empirical equity study

[eastbay]

I finally coded up the icm comparison/validation I talked about doing awhile ago. The idea was to empirically determine my equity as a function of chip stack distribution and compare that to what ICM says.

A brief look at some bubble numbers:

Overall average equity on the bubble:

ICM: .232
Actual: .277

Looks pretty good. I've got about 4% more of the prize pool than ICM says I "deserved".

Breaking it down into situations where someone has more than half the chips:

stacks: actual equity, icm predicted equity, difference, standard error of sample, number of data points in this range
0,0,0,0: 0.282925; icm = 0.239223; delta = 0.0437015; std err=0.00508585; N=1265
0,0,0,1: 0.266667; icm = 0.195756; delta = 0.0709111; std err=0.0189215; N=90
0,0,1,0: 0.244944; icm = 0.203231; delta = 0.0417128; std err=0.0190025; N=89
0,1,0,0: 0.215385; icm = 0.18758; delta = 0.0278049; std err=0.0148834; N=117
1,0,0,0: 0.426316; icm = 0.393979; delta = 0.0323365; std err=0.0221329; N=19

0,0,0,0 means no one has more than half the chips.
0,1,0,0 means the guy to my immediate left does. etc.

This is kind of interesting, if it really holds up. You can see the increasing advantage of having the big stack further away to the left. This is an effect that is completely missing from the ICM, which takes no heed of position.

Another interesting fact is that the other players are playing the big stack at this phase of the game much more often than I am (N=19). And yet I have significant equity advantage over them on average.

Discuss. Probably more to come.

eastbay

[ilya]

Quick question, for the ICM numbers did you just plug in whatever the stacks were on the 1st 4-handed hand?

[eastbay]

[ QUOTE ]
Quick question, for the ICM numbers did you just plug in whatever the stacks were on the 1st 4-handed hand?

[/ QUOTE ]

For the ICM numbers I plugged in for every 4-handed hand.

eastbay

[AleoMagus]

Interesting.

If ICM is to be trusted, I suppose your extra equity can be accounted for by difference in playing ability.

What I'm really wondering though, is if your results seem to indicate a different model would be more appropriate.

[ QUOTE ]
This is an effect that is completely missing from the ICM, which takes no heed of position.


[/ QUOTE ]

I'm thinking about this, and I am not sure that this is missing from ICM. After all, ICM assumes equal playing ability and I'm thinking that this position related difference in equity might have everything to do with you, and the fact that you have an equity edge.

I think these results might mean more if you only ran them on bubble histories that you were actually not involved in. In this way, it would always be different players and might more accurately reflect the equal skill assumption. Then again, it could be even more important to look at the advantage biased results.

...Already I'm thinking I'm wrong about the position differences being a factor of your advantage. It would make sense that these positional distances from the big stack would affect equity.

Can you run this and calculate equity actual/predicted for the other three positions as well?

I have a feeling this is going to become a very important thread. I can't wait to see where this goes. Very ambitious putting this all together. Has the potential for a major contribution to tourney theory as a whole.

Regards
Brad S

[eastbay]

[ QUOTE ]
Interesting.

If ICM is to be trusted, I suppose your extra equity can be accounted for by difference in playing ability.

[/ QUOTE ]

That's certainly in there. I originally had thought that was the purpose of the investigation, since I was approaching this from the perspective of determining my equity to the best approximation. Right now I do that with ICM, for lack of something better. That something better might be a curve fit on empirical data, that includes things like my playing advantages.

But, now you're getting me thinking about something a little more general and theoretical: an empirical distribution based on the player pool as a whole. That has some merit too, I think.

[ QUOTE ]


What I'm really wondering though, is if your results seem to indicate a different model would be more appropriate.

[ QUOTE ]
This is an effect that is completely missing from the ICM, which takes no heed of position.


[/ QUOTE ]

I'm thinking about this, and I am not sure that this is missing from ICM. After all, ICM assumes equal playing ability and I'm thinking that this position related difference in equity might have everything to do with you, and the fact that you have an equity edge.

I think these results might mean more if you only ran them on bubble histories that you were actually not involved in.


[/ QUOTE ]

Yes, that's an interesting idea.

[ QUOTE ]

In this way, it would always be different players and might more accurately reflect the equal skill assumption. Then again, it could be even more important to look at the advantage biased results.

[/ QUOTE ]

For me, yeah. In general, probably not.

[ QUOTE ]

...Already I'm thinking I'm wrong about the position differences being a factor of your advantage. It would make sense that these positional distances from the big stack would affect equity.


[/ QUOTE ]

I think it's both effects mixed together.

[ QUOTE ]


Can you run this and calculate equity actual/predicted for the other three positions as well?


[/ QUOTE ]

Yes, I could do that. Time permitting.

eastbay

[rachelwxm]

Pretty interesting. So if I read it correctly, your ROI is 18% (.277-rake) and you make extra 4% from bubble on.

I have looked at ROI boost for each stages and found for low limit, you can make 50% of you equity from bubble on. I think that high limit player are generally much better in short handed games.

What most interesting is the fact you look at the position vs big stack. Although those numbers are subjet to sample size bias since the difference of each cases is similar to the mean standard deviation. So 1000 mean you are the big stack, I was a little bit surprised that this case compared unfavorably to 0000 which is a big sample. /images/graemlins/smile.gif

I have not look at anything regarding second part in my analysis though. /images/graemlins/tongue.gif

[Irieguy]

Why wouldn't the overall average equity on the bubble according to ICM be 25%?

If you are looking at the "overall average equity," it seems like everybody would have equal equity in the ICM.

Irieguy

[eastbay]

[ QUOTE ]
Why wouldn't the overall average equity on the bubble according to ICM be 25%?

If you are looking at the "overall average equity," it seems like everybody would have equal equity in the ICM.

Irieguy

[/ QUOTE ]

They would if we played the bubble with even stacks on average. We don't. Prior play and bubble play influences the average stack distribution when it's 4-handed.

eastbay

[The Yugoslavian]

[ QUOTE ]

0,0,0,1: 0.266667; icm = 0.195756; delta = 0.0709111; std err=0.0189215; N=90


[/ QUOTE ]

Is this the situation when the big stack is to your right so you get to act just after him? (I want to make sure I've got yout 0,0,0,1 down). It is interesting that in this you far outstrip the ICM average (more than any other 4-handed situation). It seems either there is a larger advantage to being behind the big stack than I thought, you excel vs. other smallish stacks when the big stack isn't playing the hand, or that the large difference in delta could be an anomoly.

Believe it or not I have been eagerly awaiting this thread for some time and along with Aleo, would be *very* interested to see what one comes up with when analyzing 4-handed play on Party Poker with random participants.

I'm interested in what else you think your numbers may mean other than ICM cannot take into account position to the big stack. Would running this to determine equity of pushers 4-handed or folders 4-handed be meaningful in any way (i.e. on average is it better to push rather than fold, and if so how close is it)? Would this be some sort of factor of the buyin you're at (55s I assume?) where general tightness/looseness would determine the equity of each action?

