear? What does it look like? Should one player make -CEV plays?


[/ QUOTE ]

I interpret this as: if I lose the tourney I lose $10, if I win I get $11, but we still start with equal numbers of tourney chips. Then with optimal play from both players, if I have 100x% of the chips at some point, my equity is 21x-10.

In general one should only ever make -CEV plays if you think it will induce the opponent to make even worse -CEV plays (in total) in later hands of the tournament. Of course a play may be -CEV against one type of oppo while +CEV against another.
Optimal plays will always be >=0 CEV against any oppo.

Marv

[suited_ace]

Sorry to bring out the post from the dead, but I was saving it to read it when I would have enough time to go through it all.

Here's an idea: instead of contacting the poker sites, contact these guys (http://www.cs.ualberta.ca/~games/poker/) (they are the ones that developed the AI behind Poker Academy (http://www.poki-poker.com/)). I think they would be very interested in extending their research into that.

[JNash]

Sorry for posting so long after the fact-- I have not been to the site in a long time. Since I did the original S-Curve post, I thought I should comment.

Two comments: first, I now disavow the S-curve hypothesis. I believe that I was wrong. (See below.) Second, doing an empirical study of the ACTUAL chipEV-->tourneyEV function would be very interesting.

Starting with the idea of emprical testing first...In that case, rather than focusing on heads-up, you might as well test how stack size at certain points in a tourney affects tournament EV. For example, in large MTT tourneys, after about an hour the field is usually cut by 50% and there is a pretty wide dispersion between chip leaders and small stacks. With enough tournament samples, you could estimate pretty easily what the emprical tournament EV is for someone in the top 10% of stack size, with an average stack, etc. If you had enough data and the patience of a PhD student, you could even run some regressions to try to separate the effect of skill (observed win rate) and stack size.

Now, as to the S-Curve hypothesis. As my nick indicates, I am interested in game theory, so my original post was in the theory forum. So, I take as a given all the standard theoretical assumptions, i.e. equal skill, optimal game-theoretic play, etc. Under these conditions, I now believe that the following Sklansky/Malmuth assertions are absolutely correct (in theory, that is):
1) EV of winning a heads-up freezeout is proportional to the fraction of total chips you have (i.e. linear)
2) In any winner-take-all tournament (heads-up is just a special case of this), the probability of winning is proportional to the chips you have.
3)) Tournament EV is a concave fuction of chip EV--i.e. NOT the S-curve I had hypothesized.

1) In TPFAP, there is a very elegant proof "by symmetry" of the proportionality argument. It basically relies on the reasoning that if I have, say, 20% of the chips, and you have 80% of the chips, and we have equal skill, then I have a 50/50 chance of doubling up. The critical element of the proof is that, with equal skill, I always have exactly a 50/50 chance of doubling up. I started to question whether this was indeed true. Might it be possible that the optimal game-theoretic strategy actually depends on the stack sizes? In that case, all bets are off and the proof doesn't work.

I have since then convinced myself that the fact that the big stack has some "extra" chips in reserve does not affect the optimal strategy for the two players at all. The only thing that matters in determining the optimal play is the size of the blinds relative to the size of the stacks. If I have 200 chips out of 2000 in play, and you have 1800, we are currently effectively playing a game where we each have 200, since the most we can bet is allin. I don't have a "proof" proof, (of the fact that the optimal strategy for the two players is independent of any "extra" chips one of them may have), but I believe this to be true.

2) If optimal optimal play depends only on the size of the smallest stack involved in a confrontation, and the availability of extra chips for some players does not matter, then the TPFAP proof for the multi-player winner-take-all tourney also goes through without a hitch.

3) Finally, the concavity question, which is the same thing as the TPFAP assertion that "the chips you win are always worth less than the chips you lose."

I now believe that this is always true in the case of multiple payouts and more than 3 players.

I'll give a heuristic argument for this effect. First, suppose you have 80% of the chips in play, 5 places pay, and there are 5 players left. You are very, very likely to win 1st place. If you win a confrontation that busts out one of the players, you gain some EV, but some of it "leaks" to all the other players who are now assured of finishing one place higher. So, your gain in chips does not give you a linear pickup in tournament EV.

Extending this further, suppose you are an above-average stack, and you win a confrontation with a small stack. You've pushed the small stack closer to elimination, which benefits not only you, but also other players--i.e. you get concavity again.

My original S-curve argument assumed that big stacks have an advantage over small stacks because they can "bully" the short stack. While this may still be empirically true in actual play, I now believe that the game-theorically correct play for the small stack does NOT depend on the presence of extra chips for the big stack. The small stack plays strictly based on pot-odds, the size of the blinds, and the maximum amount that can be bet--i.e. his own stack size.

Anyway, sorry for sending y'all on a blind goose chase, but I now no longer believe in the S-curve...

May your flops be disguised and your rivers kind!

[maddog2030]

[ QUOTE ]
My original S-curve argument assumed that big stacks have an advantage over small stacks because they can "bully" the short stack. While this may still be empirically true in actual play, I now believe that the game-theorically correct play for the small stack does NOT depend on the presence of extra chips for the big stack. The small stack plays strictly based on pot-odds, the size of the blinds, and the maximum amount that can be bet--i.e. his own stack size.

