Testing ICM -- some questions for discussion

[eastbay]

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

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

eastbay

[Apathy]

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

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

eastbay

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

[schwza]

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No wonder this is so inaccurate

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

[Apathy]

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No wonder this is so inaccurate

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

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

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

[the shadow]

The owner -- Adam Schultz, who posts here as AdamSchultz -- has made several posts acknowledging some data collection problems and bugs, but asserting that those problems have been corrected and offering refunds to dissatisfied customers. According to Adam, at least one problem apparently concerns PP incorrectly listing 3-player games that actually have 4 players. Other posters complain about continuing incorrect information. I don't own, haven't used, and get no commission on Poker Prophecy, so I don't know from personal experience.

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

The Shadow

[hansarnic]

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


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With you so far.

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So,

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


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

[gumpzilla]

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

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

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

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

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

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

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

[poincaraux]

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

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

[gumpzilla]

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

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

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

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

[Pokerscott]

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

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

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

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

Pokerscott