I'm also interested to know if anything meaningful can be gleaned from the equity your opponents had vs. the big stack -- I'm wondering if the average player may generally outperform or significantly underperform in different situations vs. the big stack and/or with the big stack. Hmmm, perhaps this question is better answered by running ICM vs. actual equity #s for 4-handed hands with random players.

Also, do your numbers mean you're making an extra 4% ROI from bubble on and making 14% by just getting to the bubble with a playable stack?

Very intersting stuff Eastbay, major props to you.

Yugoslav

[hansarnic]

This is very interesting, but as others have said, it really only demonstrated YOUR equity at certain stages and that comes from your realtive skill level.

What would be interesting (and not that hard to do) would be to sim certain player / stack situations & compare the results to ICM.

One can do this by just getting each player to move in on every single hand (thus negating the effects of blinds & skill).

This is something I've been meaning to set-up as & when I have time, so will have a go over the next week or so.

Of course this only validates ICM which as we know doesn't take account of certain key variables in tournament structure, (position, skill and blinds) being the key ones. And as Aleo said the best way to do this would be to crunch huge amounts of actual data.

Strikes me this should be a job for Party or Stars!!

[eastbay]

[ QUOTE ]
This is very interesting, but as others have said, it really only demonstrated YOUR equity at certain stages and that comes from your realtive skill level.

[/ QUOTE ]

Of course. Whose equity do you think I am interested in?

[ QUOTE ]

What would be interesting (and not that hard to do) would be to sim certain player / stack situations & compare the results to ICM.

One can do this by just getting each player to move in on every single hand (thus negating the effects of blinds & skill).

[/ QUOTE ]

All players on all hands? I have doubts about the usefulness of that model. How does position then matter?

eastbay

[eastbay]

[ QUOTE ]
[ QUOTE ]

0,0,0,1: 0.266667; icm = 0.195756; delta = 0.0709111; std err=0.0189215; N=90


[/ QUOTE ]

Is this the situation when the big stack is to your right so you get to act just after him? (I want to make sure I've got yout 0,0,0,1 down). It is interesting that in this you far outstrip the ICM average (more than any other 4-handed situation). It seems either there is a larger advantage to being behind the big stack than I thought, you excel vs. other smallish stacks when the big stack isn't playing the hand, or that the large difference in delta could be an anomoly.

[/ QUOTE ]

I noticed this as well. It may be that the typical player misplays the big stack, sitting patiently waiting for someone to bust, and leaving my blinds alone.

[ QUOTE ]

I'm interested in what else you think your numbers may mean other than ICM cannot take into account position to the big stack. Would running this to determine equity of pushers 4-handed or folders 4-handed be meaningful in any way (i.e. on average is it better to push rather than fold, and if so how close is it)? Would this be some sort of factor of the buyin you're at (55s I assume?) where general tightness/looseness would determine the equity of each action?

[/ QUOTE ]

If I follow you - yes, that that's the point of knowing the equity as a function of stack distribution. For any decision, say, pushing, if you estimate what hands you'll be called by, you can calc. the probability of the call, and get the probability of the resulting stack distributions, and finally determine whether your push or a fold results in more equity. Until now, ICM has been the tool of choice for determining equity=f(stacks). I'm seeing if I can improve on that. Obviously, every person has a different such function, so in the first instance, I'm interested in my own function. I might also try to compute an average function for general interest, however.

[ QUOTE ]

I'm also interested to know if anything meaningful can be gleaned from the equity your opponents had vs. the big stack -- I'm wondering if the average player may generally outperform or significantly underperform in different situations vs. the big stack and/or with the big stack. Hmmm, perhaps this question is better answered by running ICM vs. actual equity #s for 4-handed hands with random players.


[/ QUOTE ]

Yes.

[ QUOTE ]

Also, do your numbers mean you're making an extra 4% ROI from bubble on and making 14% by just getting to the bubble with a playable stack?


[/ QUOTE ]

Yes, I think that's right.

eastbay

[hansarnic]

[ QUOTE ]

What would be interesting (and not that hard to do) would be to sim certain player / stack situations & compare the results to ICM.

One can do this by just getting each player to move in on every single hand (thus negating the effects of blinds & skill).

[/ QUOTE ]

All players on all hands? I have doubts about the usefulness of that model. How does position then matter?

eastbay

[/ QUOTE ]

Position doesn't matter. Nor do blinds or skill levels. But it would validate (or disprove) ICM, which would be useful in itself.

[eastbay]

[ QUOTE ]

Position doesn't matter. Nor do blinds or skill levels. But it would validate (or disprove) ICM, which would be useful in itself.

[/ QUOTE ]

I have some doubts about that. It raises an interesting question though: can a game of players who all play the same generate different equities for the same stack distribution, by the manner in which they all play? It seems there should be an easy logical proof or an easy counterexample, but neither came to me in the 10 seconds it took me to write this post.

eastbay

[eastbay]

[ QUOTE ]
[ QUOTE ]

Position doesn't matter. Nor do blinds or skill levels. But it would validate (or disprove) ICM, which would be useful in itself.

[/ QUOTE ]

I have some doubts about that. It raises an interesting question though: can a game of players who all play the same generate different equities for the same stack distribution, by the manner in which they all play? It seems there should be an easy logical proof or an easy counterexample, but neither came to me in the 10 seconds it took me to write this post.

eastbay

[/ QUOTE ]

Ok, I thought of one. Consider HU match. Strategy A is to push when you have half or more of the chips, and fold if you have less than half. Clearly equity distribution is a step function for two players both playing this strategy.

Strategy B is push everything, call everything. We know that equity distribution is basically linear for this strategy.

So the answer is yes: the equity function of a game of players all playing the same strategy does depend on the strategy employed. And this is why I am not too fond of your "validation" of ICM for all pushes. It may generate a reasonable distribution, but it also may not, and I don't think it will close the case either way.

eastbay

[eastbay]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

Position doesn't matter. Nor do blinds or skill levels. But it would validate (or disprove) ICM, which would be useful in itself.

[/ QUOTE ]

I have some doubts about that. It raises an interesting question though: can a game of players who all play the same generate different equities for the same stack distribution, by the manner in which they all play? It seems there should be an easy logical proof or an easy counterexample, but neither came to me in the 10 seconds it took me to write this post.

eastbay

[/ QUOTE ]

Ok, I thought of one. Consider HU match. Strategy A is to push when you have half or more of the chips, and fold if you have less than half. Clearly equity distribution is a step function for two players both playing this strategy.

Strategy B is push everything, call everything. We know that equity distribution is basically linear for this strategy.

So the answer is yes: the equity function of a game of players all playing the same strategy does depend on the strategy employed. And this is why I am not too fond of your "validation" of ICM for all pushes. It may generate a reasonable distribution, but it also may not, and I don't think it will close the case either way.

eastbay

[/ QUOTE ]

It is interesting to note that strategy A employs "equal skill" and yet the distribution is not linear. I understand Mr. Sklansky provides a "flawless logical proof" in TPFAP that any such distribution must be linear.