[/ QUOTE ]

Can you expand more on what you mean exactly here? An example perhaps? Because I'd disagree with what I think you're saying, but it feels like I might be misunderstanding you the way it's worded. By the way, are you talking about shortstacks compared to the blinds or compared to other stacks?

Also I think the medium stacks (in terms of the field) are the ones to be bullied more so than the short ones. There tends not to be enough in the pot in terms of blinds/antes to make up for the equity they'd lose by putting their own stack at risk in there in marginal situations.

[eastbay]

[ QUOTE ]

1) In TPFAP, there is a very elegant proof "by symmetry" of the proportionality argument. It basically relies on the reasoning that if I have, say, 20% of the chips, and you have 80% of the chips, and we have equal skill, then I have a 50/50 chance of doubling up. The critical element of the proof is that, with equal skill, I always have exactly a 50/50 chance of doubling up. I started to question whether this was indeed true. Might it be possible that the optimal game-theoretic strategy actually depends on the stack sizes? In that case, all bets are off and the proof doesn't work.


[/ QUOTE ]

Let's not confuse issues. Equal skill by no means implies both players are playing optimally (game theoretic sense). The so-called "proof" of S&M is obviously wrong by counterexample.

Both players play the following (pathological, but equally "skilled") strategy: push if you have more than half the chips, fold if you have less than half. Flip a coin if stacks are even.

You need (at least) some kind of stack independence of strategy condition before this symmetry argument has any hope of being sufficient to guarantee a linear relation. Otherwise, you're left with a family of curves which are all admissible by symmetry, from linear to step function, with the only symmetry requirement being f(x)+f(1-x)=1.

Now, I grant that S&M may have been trying to simplify the discussion for a general audience, but I would prefer they not use the word "proof" in that context.

As an aside, buried somewhere, I think in this thread, is someone's offering of a proof of what you conclude in your discussion: that the "extra" chips are strategically useless for optimal play. I never took the time to try to convince myself that it is correct, but I suspect that it is.

eastbay

[JNash]

[ QUOTE ]
You need (at least) some kind of stack independence of strategy condition before this symmetry argument has any hope of being sufficient to guarantee a linear relation.

[/ QUOTE ]

I agree completely. What I am saying is precisely that a (game-theoretically) optimal strategy will depend only on the size of the smaller of the two stacks in relation to the blinds--i.e. will be independent of the stack sizes relative to each other.

So the counter-example which you provide, in which the strategies depend on the stack sizes (and you yourself called the strategy pathetic...) is just not an optimal strategy.

So, it depends on what you think the "theorem" actually says. I believe that it says that $EV is linear in stack sizes only if they both play equally WELL in the sense of playing optimally in a game-theory sense.

As your argument shows, if they play an identical, but stack-dependent strategy, then all bets are off for the proof.

As you also (correctly) state, the proof would work even for two identical strategies that may not be game-theory optimal, as long as they are stack-size independent.

Per your suggestion, I'll go back through the posts and look for the argument that the extra chips don't matter. I can formally prove it for a very "all-in or fold" simplified [0,1] model, but can't prove it for more general cases.

[JNash]

What I mean is as follows.

First off, let me emphasize that I am speaking only about the "optimal" strategy in a game theory sense. In practice, people play all kinds of strategies that may be quite different than optimal.

I am asserting (without a formal proof, just some heuristic arguments), that a game-theoretically optimal strategy will depend only on the size of the SMALLER of the two stacks, and of course the size of the blinds relative to the size of the smaller stack.

To give my heuristic argument for this: because you can never bet more than the size of the smaller of the two stacks, that's all they're playing for, and the presence of extra chips should make no difference to how either side should play.

In a multi-player (>2) tourney, the distibution of chips among the remaining players does affect how you should play, and in fact the "linearity theorem" does not hold.

I hope that makes it a bit clearer...

I also agree with your point about medium stacks being easier to bully than shorter stacks. The reason, though, is not because they should be afraid of the "extra" chips that a big stack may have, but rather because a medium stack by definition will have a stack that is larger relative to the size of the blinds than a shorter stack. As stack size decreases relative to the blinds, the optimal strategy becomes looser and looser (i.e. you should optimally play more hands).

[maddog2030]

[ QUOTE ]
In a multi-player (>2) tourney, the distibution of chips among the remaining players does affect how you should play, and in fact the "linearity theorem" does not hold.

[/ QUOTE ]

Oh okay, I thought you were implying this was the case for >2. That's where I got confused. I agree with you HU... simply because chips are now constant values and normal cash game logic applies.

[the shadow]

[ QUOTE ]
Per your suggestion, I'll go back through the posts and look for the argument that the extra chips don't matter. I can formally prove it for a very "all-in or fold" simplified [0,1] model, but can't prove it for more general cases.

[/ QUOTE ]

Take a look at marv's post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2389072&page=2&view=c ollapsed&sb=5&o=31&vc=1).

JNash, OK, let me get this straight -- you're saying that the optimal strategy depends upon the ratio of the small stack to the big blind. Let's take two different scenarios:

A. Hero and villian each have 6750 chips and the big blind is 400.

B. Hero has 10125 chips, villian has 3375 chips, and the big blind is 200.

In both cases, the ratio of the small stack to the big blind is 16.875. Do I understand correctly that the optimal strategy is the same in both cases? In scenario B, is it further correct that the optimal strategy for the hero is the same as the optimal strategy for the villian?

Thanks for joining the thread,

The Shadow