Exploring how marginal chip value changes with stack size
 

I posted this message, in somewhat different form, as part of a thread in the Books and Publications forum. It didn't get much of a response. I'm hoping that more people will see it here, and will be interested in commenting. It's relevant to certain ongoing controversies.


General Form--Tournament Equity as a Function of Stack Size

In the earlier thread, CityFan proposed the following function, which appears to me a valid formulation, to address the question of the marginal value of gaining or losing a chip:

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In any tournament scenario S (including size of the blinds, position of players at the tables, time until next level[, the strategies, skills, and idiosyncracies of all players in the tournament,] etc.), there is a function F_r which gives player r's equity in the tournament as a function of every player's chip stack.


F_r = F_r(x_1,...,x_r,...,x_n,S)


Move the chips around, leaving all other conditions S unchanged, and F_r will change for each player.


Such a function exists whatever assumptions you make about how the players play. You don't have to assume that they play optimally, merely that each will play according to SOME strategy.


Now suppose you artificially increase player r's chip stack by an amount h, drawing the chips equally from each of the other stacks.


F_r[new] = F_r(x_1*(1-h/T),...,x_r + h,...,x_n*(1-h/T),S)


Where T = total chips - x_r


Note that the total number of chips in the tourney has not changed.


Player r's increase in equity is given by


F_r[new] - F_r
= F_r(x_1*(1-h/T),...,x_r + h,...,x_n*(1-h/T),S) - F_r(x_1,...,x_r,...,x_n,S)


Now, usually we would study the derivative of player r's equity w.r.t [with respect to] the number of chips he receives


lim(h->0) (F_r[new]-F_r)/h


I think this is valid, but it could be argued that this will often be zero, since the addition of one chip may not change his equity at all, because all bets are in multiples of the small blind.


Even then, we can look at (F_r[new]-F_r)/h for the smallest SIGNIFICANT changes in r's chip stack (or some other way of studying the "gradient" of a step function) and we then have a workable definition for the incremental value of a chip to player r.

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Note that this is a completely general formulation, which does not assume equal skill levels, as does, say, the independent chip model (ICM).

The derivative of F_r with respect to h shows the incremental value (meaning, impact on $EV) of player r gaining or losing a chip. Studying how the value of this derivative changes as x_r increases or decreases shows how "relative chip value" or "marginal chip value" changes over the range of possible stack sizes.

To give an example, it would be instructive, in a given tournament situation, to be able to determine the risk to your tournament $EV of risking 100 chips in an effort to win 100 chips, and to compare that to the potential reward to your tournament $EV of winning those 100 chips. Note that this does not entail actually calculating the derivative of F_r along every point in the range from x_r-100 through x_r+100. All that is needed is a comparison of F_r for three values of x_r: the current x_r, x_r-100, and x_r+100. Unfortunately, just stating the form of the function F_r does not enable us to solve it, to attach any actual value to F_r for any value of x_r.

CityFan defined h as the amount by which x_r grows, and specified that h is removed in equal amounts from the other stacks x. Of course removing h from the other stacks x is not the only way that h could be added to stack x_r in an actual poker tournament. In practice, chips will usually be removed from just one or a few of the other stacks. One could devise functions to model these other scenarios. However, in its broad outlines, I think the analysis of these other scenarios would parallel the analysis of CityFan's more specialized scenario.


Implications as to Range Across Which Marginal Chip Value Can Be Increasing

Even though it is not practical to calculate exact values of F_r for a full scenario S, reflection on the general F_r function quickly leads to important implications.

There has, of course, been a debate raging on the question of under what circumstances the marginal value of acquiring chips is increasing as one's stack grows, and under what circumstances it is decreasing. This question is equivalent to the question of how the value of F_r changes as the value of h changes. A slight extension of this principle is to posit that for negative values of h, the loss of chips from x_r is added to all other stacks x in equal amounts.

The rules of a tournament impose constraints on the value that F_r can take:

(1) Define Z as the total prize pool, and z_m as the prize for finishing in position m. Then Z = z_1+z_2+...+z_m. F_r can never be greater than z_1.

(2) Define X = x_1+x_2+...+x_n (the total of all chips in play). Then F_r = z_1 when x_r = X. In all other cases, F_r < z_1, necessarily, because there will always be at least a slight chance of another surviving player winning first place.

(3) F_r can never be less than zero.

(4) When x_r = 0, F_r will always be zero, unless the number of surviving players has previously been reduced to the number of payout positions, or less. (This factor would be included in the scenario S.)

(5) When x_r > 0, F_r will always be positive, because player r will retain at least a slight chance of finishing in the money.