I don't own the book. Anyone care to comment?

eastbay

[rachelwxm]

Is this in Slansky's book?

Any strategy that based on stack size instead of cards vialate this relation, like the one you created. For example push more if big stack, etc because while it's vague when you talk about same strategy but chip stack is certainly not symmetric, you either have more or less chips.

[hansarnic]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

Position doesn't matter. Nor do blinds or skill levels. But it would validate (or disprove) ICM, which would be useful in itself.

[/ QUOTE ]

I have some doubts about that. It raises an interesting question though: can a game of players who all play the same generate different equities for the same stack distribution, by the manner in which they all play? It seems there should be an easy logical proof or an easy counterexample, but neither came to me in the 10 seconds it took me to write this post.

eastbay

[/ QUOTE ]

Ok, I thought of one. Consider HU match. Strategy A is to push when you have half or more of the chips, and fold if you have less than half. Clearly equity distribution is a step function for two players both playing this strategy.

Strategy B is push everything, call everything. We know that equity distribution is basically linear for this strategy.

So the answer is yes: the equity function of a game of players all playing the same strategy does depend on the strategy employed. And this is why I am not too fond of your "validation" of ICM for all pushes. It may generate a reasonable distribution, but it also may not, and I don't think it will close the case either way.

eastbay

[/ QUOTE ]

Don't think I understand Eastbay.

Strategy A leads to the player with the most chips at the start of the HU match winning virtually 100% of the time, no?

So that player has more skill (or rather the lower stacked player has no skill as he just folds every hand giving the game to his opponent).

Strategy B takes skill entierly out of the equation.

And yes, there are a couple of proofs of $EV = Chip count in HU in TPFAP, one of which is calculated on the basis of both players going all-in every hand.

[jcm4ccc]

[ QUOTE ]
[ QUOTE ]
Quick question, for the ICM numbers did you just plug in whatever the stacks were on the 1st 4-handed hand?

[/ QUOTE ]

For the ICM numbers I plugged in for every 4-handed hand.

eastbay

[/ QUOTE ]

If I understand you correctly, you are plugging in data points for every 4-handed hand. So, for example, some tournaments might have 12 data points, and some might have only one. The data might look like this:

tourney stack icm actual
1111111 1500 $10 $15
1111112 1500 $10 $25
1111112 1500 $10 $25
1111112 1450 $9.5 $25

if that's true, I would say that this is a flaw in your research design. Your data points are not independent. They would be correlated with data points from the same tournament. Tournaments that lasted longer on the bubble would have more influence on your data than other tournaments.

This is a signficant problem, because there is probably a correlation between the size of the big stack and the length of the bubble. In other words, tournaments in which one stack has more than half the chips would have fewer hands on the bubble, and therefore have less influence on your data (fewer data points from those tournaments).

I would do as irieguy suggested. Use only the first hand in which you are four-handed. Alternatively, you could use the first four hands of a tourney (so that everybody gets the SB and BB once), average the results, and use that as your data point. Tournaments that lasted less than 4 hands would not be included in the data (though that adds a confounding factor--you would want to track how many tournaments are not included).

[Phil Van Sexton]

Apologies if this has been done 1 millions times already...

Have you run this analysis based on the position of the button rather than the position of the big stack? Clearly, if 4 players have equal stacks, the ICM will be .25 for all, but the button's equity is really higher and the BB's lower.

How much, I'm not sure, but likely significant if the blinds were 300/600 since the BB will need to post 30%+ of his stack over the next 2 hands. Meanwhile, the button posts nothing and can steal if he chooses.

[eastbay]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Quick question, for the ICM numbers did you just plug in whatever the stacks were on the 1st 4-handed hand?

[/ QUOTE ]

For the ICM numbers I plugged in for every 4-handed hand.

eastbay

[/ QUOTE ]

If I understand you correctly, you are plugging in data points for every 4-handed hand. So, for example, some tournaments might have 12 data points, and some might have only one. The data might look like this:

tourney stack icm actual
1111111 1500 $10 $15
1111112 1500 $10 $25
1111112 1500 $10 $25
1111112 1450 $9.5 $25

if that's true, I would say that this is a flaw in your research design. Your data points are not independent. They would be correlated with data points from the same tournament. Tournaments that lasted longer on the bubble would have more influence on your data than other tournaments.

This is a signficant problem, because there is probably a correlation between the size of the big stack and the length of the bubble. In other words, tournaments in which one stack has more than half the chips would have fewer hands on the bubble, and therefore have less influence on your data (fewer data points from those tournaments).

I would do as irieguy suggested. Use only the first hand in which you are four-handed. Alternatively, you could use the first four hands of a tourney (so that everybody gets the SB and BB once), average the results, and use that as your data point. Tournaments that lasted less than 4 hands would not be included in the data (though that adds a confounding factor--you would want to track how many tournaments are not included).

[/ QUOTE ]

I am not sure what you are objecting to. Are you objecting to how I computed the icm average for the region in stack-space, or how I am computing the empirical equity? Or both?

You may have a point or you may be confused, depending.

eastbay

[eastbay]

[ QUOTE ]
Apologies if this has been done 1 millions times already...

Have you run this analysis based on the position of the button rather than the position of the big stack?


[/ QUOTE ]

It is not either/or, it is an additional variable, and no, it's not one I've separated out in the analysis yet.

But I agree with you that it's an interesting angle to explore. To do so may require more data than I currently have, though.

eastbay

[eastbay]

[ QUOTE ]
Is this in Slansky's book?

[/ QUOTE ]

I understand he makes a "purely logical" proof that for "equal skill" the distribution must be linear, yes. I am also convinced this is wrong without further qualification, as my counterexample shows.

[ QUOTE ]
Any strategy that based on stack size instead of cards vialate this relation, like the one you created.


[/ QUOTE ]

Yes but that's a fairly huge restriction on the validaty of the 'proof'. Most players do incorporate stack size in their nominal, realistic strategies - as they should. I've simply taken that to an extreme to illustrate the effect in the clearest of terms.

[ QUOTE ]

For example push more if big stack, etc because while it's vague when you talk about same strategy but chip stack is certainly not symmetric, you either have more or less chips.

[/ QUOTE ]

I don't follow this sentence. Clearly two identical strategies must be considered to have "equal skill."

eastbay

[eastbay]

[ QUOTE ]

Don't think I understand Eastbay.

Strategy A leads to the player with the most chips at the start of the HU match winning virtually 100% of the time, no?

[/ QUOTE ]

Not virtually - exactly.

[ QUOTE ]


So that player has more skill (or rather the lower stacked player has no skill as he just folds every hand giving the game to his opponent).


[/ QUOTE ]

No. Both players are playing identical strategies. Clearly then both strategies must have equal skill - because they are the same.

[ QUOTE ]

Strategy B takes skill entierly out of the equation.

[/ QUOTE ]

Any two identical strategies takes skill out of the game.

[ QUOTE ]


And yes, there are a couple of proofs of $EV = Chip count in HU in TPFAP, one of which is calculated on the basis of both players going all-in every hand.

