Did FGS/PSM show that ICM overvalued or undervalued bigger stacks?

And what is PSM exactly?

But isn't it risk-averse only for calling all-in? For the same reason, doesn't it encourage risk-taking when you are the one pushing? But if that is the case, these two balance each other out (that's not to say that ICM necessarily does the balancing perfectly).

You give a good theoretical example to demonstrate your point, but I doubt anyone will be using ICM in that kinds of situation.

I can kinda see your argument working in the sense that short stacks have a much greater chance of their stacks decreasing even a little bit, causing their all-in moves to lose significant fold equity – something bigger stacks never have to worry about.

Not as part of this simulation. We only simulated the games and recorded the outcomes / scores for each model. (PSM is a proprietary model used in "SNG Solver". Details on the algorithm were never published as far as i know, but it seems to be fairly similar to FGS-1.)

Regarding your general question though, i believe that really depends on the payouts. ICM assigns the finishing-percentages for each place based only on the stacks, not the payouts.

e.g.:
Stacks 1bb, 100bb, 100bb

1) Payouts 50/50/0
2) Payouts 95/5/0

ICM assumes that the short-stack will finish 3nd with the same frequency for both scenarios (~98.5%). This is very clearly not the case in practice with competent players, since the big stacks should never collide in 1). The big-stacks are likely under-evaluated in 1) and/or over-evaluated in 2).

This is a great example. On the other hand, it is a rare scenario. Do you think your last sentence generalizes to all (and in particular more common) cases of stack sizes ranging from what you gave to when they are equal, i.e. 1BB/100BB/100BB, 3BB/99BB/99BB, 11BB/95BB/95BB, etc., just to lesser degrees as the stacks sizes become close?

It's also risk averse for pushing all-in if you have good reason to expect you will be called, especially by someone who covers you. And the shorter your own stack, plus the bigger your opponents' stacks, the more able they are to call you, thus reducing the number of hands you can play aggressively.

But ICM does not take any "good reason to expect you will be called" into account, so we can't use that to say ICM undervalues bigger stacks.

Yes it does. Playing an ICM strategy, a big stack can call you with a weaker hand than a short stack can. So you can expect to be called more often when you shove into a big stack than when you shove into a short stack if you know they are playing an ICM-optimal strategy. And this knowledge will change your own ICM-optimal strategy.

My point was that ICM does not take strategy into account; it merely maps chip EV to $ EV. So you're right, assuming everyone is playing an ICM strategy. But ICM does not assume this. It "doesn't care" what strategy anyone is using because ICM doesn't tell you what action to take. Rather, given the stack sizes and payout structure, it performs a mapping, enabling you to make a decision. If we assume that everyone is playing an ICM strategy, that's not going to affect our ICM calculations. Rather, it's going to change our chip EV calculations of jamming/calling. So in the end it will affect our decision-making, but it's irrelevant to whether ICM (i.e. the formula) is undervalues or overvalues bigger stacks.

By mapping chip EV to $EV, ICM is implicitly assuming that other players in the future play a strategy that maximizes their chip EV. It assumes that each of a player's chips contributes equally to the chance that a player will win the tournament.

I am suggesting that if we instead assumed that they were playing an ICM strategy going forward, we would map their chips into $EV differently, because ICM allows big stacks to leverage the chips they have more easily than small stacks. Increasing your stack would not increase your chance of winning the tournament in a linear fashion.

So yes, if we assumed everyone was playing an ICM strategy, that would change the calculations (but those calculations would no longer be calculations based on ICM).

Hmm, still not sure where you're coming from.

ICM explicitly assumes that a players chances of taking first is the ratio of his stack to the total number of chips in play. Thus, it implicitly assumes that your opponents' strategies do not affect your chances of taking first. You wrote, "ICM is implicitly assuming that other players in the future play a strategy that maximizes their chip EV." I'm not seeing that. I would consider ICM to still be valid even if we assumed that players won't play a strategy that maximizes their chip EV.

I am talking about the ICM's formula, but I feel like you might be talking about the ICM process, i.e. the process of calculating the chip EV of various actions, calculating the $ EV of the chip EV of those various actions, then comparing the $ EVs to decide what action to take.

The first sentence here is correct, but the others are not.
Yes, IC assumes your chance of taking first is proportional to your chip stack.

Now, we have to ask, under what combination of strategies of you and you opponents is it actually TRUE that your chance of taking first is proportional to your chip stack?

Clearly, you opponents could play a strategy where this is not true, and thus ICM would not be a valid assumption. For example, if all your opponents were playing the "fold every hand" strategy, you chance of winning would not be proportionate to your chip stack, it would be 100%.

Now, everyone's chance of winning WOULD be proportionate to their stack if they were all playing a strategy where they only cared about maximizing their chance of winning, ignoring other places. In this case where the tournament is winner-take-all, chip EV always equals tournament EV. So ICM is only exactly valid if we assume people are playing a chip EV=$EV strategy. If they deviate from this, their chance of winning the tournament is no longer proportionate to their stack.

