Testing ICM -- some questions for discussion

[Degen]

let there be flame


Andre

[marv]

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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?


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(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.

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Assume that one player gives another player odds 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?


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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 (they are the ones that developed the AI behind Poker Academy). 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]

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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.

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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]

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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.


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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]

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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.

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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]

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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.

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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]

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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.

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Take a look at marv's post.

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