[/ QUOTE ]

Right, but there is also a 'purely logical proof.' Can you quote it for us? It seems clearly wrong, at the very least through insufficient qualification.

eastbay

[jcm4ccc]

do all of your tournaments have the same number of data points? for example, does a tournament that lasts one hand on the bubble have the same number of data points (rows on an Excel worksheet) as a tournament that lasts 20 hands on the bubble?

if not, there is a serious flaw in your design. you are overweighting the tournaments that last longer on the bubble. all of your calculations are flawed, if that is true.

[Phil Van Sexton]

[ QUOTE ]
It is not either/or, it is an additional variable

[/ QUOTE ]

Why is it either/or? A break down by your position just like your break down by big stack position would be interesting. Whether or not there is a big stack is not important, just the % difference between ICM and actual.

[rachelwxm]

[ QUOTE ]
[ QUOTE ]
Is this in Slansky's book?

[/ QUOTE ]

I understand he makes a "purely logical" proof that for "equal skill" the distribution must be linear, yes. I am also convinced this is wrong without further qualification, as my counterexample shows.

[ QUOTE ]
Any strategy that based on stack size instead of cards vialate this relation, like the one you created.


[/ QUOTE ]

Yes but that's a fairly huge restriction on the validaty of the 'proof'. Most players do incorporate stack size in their nominal, realistic strategies - as they should. I've simply taken that to an extreme to illustrate the effect in the clearest of terms.

[ QUOTE ]

For example push more if big stack, etc because while it's vague when you talk about same strategy but chip stack is certainly not symmetric, you either have more or less chips.

[/ QUOTE ]

I don't follow this sentence. Clearly two identical strategies must be considered to have "equal skill."

eastbay

[/ QUOTE ]

Point all accepted, all that I want to point out is just stack based strategy does not create linear relationship, like the one you did. Particularly stack size is of opposite for two different people. The fact that one play big stack aggressively and the other short stack play passively lead to the one direction, not random equilibrium (linear relationship). /images/graemlins/cool.gif

[eastbay]

[ QUOTE ]
do all of your tournaments have the same number of data points? for example, does a tournament that lasts one hand on the bubble have the same number of data points (rows on an Excel worksheet) as a tournament that lasts 20 hands on the bubble?

if not, there is a serious flaw in your design. you are overweighting the tournaments that last longer on the bubble. all of your calculations are flawed, if that is true.

[/ QUOTE ]

I see your objection and I think it's legitimate in the coarse-grained results I've shown so far, especially for the fully averaged case. That wasn't my original intention for how to compute it - I wanted to make the discretization fine enough that you wouldn't have more than one instance of any given distribution in a single SnG except as a rare coincidence.

However, it's not clear to me now if there's a problem in using the same tournament results multiple times so long as the chip stack distribution for which you're recording it doesn't occur more than once in the same tournament. Thoughts? My inclination is that it's fine to do that. You just have to not double count in the same range of stack-space.

Which, in the case of lumping all of them together, means no double-counting at all. Oops.

eastbay

[eastbay]

[ QUOTE ]

Point all accepted, all that I want to point out is just stack based strategy does not create linear relationship, like the one you did. Particularly stack size is of opposite for two different people. The fact that one play big stack aggressively and the other short stack play passively lead to the one direction, not random equilibrium (linear relationship). /images/graemlins/cool.gif

[/ QUOTE ]

Agree. The point is mostly academic, but I do think that there is no "purely logical proof" of linearity once reasonable, stack-dependent strategies are considered. The magnitude of the nonlinearity in real-world play is another question.

eastbay

[eastbay]

jcm4cc discovered an error in my initial calculational procedure, so wait on corrected results. Until then...

I wanted to lay out my thoughts about grouping the stack space for a more detailed look at the function equity=f(stacks_k) where stacks_k are the possible distributions of stacks for k=1..N for N-handed play.

Starting with the case 3-handed play because it's easier to visualize:

It's a geometrical property of an equilateral triangle that the normal distances from any point in the triangle to the sides of the triangle sum to a constant, the height of the triangle. So you can visualize the possible combinations of 3 stacks as the points in the interior of an equilaterial triangle, with height equal to the number of chips in the tournament.

Now, you could never in a practical way collect enough data for each of the possible sets of 3 chip stacks. There are too many combinations to ever get a big enough sample for each one to determine its expected equity. So you have to lump certain ranges together to get a reasonable sample. You can visualize this as dividing up the interior of the triangle in some way.

A logical way to do this is to divide the triangle parallel to each side into M equal intervals. For M=2:

http://rwa.homelinux.net/poker/M2.png

I've labeled these regions of the space according to the interval for players (A,B,C). Choosing higher M you can get a finer-grained description of the function over this space.

For N=2 you get a line segment, and for N=4 it's a tetrahedron. In math speak I think in general this is called an n-dimensional simplex.

So when I present results for (1,0,0,0) it's not so much about the "presence of the big stack" but a region of the stack space so described.

My immediate goal is to describe the equity function over this space for me in particular as player A, and compare that to what ICM says the function looks like (which is what I normally use for valuing various stacks). ICM has 3-way symmetry, whereas my function ought to only have 2-way symmetry for players A and B who are chosen at random from a large sample.

There's lots of additional ideas popping up in the thread for other interesting things to try to measure. Keep 'em coming.

eastbay

[parappa]

Thanks not only for your work, but also for this quite lucid explanation of what you're on about. As someone with only small amounts of college math, I can pretty easily follow what you're saying thanks to the explanation.

[hansarnic]

[ QUOTE ]

Don't think I understand Eastbay.

Strategy A leads to the player with the most chips at the start of the HU match winning virtually 100% of the time, no?

[/ QUOTE ]
[ QUOTE ]

Not virtually - exactly.


[/ QUOTE ]


Yes, except where starting BB size > stack size, i.e. where small stack cannot fold his BB, wins showdown & becomes big stack

[ QUOTE ]

So that player has more skill (or rather the lower stacked player has no skill as he just folds every hand giving the game to his opponent).


[/ QUOTE ]
[ QUOTE ]

No. Both players are playing identical strategies. Clearly then both strategies must have equal skill - because they are the same.


[/ QUOTE ]

But in reality they are not playing the same strategy. One stack is folding 100% of the time the other is pushing 100% of the time. The smaller stack's strategy will always lead to him losing. It's a bit like saying to both players - when you are sitting in the 1 seat you should push and when you're sitting in the two seat you should fold. This is 'the same' strategy but the actual skill level is pre-determined by events outside the players control (i.e. seat number or starting stack size).

[ QUOTE ]

Strategy B takes skill entierly out of the equation.

[/ QUOTE ]

[ QUOTE ]

Any two identical strategies takes skill out of the game.


[/ QUOTE ]


Not in the examples above

[ QUOTE ]

And yes, there are a couple of proofs of $EV = Chip count in HU in TPFAP, one of which is calculated on the basis of both players going all-in every hand.