If everyone is playing an ICM strategy, they are not solely trying to win, they are taking other places into account, and thus that strategy does not always lead them to winning a stack-proportiate amount of time. So if everyone is playing ICM, then ICM itself does not estimate an accurate chance of winning for each of the players.

How do you know this? What if having a bigger-than-average stack provides an advantage greater than what is accounted for by the difference in stack sizes, or vice versa? We don't know. No one is even close to knowing what such a strategy would look like. And that's precisely why ICM takes it as an assumption, an unproven assertion. Moreover, it doesn't make any statement about what situation it applies to (in other words, that's up to debate).

I'm fairly certain this HAS been proven, and is rather trivially easy to prove. It's the whole reason that ICM adopts this assumption.

As a shorthand:
- Playing a chip maximizing strategy is identical to playing a cash game strategy.
- In a heads-up pot in a cash game, if me and my bigger-stacked opponent are playing identical, chip-maximizing strategies, my chance losing $x in a hand will be equal to my chance of winning $x. And my chance busting in a particular hand will be equal to my chance of doubling up.
- Obviously, if two players in a heads-up tournament each have 50% of the chip and play identical strategies, they each have a 50% chance of winning.
- Now, assume one player has 75% of the chips, and the other player has 25%.
- If both players play cash game strategy, the short stack has an equal chance of doubling up or busting.
- This means that the short stack has a 50% of getting the tournament to a situation where the stacks are even, and a 50% chance of busting before that situation ever occurs.
- The 50% of the time the short stack doubles, he now has a 50% chance to win the tournament, so his total chance to win is 50%*50% = 25%, which is exactly proportional to his chip stack.
- You can run the same exercise with any set of stacks; if they all play cash game strategy, they will always have a chip-proportionate chance of winning.

Now, playing an ICM strategy in tournament paying multiple places is not a chip-maximizing strategy. The assumption that one will have an equal chance of winning $x as losing $x for all x up to one's stack size is no longer true, because chips will have diminishing marginal value, and I will not be willing to risk x to win x. So if you play an ICM strategy (or any of your opponents are playing an ICM strategy) your chance of winning is no longer proportionate to your chip stack.

The math works out when, as you did, the outcomes are limited to either a bustout or a doubleup. But what about outcomes such as when the 25% stack first gains an additional 5% and then doubles up, now having 60%. Notice that this situation was accounted for in your calculations (it is among the occurrences of doubling up), yet it's result is mistaken (it should be 60%, not 50%). So in reality, some (most) of the times among the 50% of times that the 25% stack "doubles up," he actually more than doubles up. This adds a complexity to the calculation not accounted for by the simplified analysis you gave.

Actually, i'm fairly certain this has only been proven for rather simple abstractions of poker, but not the full game.

In the full game we have a pretty good idea that this does not actually hold. For instance there is basically consensus that a 10bb shortstack on a full table of 100bb stacks does have an edge in the game and will be a favorite to double up.

Iirc the original Malmuth/Harville "ICM" formula was initially published for horse betting and got picked up for poker years later. It happens to work pretty well, but i doubt the initial assumptions had anything to do with poker in particular.

This is true...the 10bb stack isn't exactly an absolute favorite to double up, but they are a chip equity favorite. This is because they once they are all in, they can earn dead money from people who have been bet out of multiway protected pots. So they will frequently more than double-up when they win a heads-up hand. If every other player just check/called every street once anyone was all-in, they wouldn't have this advantage.

Some of the time, the 25% stack will win some chips and double up, like in your example, and his probability of winning from that point forward will be more than 50%. But some of the time, the 25% stack will lose some chips and double up, and his probability of winning from that point forward will be less than 50%. Playing a strategy to maximize chip equity, his probability of winning a little will be equal to his probability of losing a little, so this washes out in the calculations.

If he wins 5% and doubles, he has 60% of the chips. If he loses 5% and doubles, he has 40% of the chips. Obviously, his probability of winning with 60% is equal to his probability of losing with 40%. And since he is equally likely to win 5% as lose 5%, his average probability of winning after doubling is still 50%.

Now, as I discussed in the previous post, he is a bit more likely to more than double up in a multiway tournament, because a lot of pots will have other players' dead money in them, while he will never have dead money in a pot once he is all-in. This is indeed not accounted for by ICM. The amount by which this should affect the calculations depend on what strategy others are using (in particular, how loose and aggressive they are, and how often they bet into side pots). And it is one of many things ICM does not account for. It also doesn't account for other players playing an ICM strategy (or any strategy that deviates from chip EV).

Does != mean not equal to? I usually do that with /= but only cause in math class you draw an equals with slash going through it, and my use is the closest thing to what I do in math class.

Or you can play higher level kick ass games using http://www.irongeek.com/alt-numpad-a...and-chart.html

and express it as Alt+216 (using num key pad) --> ╪

or even better while still evading LATEX

http://math.typeit.org/

that gives you this; ≠


Or if totally bored and in a hurry try =/= or <>.


hehehe!

!= is used in C-like programming languages as "not equal", /= has a different meaning there.

