Testing ICM -- some questions for discussion

[eastbay]

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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).


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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

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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]

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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

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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, 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]

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[ 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

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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

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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]

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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.

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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]

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

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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. It also seems to be a logical inference from what eastbay was suggesting.

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]

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

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

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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:



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]

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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.

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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

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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]

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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.

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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.

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. 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