[/ QUOTE ]
[ QUOTE ]

Right, but there is also a 'purely logical proof.' Can you quote it for us? It seems clearly wrong, at the very least through insufficient qualification.

eastbay

[/ QUOTE ]

p 105 "There is however a more elegant,purely logical proof that equal players in a symetrical situation must win exactly in proportion to the size of their stacks. Supose these two players agreed to play the same freezout everyday for ten years. Given they are equal players, their long term reults would necessarily be to break even. Well in order for them to break even, the smaller stack must win exactly the same fraction of times as the fraction of the total chips he possesses. This reasoning extends to any number of players as long as they play equally well, and there is no positional advantage for one compared to the other."

[eastbay]

[ QUOTE ]

But in reality they are not playing the same strategy.

[/ QUOTE ]

I think you have a strange notion of strategy, then. A strategy is a set of rules for how to act according to context, i.e., if I have AA I am going to limp and move in on any flop. If I have AK and less than 10BB, I am going to move in. Any such set of rules constitutes a strategy. The idea that a strategy must be stack independent is absurd - that's bad poker. In my example, each player is playing according to exactly the same set of rules.

[ QUOTE ]

One stack is folding 100% of the time the other is pushing 100% of the time.

[/ QUOTE ]

In the same context, they make the same move. That's playing the same strategy.

[ QUOTE ]

The smaller stack's strategy will always lead to him losing. It's a bit like saying to both players - when you are sitting in the 1 seat you should push and when you're sitting in the two seat you should fold. This is 'the same' strategy but the actual skill level is pre-determined by events outside the players control (i.e. seat number or starting stack size).


[/ QUOTE ]

There's a big difference between your example and mine, and that is that a reasonable poker strategy will never be absolute seat number dependent - there's never any advantage to considering that as a variable in a strategy, but a reasonable poker strategy WILL be dependent on relative stack size - there's advantage to be gained (or lost if misapplied) by considering relative stack size. So it is reasonable to consider the space of stack-dependent strategies, but it is not reasonable to consider the space absolute seat number dependent strategies.

eastbay

[eastbay]

[ QUOTE ]

p 105 "There is however a more elegant,purely logical proof that equal players in a symetrical situation must win exactly in proportion to the size of their stacks. Supose these two players agreed to play the same freezout everyday for ten years. Given they are equal players, their long term reults would necessarily be to break even. Well in order for them to break even, the smaller stack must win exactly the same fraction of times as the fraction of the total chips he possesses. This reasoning extends to any number of players as long as they play equally well, and there is no positional advantage for one compared to the other."

[/ QUOTE ]

Sklansky either means something strange by "equal players" or "in a symmetrical situation", or "the same freezout" or he's simply wrong.

The two players break even in my scenario, and they do not win in proportion to the size of their stacks. He says "the same reasoning" except that he supplies none.

There is only one strict symmetry requirement for the equity function and that is that a player must win half the time if he has half the chips. Any curve which passes through (0,0), (0.5,0.5), and (1,1), and has the property that eq(x) +eq(1-x) = 1. is admissible. This includes an infinite number of variously skewed "S" curves, ranging from fully linear to a step function for the degenerate case I provided.

eastbay

[jcm4ccc]

[ QUOTE ]
it's not clear to me now if there's a problem in using the same tournament results multiple times so long as the chip stack distribution for which you're recording it doesn't occur more than once in the same tournament

[/ QUOTE ]

It is wrong to use the same tournament results multiple times in your analysis, regardless of the chip stack distribution. Here’s an example of why it’s wrong:

Suppose you have played in 2 tournaments in a $100 tourney. When it got down to 4-handed, you had 2000 chips (both times). According to ICM, you should win 20% of the prize pool in each tournament. $1000 * 2 * 20% = $400 you are expected to win.

Let’s say that you win one tournament, and you lose one tournament. Your total prize is $500, or 25% of the prize pool. The ICM model predicted you would win 20% of the prize pool, and you won 25%. You did better than expected. This is pretty straightforward.

Now, let’s suppose that the bubble lasted 4 hands in the tournament that you won, and the bubble lasted 1 hand in the tournament that you lose. You decide to include each hand in your analysis (as long as the chip stack distribution is different). Here’s what your data might look like (I don’t know how to do tables):

Tourney__Hand__Chip Count__ICM Pred__Actual
1________1_____2000________20%_______50%
1________2_____2100________21%_______50%
1________3_____1900________19%_______50%
1________4_____1500________15%_______50%
2________1_____2000________20%_______00%
TOTAL_____________________19%_______40%


By this analysis, it looks like you won 40% of the prize pool, when actually you only won 25% of the prize pool. You look a lot better than you actually are. That is because you overweighted the tournament in which you won.

We can reverse this, of course. Let’s suppose that the bubble lasted 4 hands in the tournament that you lost, and the bubble lasted 1 hand in the tournament that you won. Here’s the analysis:

Tourney__Hand__Chip Count__ICM Pred__Actual
1________1_____2000________20%_______00%
1________2_____2100________21%_______00%
1________3_____1900________19%_______00%
1________4_____1500________15%_______00%
2________1_____2000________20%_______50%
TOTAL_____________________19%_______10%


Now you look worse than you actually are. That is because you overweighted the tournament in which you lost. You need to weight each tournament equally, or your analysis doesn’t work.

[eastbay]

[ QUOTE ]
[ QUOTE ]
it's not clear to me now if there's a problem in using the same tournament results multiple times so long as the chip stack distribution for which you're recording it doesn't occur more than once in the same tournament

[/ QUOTE ]

It is wrong to use the same tournament results multiple times in your analysis, regardless of the chip stack distribution. Here’s an example of why it’s wrong:

Suppose you have played in 2 tournaments in a $100 tourney. When it got down to 4-handed, you had 2000 chips (both times). According to ICM, you should win 20% of the prize pool in each tournament. $1000 * 2 * 20% = $400 you are expected to win.


[/ QUOTE ]

I think you're missing something. My equity is not only a function of my stack, it's a function of the remaining stacks as well. If I have 2000 chips and the chips are evenly distributed amongst the other players, clearly I have less equity than if for my same 2000 chips, one other player only has 1 chip left. This is what I mean by chip stack distribution.

My question relates to using data from the same tournament to compute the expectation for _different_ chip stack distributions. Your example was lumping all the data into one equity expectation. I am asking about the case where you are computing equity _for each distribution_.

eastbay

[jcm4ccc]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
it's not clear to me now if there's a problem in using the same tournament results multiple times so long as the chip stack distribution for which you're recording it doesn't occur more than once in the same tournament

[/ QUOTE ]

It is wrong to use the same tournament results multiple times in your analysis, regardless of the chip stack distribution. Here’s an example of why it’s wrong:

Suppose you have played in 2 tournaments in a $100 tourney. When it got down to 4-handed, you had 2000 chips (both times). According to ICM, you should win 20% of the prize pool in each tournament. $1000 * 2 * 20% = $400 you are expected to win.


[/ QUOTE ]

I think you're missing something. My equity is not only a function of my stack, it's a function of the remaining stacks as well. If I have 2000 chips and the chips are evenly distributed amongst the other players, clearly I have less equity than if for my same 2000 chips, one other player only has 1 chip left. This is what I mean by chip stack distribution.