"x/=y" is equivalent to "x=x/y", ie: assign to x x's old value divided by y

NickMPK, I am now your faithful convert. Thanks for the patient explanations

You wrote earlier,

Any ideas for how to implement this formulaically?

It's always referred to as the "Harville" formula in relation to horse racing literature (even very recently published papers and books call it this) and it appears to have first been presented in this 1973 paper: Assigning Probabilities to the Outcomes of Multi-Entry Competitions in which Harville gives no references apart from "Fabricand, Burton P., Horse Sense, New York: David McKay Company, Inc., 1965."

But: by pure chance last year I was looking for stuff on ranking/rating systems and I noticed Plackett presented the very same formula 5 years earlier in this 1968 paper: Random Permutations (see formula #57 on pp533):



Juk

There was a brief discussion if the probability of winning a winner takes all tourney would be equal to the percentage of chips they have, if everyone wasplaying a perfect cash game strategy. The answer is clearly no as long as there are more than 2 players involved:
Let's say player 1 has 15bb, player 2 has 10bb and player 3 has 5bb. The small stack has clearly the edge as he can play like everyone would have 5bb. The other player have to worry to loose 10bb.
e.g. bigstack is in the button and wants to push. Since he can loose 10bb to the medium stack he can push less hands profitable than the small stack could in the same scenario.
When ICM comes into play things change as loosing the 5bb for the small guy means more than it means to the big stack, therefore he has to tighten up as well. I will go more into detail about that in the next post.

Since we said everybody should play the same strategy, why not assume everybody is playing the nash ranges provided by holdemresources.net?
I don't think we should give our players more than 10bb effective in the simulations as this would include a mixed strategy with minraises which is really, really hard to solve.
I simulated a bubble round with a bigstack (BS) that has 4500, a mediumstack (MS) 3000and a smallstack (SS) 1500. To account for position I ran 2*3 simulations in which every player is in every position exactly 1 time.
In simulation1 the BS is on the BUT, the MS is in the SB and the SS in the BB. Than the blinds go around like in a real game. Stack sizes are the same for all simulations.
In simulation 2, I switched the position of the SS and the MS.
The results of simulation 1 are listened here:
http://www.holdemresources.net/hr/sn...2=3000&s3=1500
http://www.holdemresources.net/hr/sn...2=1500&s3=4500
http://www.holdemresources.net/hr/sn...2=4500&s3=3000
The results for simulation 2 are here:
http://www.holdemresources.net/hr/sn...2=1500&s3=3000
http://www.holdemresources.net/hr/sn...2=3000&s3=4500
http://www.holdemresources.net/hr/sn...2=4500&s3=1500

In the following I list how much EQ each player gains depending on position. In the middle column I have the results my downloadable tool provided me for simulation 1. As you see in the end the EQ the bigstack gains in this round equals the EQ the other player loose in the online tool. The downloadable version is a little bit off, so I don't know why I payed 99$ when Holdemresources offers a better service for free.

Simulation 1 Simulation 2
EQ BS: BU: +1,257 1,216 +1,121
SB: +,073 +0,128 +0,020
BB -0,429 -0,472 -0.485%
Total: 0,901 0,917 Total 0,656

EQ SS BU: +1,815 +1,741 +1,207
SB: -0,537 -0,491 -0,291
BB -1,345 -1,348 -1,520
Total: -0,067 -0,098 Total: -0,604

EQ MS BU: +0,966 +0,962 +1,501
SB: +0,088 +0,132 -0.721%
BB: -1,888 -1,875 -0,83
Total: -0,834 -0,781 Total: -0,05

Total Simulation 1 downloadable: +0,038
Total Simulation 1 Online: 0,000
Total Simulation 2 Online: 0,002

We clearly see that the BS gains EQ in the bubble, which we would expect, therefore yes ICM seems to underetimate big stacks.
The middle stack in this example looses more EQ than the bigstack. However the positions on the table seem to be more important than having a big or medium stack. It's hard to win if the BS can profitable push ATC against the middle stack and 87,3% against the short stack when they are blind on blind.

Unfortunately these results are valid only for this exact situation. For different stack sizes outcomes will be different.
Also we don't know how long the bubble will last on average and how stack sizes will evolve. To do this we would have to make 6 different game trees. Finding a formula for that shouldn't be too hard but the machine would run like forever.

Interpretation or What does that mean to us in the game?
If we have the chance to go into the bubble as a bigstack we should take it even if we have a slightly -EV call. We will have a chance to make up for it in future rounds.
If we are in the same situation and would be the middle stack after doubling up we shouldn't call since we will be the bubble bitch.
In the bubble, the player who looses the most EQ should look to make a move even it is -EV to prevent himself from bleeding even more EQ.
Conversely the big stack should be more risk averse than ICM suggests because he will not longer be able to abuse the bubble after he calls somebody or gets called. When he looses he is not chip leader anymore and if he wins they will be HU with no edge for the bigstack.

If you guys are planning to write a program that makes these simulations please let me know, I would be very interested to participate in such a project and/or the results.

Great point, and awesome analysis in the followup post. Thanks, Sinothoras1