Assuming that no rebuys or add-ons are allowed, and assuming that player r's skills are equal to the skills of the field, or are superior, we would ordinarily expect the following to be true at the start of the tournament: F_r >= Z/n (that is to say, the amount that player r paid for his stack), because player r's prospects should be at least as good as those of the average player. Under any plausible assumptions as to the characteristics of the players and the entire scenario S (and unless the tournament field is very small in comparison to the number of paying finish positions), this leads to the following conclusion: in the neighborhood of the starting value of r_1, the slope of the curve plotting F_r against x_r is close to Z/X or is greater. In other words, the value of the first derivative, F'_r, is close to Z/X or is greater.

The reason for this is that we know that the starting point of the curve is the point 0,0, and we do not believe the rate of change in F_r/x_r to be huge in the range between zero and the initial value of x_r = Z/n. This suggests that the curve in that range plotting F_r against r_1 is close to a straight line segment. Since we are discussing an average or superior player, whose initial F_r >= Z/n, we know that the average initial slope of the curve over the range of values of r_1 from 0 to X/n is greater than or equal to (Z/n - 0)/(X/n - 0) = Z/X.

If the curve were to continue in a straight line with the same slope, when the value of x_r reached X, the value of F_r would be equal to or greater than Z. However, this is impossible, because F_r cannot exceed z_1, which is less than Z. Therefore, the slope must decline at some point, and must decline very substantially.

This is a formal proof of a point that a lot of people consider intuitively apparent, and that could be adequately explained in many fewer words. But it lays the groundwork for what follows.


Further Conclusions as to Superior and Inferior Players

If on first entering the tournament, player r has an overall positive tournament EV of 200% of his entry fee, this translates to an initial F_r = 3*Z/n. This implies an initial slope of 3*Z/X, which extrapolates to F_r = 3*Z when x_r reaches X. Again this is impossible, by an even greater margin than would be the case for the player with no initial positive tournament EV. So the ultimate decline in the slope of the curve will be even sharper. What goes up, must come down, and the higher you fly....

In terms David Sklansky has previously used, for an average or superior player, the curve must at some point become convex upward. For a greatly inferior player, this is not necessarily true. It is conceivable that the curve could start at such a low angle that it would be concave through its full path. "Convex" translates to declining marginal chip value, while "concave" translates to increasing marginal chip value. Whether marginal chip value is increasing or decreasing answers the question of the utility of a "coin flip."

This is by no means a full solution to marginal chip value problems, because the curve could conceivably follow many paths to its ultimate destination.


Approximations Using Random Decision Models

In practice, in order to do marginal chip value analysis, one has to use greatly simplified models like the ICM. These can be informative, but they have definite limitations. First, these models usually have to assume that all players have equal skill levels. They are random decision models, in the sense that they assume that the tournament will be decided by random events that are not biased in favor of one player or another. Second, the simple random decision models do not correspond in any exact way to how a poker tournament is actually decided. An important example is that they generally fail to take account of the effect of increasing blinds. The ICM is one such specialized model for solving the function F_r. This offers a certain amount of insight into the impact of the payout structure on the utility of various stack sizes.

Others have created calculators for applying the ICM to final-table problems with only three prizes. It appears that solving the ICM in its general form becomes intractable with increasing numbers of prizes and increasing numbers of players remaining active. However, I have created a spreadsheet for calculating one flavor of the ICM in situations involving large numbers of active players. This specialized form of the ICM assumes that the stacks of all remaining players, except the player being analyzed (player "r"), are equal to one another. Note that this is consistent with CityFan's version of F_r, which assumes that chips acquired by player r are taken from the stacks of all other player in equal shares.

Using this model, I ran a calculation of F_r given the situation at the start of a tournament with the following payout structure:

Finish rank: % of pool awarded:
1 29.00%
2 18.50%
3 12.00%
4 10.00%
5 8.00%
6 6.50%
7 5.50%
8 4.50%
9 3.50%
10 2.50%

100.00%

The tournament has 100 players, the buy-in is $1,000, and each player receives 1,000 tournament chips. This results in a $100,000 prize pool, and a $29,000 first prize.

I then took the results, and plotted them as a graph. The results look like this:

http://home.att.net/~wrx/icm01.jpg

Again, this is a rather specialized model, one of many in the universe of models that could be true for a particular tournament structure, and just one of many that assume no skill advantage for any one player. Nevertheless, it may be observed:

(1) For a player with a skill advantage over the field, we would expect the value of F_r to be above the ICM curve, for any value of x_r. This is just another way of saying that, for any stack size, skilled player r's tournament expected value should be greater than that of an average player holding the same stack. From this, one can theorize as to where the curve begins to be convex for a skilled player. (The observation that the skilled player's curve is above the average player's curve at all points might not be true if the generally-skilled player were extremely deficient in the play of stacks of a certain size--very small, very large, or even medium.)

(2) For a player with skills below those of the field, we would expect the value of F_r to be below the ICM curve at all points. Note that this could result in a curve that was still convex at all points, was flat, was concave at all points, or had a wavy form, partially concave and partially convex.