My question relates to using data from the same tournament to compute the expectation for _different_ chip stack distributions. Your example was lumping all the data into one equity expectation. I am asking about the case where you are computing equity _for each distribution_.

eastbay


[/ QUOTE ]

Plain and simple, your "actual equity" in these charts is wrong if you are using more than one data point from the same tournaments, for the reasons I outlined above.

stacks: actual equity, icm predicted equity, difference, standard error of sample, number of data points in this range
0,0,0,0: 0.282925; icm = 0.239223; delta = 0.0437015; std err=0.00508585; N=1265
0,0,0,1: 0.266667; icm = 0.195756; delta = 0.0709111; std err=0.0189215; N=90
0,0,1,0: 0.244944; icm = 0.203231; delta = 0.0417128; std err=0.0190025; N=89
0,1,0,0: 0.215385; icm = 0.18758; delta = 0.0278049; std err=0.0148834; N=117
1,0,0,0: 0.426316; icm = 0.393979; delta = 0.0323365; std err=0.0221329; N=19

Actually, I can't see how the "actual equity" could ever be carried out to six decimal places. In any tournament, you win either 0.5, 0.3, 0.2, or 0.0 of the prize money. No matter how you add these numbers, you would need at most one decimal place. How do you ever manage to win 28.2925% of the prize pool?

[jcm4ccc]

[ QUOTE ]
Actually, I can't see how the "actual equity" could ever be carried out to six decimal places. In any tournament, you win either 0.5, 0.3, 0.2, or 0.0 of the prize money. No matter how you add these numbers, you would need at most one decimal place. How do you ever manage to win 28.2925% of the prize pool?

[/ QUOTE ]

Actually, that's a stupid statement on my part. Please ignore.

[eastbay]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
it's not clear to me now if there's a problem in using the same tournament results multiple times so long as the chip stack distribution for which you're recording it doesn't occur more than once in the same tournament

[/ QUOTE ]

It is wrong to use the same tournament results multiple times in your analysis, regardless of the chip stack distribution. Here’s an example of why it’s wrong:

Suppose you have played in 2 tournaments in a $100 tourney. When it got down to 4-handed, you had 2000 chips (both times). According to ICM, you should win 20% of the prize pool in each tournament. $1000 * 2 * 20% = $400 you are expected to win.


[/ QUOTE ]

I think you're missing something. My equity is not only a function of my stack, it's a function of the remaining stacks as well. If I have 2000 chips and the chips are evenly distributed amongst the other players, clearly I have less equity than if for my same 2000 chips, one other player only has 1 chip left. This is what I mean by chip stack distribution.

My question relates to using data from the same tournament to compute the expectation for _different_ chip stack distributions. Your example was lumping all the data into one equity expectation. I am asking about the case where you are computing equity _for each distribution_.

eastbay


[/ QUOTE ]

Plain and simple, your "actual equity" in these charts is wrong if you are using more than one data point from the same tournaments, for the reasons I outlined above.


[/ QUOTE ]

You're missing the point. I know this because the reasons you gave are not relevant to the question at hand. In your example you lumped all of the data together. My initial results did this too, and I agree that this is wrong for the reasons you gave.

But now there's a new question, which I've stated above but which I think you aren't understanding yet. Do you understand what ICM is? I'm not saying that to be condescending, you've made some statements which seem to indicate that you don't.

eastbay

[jcm4ccc]

I understand ICM.

It's your study. I'll step aside for now.

[eastbay]

[ QUOTE ]
I understand ICM.

It's your study. I'll step aside for now.

[/ QUOTE ]

I'm interested in constructive criticism. It's just not clear that you're understanding the question.

Here's a relevant example:

I have two data points each from two tournaments:

1000 2000 1000 0.3
1000 1000 2000 0.3

1000 2000 1000 0.2
1000 1000 2000 0.2

I want to calculate the expectation of (1000,2000,1000) and (1000,1000,2000). Are you contending that I cannot use all four data points - two for each of the two distributions - to do so? If not, why not?

Edit: make example clearer that distributions are about other stacks, not mine.

eastbay

[jcm4ccc]

[ QUOTE ]
I'm interested in constructive criticism. It's just not clear that you're understanding the question.

Here's a relevant example:

I have two data points each from two tournaments:

1000 2000 1000 0.3
1000 1000 2000 0.3

1000 2000 1000 0.2
1000 1000 2000 0.2

I want to calculate the expectation of (1000,2000,1000) and (1000,1000,2000). Are you contending that I cannot use all four data points - two for each of the two distributions - to do so? If not, why not?

Edit: make example clearer that distributions are about other stacks, not mine.

eastbay

[/ QUOTE ]

Independence of observations is an assumption of most statistical procedures. Some assumptions can be violated and the results can still be interpreted (for example, the normal distribution assumption). But that one can't.

If I understand your variables, they are:

Variable 1: Chip count of Player 1
Variable 2: Chip count of Player 2
Variable 3: Chip count of Player 3
Variable 4: Your results in the tournament

The problem is that your method may actually obscure your findings. Take a look at your first two observations. In observation #1, player 2 has 2000 chips. In observation #2, player 3 has 2000 chips. And yet the outcome is the same: you ended up in 2nd place. So if we just went by the first two observations, we would have to say that it makes no difference whether player 2 or player 3 has the most chips. But I think that's a mistake in your method, not an actual finding.

I may still be misinterpreting your study, but that doesn't really matter. If you are violating the independence of observations assumption, you are creating problems.

If you are interested in chip count distribution and its effect on your outcome in the tournament, I would pick one particular spot in a tournament (say, the first hand of the bubble, or the last hand of the bubble). Use the variables above, plus your own chip count (you have to include your own chip count, since that will have the most effect on your outcome). Do a regression analysis. The variable with the most effect, without a doubt, will be your own chip count. But what you could find is whether the chip count distribution has an effect, beyond the effect of your chip count.

You could even do multiple, separate studies. Look at the findings at different points in the tournament:

1) first six-handed hand
2) first five-handed hand
3) first four-handed hand

etc. You may find that chip count distribution is important when it is four-handed and five-handed, but not when it is six-handed. Or something interesting like that.

Again, I may be misinterpreting what you are trying to do. But nonetheless, you can't violate the assumption of independence of observations.

[eastbay]

[ QUOTE ]

If you are interested in chip count distribution and its effect on your outcome in the tournament, I would pick one particular spot in a tournament (say, the first hand of the bubble, or the last hand of the bubble). Use the variables above, plus your own chip count (you have to include your own chip count, since that will have the most effect on your outcome). Do a regression analysis. The variable with the most effect, without a doubt, will be your own chip count. But what you could find is whether the chip count distribution has an effect, beyond the effect of your chip count.



[/ QUOTE ]

Comment 1: Are you sure you understand ICM? This discussion implies you don't.

