Existence of -cEV yet +Equity Plays?

[Slim Pickens]

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If I'm BB with AA on the first hand of a SNG and everyone goes all in I fold. I think that's a +$ev fold.

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Can you show this without ignoring chops? You're certainly not 60% to win by calling, which in a 50/30/20 ignoring chops is about what you'd need.

[Galwegian]

This is a really interesting question. I don't have the answer but I do remeber that pzhon made a post about this a few months back - I think it was in science, math philosophy. It was quite a mathematical post but the central point was that ICM is a convex function. That is, if we think of ICM as providing a function from chip count to $ equity then this function is convex. In particular, this would imply that the answer to your question is no (at least under ICM). Sorry I don't have the link to pzhon's post (I am 2+2 search illiterate).
Of course ICM is only a model - in particular it assumes equal skill level. I imagine that it would be fairly easy to come up with examples if you allow players of different skill levels (e.g. fold to an all in raise by an inferior player where you are only slightly ahead).

[Galwegian]

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For now, I'll just say that assuming everyone is equally skilled, ICM is not just an "accepted mathematical model." Granted, it assumes that everyone is equally skilled, but given the push or fold end game of most tournaments...especially STTs...that assumption is highly valid, and proving the general formula isn't too bad (writing it down, on the other hand, is a notational nightmare, and I'm not sure I can do it on this forum without Equation Editor).



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Hi Tony
I have to disagree with this statement. I think that ICM is precisely a mathematical model (given equally skilled players). In fact, even though experience shows that it gives accurate predictions of equity I have yet to see any convincing argument as to why it is the "right" model. There is a good argument for saying that proportion of total chips = prob of finishing first. However, ICM goes further than that and basically claims that event [player A finishes in position i] is independent of the event [player B finishes in position j]. I see no convincing argument for this assumption.

[1p0kerboy]

What about late in a tournament a shortstack goes all-in. He gets a caller and you call out of the big blind under the notion that you and the other player are going to "check-it-down" to increase the chances of eliminating the shortie and moving up the payscale.

[Slim Pickens]

[ QUOTE ]
There is a good argument for saying that proportion of total chips = prob of finishing first. However, ICM goes further than that and basically claims that event [player A finishes in position i] is independent of the event [player B finishes in position j]. I see no convincing argument for this assumption.

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It does? How does P(A finishes ith given B finishes jth) make any assumptions other than those needed to say P(A,i)=S_i/sum(S_j)? I can see arguments, but they all involve some "skill" event, and ICM already assumes chips just sort of wander around the table randomly.

[Galwegian]

here is pzhon's post on the convexity of ICM (it was in Poker Theory)

http://forumserver.twoplustwo.com/showfl...rue#Post7952501

[Galwegian]

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[ QUOTE ]
There is a good argument for saying that proportion of total chips = prob of finishing first. However, ICM goes further than that and basically claims that event [player A finishes in position i] is independent of the event [player B finishes in position j]. I see no convincing argument for this assumption.

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It does? How does P(A finishes ith given B finishes jth) make any assumptions other than those needed to say P(A,i)=S_i/sum(S_j)? I can see arguments, but they all involve some "skill" event, and ICM already assumes chips just sort of wander around the table randomly.

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Not sure I understand your first statement/question. However, I will say this - ICM is not a model based on the assumption that the chips wander arouns the table randomly. In fact the random walk model has been studied and is demonstrably (at least for 3 players)different from ICM i.e it gives different values for $ equity. The problem with the random walk model is that it is very difficult to calculate accurately for more that 3 players whereas ICM is very easy to calculate (that is the great advantage of ICM)

[Slim Pickens]

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What about late in a tournament a shortstack goes all-in. He gets a caller and you call out of the big blind under the notion that you and the other player are going to "check-it-down" to increase the chances of eliminating the shortie and moving up the payscale.

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This assumes future action (although it's not terrible to treat it as having the action closed) and contains a microstack. The thing with the microstack is that the shorter you make him, the larger the relative error in his equity, and thus the larger error in the equity you can take by knocking him out. Saying a play that's -0.00001% cEV of the total chips in play is +0.00001% $EV is fairly meaningless when the short stack has 0.5% of the prize pool by ICM and the error is something like 0.1%. I'm not saying I know that's what will happen but that's my guess.

[Dima2000123]

Did you consider the value of having a big stack?

[Slim Pickens]

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[ QUOTE ]
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There is a good argument for saying that proportion of total chips = prob of finishing first. However, ICM goes further than that and basically claims that event [player A finishes in position i] is independent of the event [player B finishes in position j]. I see no convincing argument for this assumption.

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It does? How does P(A finishes ith given B finishes jth) make any assumptions other than those needed to say P(A,i)=S_i/sum(S_j)? I can see arguments, but they all involve some "skill" event, and ICM already assumes chips just sort of wander around the table randomly.

[/ QUOTE ]

Not sure I understand your first statement/question. However, I will say this - ICM is not a model based on the assumption that the chips wander arouns the table randomly. In fact the random walk model has been studied and is demonstrably (at least for 3 players)different from ICM i.e it gives different values for $ equity. The problem with the random walk model is that it is very difficult to calculate accurately for more that 3 players whereas ICM is very easy to calculate (that is the great advantage of ICM)

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I didn't mean to be confusing. My only point was that any argument saying that conditional finish probabilities used in the ICM are just as valid as the 1st place finish probability. The ways to argue against it involve claiming the players are skilled rather than mindless drones. Also, doesn't the ICM match the random walk just fine enough as long as there isn't a microstack involved? I thought that was the origin of that requirement.