Mathematical Hand Analysis (the EV of pushing 99 from MP)

[A_PLUS]

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it is positive EV.
but that doesn't mean it is the best course of action.
I would much prefer doing a standard raise here.

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How about a little color?

[sdplayerb]

well you call your post a mathematical hand analysis.
you only did it on push or fold.
if it is an analysis on the hand, you should do the other scenario, raise.

[A_PLUS]

I fully realize that. I thought I made this clear in my post, but I forgot that this was kind of done in two posts.

What I was trying to do, was work through a simple problem. (push or fold), to see if my math was correct for the simple example. I was hoping for feedback as to the accuracy of my analysis. I 100% fully realize that the anlysis is incomplete, but I did not want to move on unless I was sure the 1st step was correct.

I only disagreed with the "you should have raised" posters, b/c I am not sure they are correct (they may be), I was hoping they would provide some insight into why they thought what they did so that we could work through the problem that way.

[SeriousStudent]

The problem with "other plays will be more EV" is that you will lose 1.5BB each orbit 80% of the time if you do nothing, which not creates a smaller stack, but also

- affects the hands that others call with, making it more likely that you will be called, and be busted out
- if you are called and doubled up, then you are 3BB down from where you would have been and 6BB on the next double up

So, my question is how to best account for these 3 effects in a mathmatical model - smaller absolute stack, more chances of being called on future all ins, smaller stack even when doubled up

Possibly the best approach would be to write a simple software simulator which would play out these scenarios over several orbits.

[AlwaysWrong]

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You can not treat a MTT with 110 people left 40 paying the same as an with 11 people and 4 paying.

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I didn't. I said it is somewhat similar, as an illustration.

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Also b/c if it works for 110 people, where do you draw the line and why?

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There is no line. There is only one situation in a tournament where CEV = $EV. That is when there are two people left. (And a tournament where only first place gets a prize.) At ANY OTHER TIME, CEV doesn't = $EV. The only question is how much they differ by. ICM is a decent way of guessing that - there is no exact way.

Intuitively, CEV gets closer to $EV the further away you are from the money. In your example you are close to the money, so CEV will be significantly different from $EV. How much exactly? Can't say, nobody can. But ICM gets you a rough estimate.

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I am giving him 3 cards to beat my hand.

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No, he paid for those 3 cards. You didn't give him anything. If you feel you're really bad at playing in these spots, maybe pushing in these spots is the right course of action. But that's a seperate thread and somehting you should be working on improving in your game.

[A_PLUS]

Alwayswrong, I appreciate your comments. My arguements are only half rooted in my opinions. Mostly I just dont know, you have some interesting points, I just want to understand your underlying reasoning.

You keep alluding to using ICM. This is a quote from Fnord_too's post on CEV ~ EV. He looked into the ICM code.

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Just read their code. It looks like what it is doing is the following:

Everyone has a their chips/total chips chance of winning.

For each winning person, it calculates the probability of each other person coming in second by removing the winning players chips from the pool and using to above formula to see who wins amongst the remaining players. You multiply the prob of the assumed winner winning times the prob of the each player winning (in the now reduced field) to get the prob that that 1/2 scenario occurs and for each player sum the the probabilities of all 1/2 combinations where they are two.

Then basically repeat for each 1/2/3 combination.


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I'll admit, I dont fully follow his description, but it looks like CEV = EV when you are not in the money. I am not quoting this to say that your assertations are wrong, only to say that I dont think the ICM agrees.

What I think would be very good to discuss is how you have come to believe that CEV does not closely approximate EV in the middle-late stages of a MTT, when you still have 70 players remaining before payouts start. Im not saying its not true, I just dont like believing things b/c someone says so.

2nd. I really dont know if pushing 99 in this spot is max EV. maybe, maybe not. I was just playing devils advocate to your post. SO, you believe a standard raise is the best play. Could you please explain in detail why? I think most people who have made it to the bottom of this geeky ass thread are beyond needing "this way is the best" posts without an explanation of your thought process.

thanks

[AlwaysWrong]

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Everyone has a their chips/total chips chance of winning.

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At an arbitrary point in the tournament, in or out of the money. You just have a list of eg 110 people's chip stacks.

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For each winning person

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What you're doing here is the traditional fraction of total chips in play = probability of coming in first. So if there are 110 people and you have an exactly average stack size, you have a 1/110 chance of winning. So what we're going to do is have 110 scenarios, 1 for each person winning. At the end we're going to multiply the result from each person winning by the probability that they will win. That's what he means by "for each". Obviously there cannot actually be 110 winners, there can only be one. But we consider each possible case and see how likely it is.

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it calculates the probability of each other person coming in second by removing the winning players chips from the pool and using to above formula to see who wins amongst the remaining players.

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So in each of those 110 scenarios, it removes the winner from the tournament and pretends everyone is playing for second. Your chance of second = the fraction of the chips you have in play. So in our scenario if you had an average stack you'd now have a 1/109 chance of winning.

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You multiply the prob of the assumed winner winning times the prob of the each player winning (in the now reduced field) to get the prob that that 1/2 scenario occurs and for each player sum the the probabilities of all 1/2 combinations where they are two.

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So 1/2 here means 1st place / 2nd place. For 110 people there are 110*109=11990 possible combinations. We're going to assign a probability to each of those combinations, and then sum all the results.

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Then basically repeat for each 1/2/3 combination.

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1/2/3 because SNGs pay 3 places. In a tournament that pays 40 places, you'd have to iterate it for each spot.

Hope that's fairly clear. Trust me on the fact that I know what I'm talking about and ICM does apply to an arbitrarily sized tournament, at any time during that tournament.

[sdplayerb]

your math is correct.
i have a model created that does the math for me. the win% when called on mine isn't perfect, but close enough.
mine says the ev is 141, which is close to yours.
yours is correct.

[sdplayerb]

there is no need to worry about this, because if nobody folded an A, then the chance somebody else has one goes up.
they completely cancel each other out.

as per the more than one caller, it really happens so incredibly little that you don't need to worry about it.


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For instance, someone will fold A7, but that will affect the number of Aces available for hands like AK that will call.

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