The whole point of an equity model like ICM is to find deviations from chipEV, which considers your stack only, and $EV, which considers all stacks. Of course you'll find that bubble play has the largest deviations, with chipEV approaching $EV as you move away from the bubble. We already know this.

eastbay

[eastbay]

[ QUOTE ]
[ QUOTE ]
I'm interested in constructive criticism. It's just not clear that you're understanding the question.

Here's a relevant example:

I have two data points each from two tournaments:

1000 2000 1000 0.3
1000 1000 2000 0.3

1000 2000 1000 0.2
1000 1000 2000 0.2

I want to calculate the expectation of (1000,2000,1000) and (1000,1000,2000). Are you contending that I cannot use all four data points - two for each of the two distributions - to do so? If not, why not?

Edit: make example clearer that distributions are about other stacks, not mine.

eastbay

[/ QUOTE ]

Independence of observations is an assumption of most statistical procedures. Some assumptions can be violated and the results can still be interpreted (for example, the normal distribution assumption). But that one can't.

If I understand your variables, they are:

Variable 1: Chip count of Player 1
Variable 2: Chip count of Player 2
Variable 3: Chip count of Player 3
Variable 4: Your results in the tournament

The problem is that your method may actually obscure your findings. Take a look at your first two observations. In observation #1, player 2 has 2000 chips. In observation #2, player 3 has 2000 chips. And yet the outcome is the same: you ended up in 2nd place. So if we just went by the first two observations, we would have to say that it makes no difference whether player 2 or player 3 has the most chips. But I think that's a mistake in your method, not an actual finding.


[/ QUOTE ]

The issue here is that we are computing several different things from the same pool of data which has some non-independent data. The question is, does the data have to be independent for each quantity that we are computing, or does it have to be independent between quantities as well. Do you see? I think I am going to have to do an experiment to settle it.

eastbay

[jcm4ccc]

[ QUOTE ]
Comment 1: Are you sure you understand ICM? This discussion implies you don't.

[/ QUOTE ]

i barely understand it. probably not very well. but i understand statistical research very well.

[jcm4ccc]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I'm interested in constructive criticism. It's just not clear that you're understanding the question.

Here's a relevant example:

I have two data points each from two tournaments:

1000 2000 1000 0.3
1000 1000 2000 0.3

1000 2000 1000 0.2
1000 1000 2000 0.2

I want to calculate the expectation of (1000,2000,1000) and (1000,1000,2000). Are you contending that I cannot use all four data points - two for each of the two distributions - to do so? If not, why not?

Edit: make example clearer that distributions are about other stacks, not mine.

eastbay

[/ QUOTE ]

Independence of observations is an assumption of most statistical procedures. Some assumptions can be violated and the results can still be interpreted (for example, the normal distribution assumption). But that one can't.

If I understand your variables, they are:

Variable 1: Chip count of Player 1
Variable 2: Chip count of Player 2
Variable 3: Chip count of Player 3
Variable 4: Your results in the tournament

The problem is that your method may actually obscure your findings. Take a look at your first two observations. In observation #1, player 2 has 2000 chips. In observation #2, player 3 has 2000 chips. And yet the outcome is the same: you ended up in 2nd place. So if we just went by the first two observations, we would have to say that it makes no difference whether player 2 or player 3 has the most chips. But I think that's a mistake in your method, not an actual finding.


[/ QUOTE ]

The issue here is that we are computing several different things from the same pool of data which has some non-independent data. The question is, does the data have to be independent for each quantity that we are computing, or does it have to be independent between quantities as well. Do you see? I think I am going to have to do an experiment to settle it.

eastbay

[/ QUOTE ]

your observations have to be independent. no ifs, ands, or buts, regardless of my poor understanding of ICM. your variables do not have to be independent.

[mistaken]

[ QUOTE ]
Ok, I thought of one. Consider HU match. Strategy A is to push when you have half or more of the chips, and fold if you have less than half. Clearly equity distribution is a step function for two players both playing this strategy.

[/ QUOTE ]

Is this a generally accepted strategy?

I usually apply the "Q higher or better in the SB = push against the average heads up player when blinds are 250/500 or higher... BB play varies"

[mistaken]

Please disregard this post...
I must have been high or daydreaming when writing it /images/graemlins/tongue.gif.

Mistaken

[PrayingMantis]

Interesting post.

I'm thinking however that this kind of empirical $EV model, might create some paradoxical implications. I'm not a mathematician, so please forgive me for the language and terms in which I'll make my point.

Let's assume you're on the bubble with whatever stacks and blinds, you're on the BB and button pushes. Folded to you. Now you put button on a specific range of hands, and lets say you use normal ICM to decide whether this is a call or not with your hand. This is a common pratice for you in such spots.

Now, let's say you have played enough SNGs to come out with your "true" empirical $EV (which is obviously higher than the what ICM tells you). Now you find yourself at the same spot described about. But instead of using the normal ICM to make your decision, you now base it upon your "empirical EV" model. Not surprisingly, your decision is now different. with the same range of hands you put button on, you now probably need a somewhat better hand to call with. Suppose you fold, where you used to call earlier, based on pure ICM.

Do you see the paradoxical side of this? Your $EV for this situation was based upon making moves which now you will not do (or do differently), and therfore your equity might actually go down from now on.

This brings to mind, in a way, the way Helmuth plays in the last few years. His big success in the past ("empirical equity") made him feel he is able to pass up on situations in which he is a big favorite, in order to "preserve" his equity. It is probably not a mistake to assume that his true equity has really dropped down because of this.

Any thoughts about this problematic side of what you call empirical equity?

[Slim Pickens]

I'm sorry. I walked in late and I need a clarification. Why do you argue in this thread that your data points within the same tournament can be treated as independent trials, but in The Shadow's recent thread (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=singletable&Number=2345875 &page=2&view=collapsed&sb=5&o=14&fpart=2), you seem to make the opposite argument?

[ 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.

[/ QUOTE ]

This case is four-handed, so there is a chip distribution, and the other is two-handed, so only your chip count is variable. Is that it? If so can you explain why?

Slim

[eastbay]

[ QUOTE ]
I'm sorry. I walked in late and I need a clarification. Why do you argue in this thread that your data points within the same tournament can be treated as independent trials,


[/ QUOTE ]

I don't think I did. From thew few posts I re-read, I asked a question, and then the thread went sideways about something else and we never resolved it.

[ QUOTE ]

in The Shadow's recent thread (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=singletable&Number=2345875 &page=2&view=collapsed&sb=5&o=14&fpart=2), you seem to make the opposite argument?

[ 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.

[/ QUOTE ]

This case is four-handed, so there is a chip distribution, and the other is two-handed, so only your chip count is variable. Is that it? If so can you explain why?

Slim

[/ QUOTE ]

Truth is I'm not certain which methodology is better: use all the data points or use one.

Certainly using one is "safer" as it removes the independence question, but it severely compounds the sample size problem.

eastbay

[eastbay]

[ QUOTE ]
Interesting post.

I'm thinking however that this kind of empirical $EV model, might create some paradoxical implications.

Any thoughts about this problematic side of what you call empirical equity?

[/ QUOTE ]

I've thought about this, and concluded that I don't think it's paradoxical. It will simply take awhile (possibly a very long time) for the system to feed back on itself to reach the best strategy which is self-consistent, where the results which influence your strategy are consistent with the strategy you are currently using, and not "lagged" to the strategy you were using before.

In your example, say you tighten up too much based on your results which say you can pass some edges. Well, if this is indeed wrong, your results will begin to converge back to the mean, which will take you back towards your original strategy. Presumably there is a happy medium where the strategy and the results are consistent.

There is actually an algorithm in game theory, I've forgotten the name of it, which is guaranteed to find optimal strategies that works pretty much exactly like this.

eastbay

[the shadow]

[ QUOTE ]
There is actually an algorithm in game theory, I've forgotten the name of it, which is guaranteed to find optimal strategies that works pretty much exactly like this.


[/ QUOTE ]

Trial and error? /images/graemlins/smile.gif



Seriously now, were you thinking of backwards induction (http://www.economics.laurentian.ca/Strategic_Think.27/Modules/Course_Schedule.98/dictionary2.htm#B) a/k/a rollback a/k/a Zermelo's algorithm? If so, my recollection is that it applies to games of perfect information. Cite (http://plato.stanford.edu/entries/game-theory/). If not, I'd like to hear more about what you were thinking of, since I'm trying to learn some more game theory.

The Shadow

[eastbay]

[ QUOTE ]
[ QUOTE ]
There is actually an algorithm in game theory, I've forgotten the name of it, which is guaranteed to find optimal strategies that works pretty much exactly like this.


[/ QUOTE ]

Trial and error? /images/graemlins/smile.gif



Seriously now, were you thinking of backwards induction (http://www.economics.laurentian.ca/Strategic_Think.27/Modules/Course_Schedule.98/dictionary2.htm#B) a/k/a rollback a/k/a Zermelo's algorithm? If so, my recollection is that it applies to games of perfect information. Cite (http://plato.stanford.edu/entries/game-theory/). If not, I'd like to hear more about what you were thinking of, since I'm trying to learn some more game theory.

The Shadow

[/ QUOTE ]

Ha. No, that's not it. It's in every game theory 101 book I've ever read, though. The analogy may not be perfect, but the basic idea of iterating strategies and adjusting based on past results is there.

eastbay

[Slim Pickens]

[ QUOTE ]
[ QUOTE ]
I'm sorry. I walked in late and I need a clarification. Why do you argue in this thread that your data points within the same tournament can be treated as independent trials,


[/ QUOTE ]

I don't think I did. From thew few posts I re-read, I asked a question, and then the thread went sideways about something else and we never resolved it.

[ QUOTE ]

in The Shadow's recent thread (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=singletable&Number=2345875 &page=2&view=collapsed&sb=5&o=14&fpart=2), you seem to make the opposite argument?

[ 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.

[/ QUOTE ]

This case is four-handed, so there is a chip distribution, and the other is two-handed, so only your chip count is variable. Is that it? If so can you explain why?

Slim

[/ QUOTE ]

Truth is I'm not certain which methodology is better: use all the data points or use one.

Certainly using one is "safer" as it removes the independence question, but it severely compounds the sample size problem.

eastbay

[/ QUOTE ]

It seems like jcm gave a compelling argument that multiple points from the same tournament violates the necessary condition that observations be independent of each other. I have only basic training in statistics, but it seems to me that since having X chip distribution necessarily leads to a distribution of the number of observations of finish A that is different than Y chip distribution, then without removing this effect from ther results, they are not independent. I know that's nothing new but I guess I cast my vote on the side of one point per tournament.

Slim

[eastbay]

[ QUOTE ]

It seems like jcm gave a compelling argument that multiple points from the same tournament violates the necessary condition that observations be independent of each other. I have only basic training in statistics, but it seems to me that since having X chip distribution necessarily leads to a distribution of the number of observations of finish A that is different than Y chip distribution, then without removing this effect from ther results, they are not independent. I know that's nothing new but I guess I cast my vote on the side of one point per tournament.

Slim

[/ QUOTE ]

First problem is that independence is not a binary thing. It's shades of gray. If you want to be a purist, you could say that no two games where two players faced each other twice were permissible, since they may have learned about each other, and the resulting data is not independent in that case. The second result depends on what was learned in the first. But, at some point you have to draw the line, and do some computing rather than just disposing.

Second issue is that even two most certainly dependent data points may get used in such a way that dependence doesn't matter.

In any case, we vote the same way about the first thing to try.

eastbay

[the shadow]

I'm glad that a consensus is forming about the data dependence/independence issue.

However, the more I think about it, the bigger the selection bias issue appears to be. Let's analogize a SNG to a basketball game or a football game. If we take a look at the score at halftime or the start of the fourth quarter, I'll bet that the team with the greater points is more likely to win the game. But that lead is due, at least in part, to a difference in skill between the two teams and their coaching staffs.

Now, if we take a look at the chip count in a SNG anytime after the starting position, we have to recognize that at least some of the difference is due to relative differences in skills. After all, it's a fact that some players have higher ITM%s/ROIs than others. That's one of the "shortcomings" of ICM and why eastbay has considered modifying ICM with a skill factor (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=singletable&Number=183 8524&Forum=f22&Words=skill&Searchpage=0&Limit=25&M ain=1838524&Search=true&where=bodysub&Name=5671&da terange=1&newerval=&newertype=w&olderval=&oldertyp e=&bodyprev=#Post1838524).

If we use chip counts from the middle of actual HU tourneys, I cannot see a way to get around selection bias. The same applies with even greater force if we use chip counts once a SNG has become heads up. After all, the two players accumulated their chips at least in part through their relative skill over their opponents and those skill differences had more time and chips with which to express themselves.

As a result, the only way that I can see to use data from live tourneys to negate the null hypothesis (i.e., that the equity function is linear in a HU freezeout) is to use random starting chip stacks. That way the differences in the initial condition don't reflect skill differences. Of course, that pretty much rules out using data from traditional online tournaments.

If data from actual SNGs were used, the results may not be sufficient to negate the null hypothesis, but if that data, notwithstanding the selection bias, was still consistent with the null hypothesis, it might tend to make the hypothesis more likely. Given the difficulty of data collection and selection bias issue, it seems to me that the most fruitful approach at this point is to double-check gumpzilla's argument (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=2362743&page=1&view=c ollapsed&sb=5&o=31&vc=1).

The Shadow (who's more impressed by gumpzilla's argument than David Sklansky's proof)

[holdem2000]

I haven't read every reply so maybe this was already discussed...

ICM assumes players have the same strategies, which whether or not it is true, increasing blinds ruins. If every players plays a shorstack tight and a tallstack loose on the bubble at 25/50, but is loose/agressive on any stack at 200/400, then for a given tournament with rapidly rising blinds, the player on a large stack will have a "different" strategy I think.

If the blinds stay constant ICM claims the symmetry of strategies ensures linear equity functions; however, I don't think this applies to constantly rising blinds.