Quite independent of the results of the ICM, I earlier stated the conclusion that for a superior player, across the initial range of x_r, between zero and the buy-in amount, here 1,000 chips, the slope of the curve plotting F_r against x_r should have a slope of $1/1 chip or greater. A curve, continuing from this initial segment, either concave or linearly, reaches the maximum value of F_r, $29,000, at the point at which x_r = 29,000, or earlier. This apparently places an outside limit on the range of values of x_r for which the curve can be concave--in other words, the range for which marginal chip value can be increasing. More generally, if z_1 is first prize, marginal chip value cannot increase past the point at which x_r = (z_1/Z)*X. In practice, there would not be a sharp break from positive to zero marginal chip value at such a point, so marginal chip value must begin to decline at some point before the stack reaches (z_1/Z)*X, or 29,000 chips in the example given.

(This will hold true at all points of the tournament until some of the prize money has been awarded. Thereafter, marginal chip value could be constant or increasing until some larger stack size was reached.)


Questions Remaining Unanswered

This still doesn't prove or disprove the principle that has been stated as, "Chips gained will usually increase your equity less than chips lost will decrease it." That would depend a great deal not only on the shape of the curve, but also on where on the curve the player usually spends the most time.

The modest conclusions just stated still leave a considerable range of stack sizes over which marginal chip value may be increasing for a superior player, due to the chip utility effect. In the example given, it would be a major achievement to increase one's stack from 1,000 to, say, 6,000--and it's entirely plausible that marginal chip value could be increasing up to that point, or much higher.

F_r is a function of multiple variables. Note that the curve plotting F_r against x_r, and hence player r's location on that curve, can and routinely does change substantially as a result of changes to the other stacks x, or the scenario S. Events leading to this include, as examples only, other players winning chips from each other, players busting out, blinds increasing, or other players wising up and starting to play better poker. All of this has the potential for placing player r in a position in which marginal chip value is now increasing, although it was previously decreasing for him. However, the range across which marginal chip value can potentially be increasing remains subject to constraints as discussed above.

As CityFan has noted, "In tournaments we all find ourselves rooting for one or other player in an all-in situation that doesn't involve us, precisely because the result will have an effect on our tournament equity."

I welcome comments on these observations. I think that it may be possible to extend these ideas, to further quantify the situations in which increasing chip utility has a potential for putting a player in a situation of overall increasing marginal chip value. Any thoughts along those lines would be helpful.

Re: Exploring how marginal chip value changes with stack size
 

this looks cool, wish I could understand it.

Re: Exploring how marginal chip value changes with stack size
 

Oh yeah, um, I skimmed it and decided to come back to it later.

I don't think my function F has any mystical significance, except that I wrote it in response to the challenge that "chip value is not a meaningful concept".

Arguing about the properties of F is precisely what Sklansky, Snyder and people on here have been doing to death for the past x weeks.

It's interesting, but well known (whisper it around here...), that F is generally convex - that is, incremental chip value decreases with the size of your stack. I like the observation that for an expert player F should always be ABOVE the curve you've plotted which assumes equal skill - though that might not be true if that player simply couldn't be bothered grinding with a short stack and therefore played poorly.

That's a situation in which F for a skilled player might not be convex.

I think though, unless we're going to do some serious maths, bringing all kinds of multivariate functions into play is probably not going to advance any debate. We're merely obscuring the wood by identifying a few trees.

So, sorry, I'm not going to be drawn into any great discussion about this F, unless there are questions that really need the maths*.

*which there might be

Re: Exploring how marginal chip value changes with stack size
 

Except to say, if F is convex everywhere, then chips gained will ALWAYS be worth less than chips lost.

Re: Exploring how marginal chip value changes with stack size
 

Here is a quote from Mason on chip stacks:

"I really don't think that anyone here believes that having a small stack is superior than having a big stack. If that was the case, people would be trying to get broke, not accumulate chips."

Re: Exploring how marginal chip value changes with stack size
 

I skimmed and did not carefully read, but let me make a couple of points.

(1) The derivative you mention is a particlar directional derivative, but you could consider any direction (or consider all partial derivatives).

(2) But you cannot generally use derivatives, since F may have discontuities.

Just consider the case where 3 (or more) players (on one table) each have the strategy of going all in on every hand. (Or suppose the blinds/antes had doubled every nanosecond so that now everyone is all in on every hand.) Then clearly, treating x_1,...,x_r,...,x_n (n at least 3) as real variables, each F_r will have a discontinuity wherever two variables are equal, chopping the n-simplex into n! pieces.

ICM problems
 

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It appears that solving the ICM in its general form becomes intractable with increasing numbers of prizes and increasing numbers of players remaining active.

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This is an interesting point, and I'd like to know how much this has been discussed before.

In the course of other consultation, I recommended the use of the ICM as a settlement for stopped tournaments to a major poker server. They seemed interested until I mentioned that the exact calculation is complicated, and they may need to use an approximation if there are hundreds of players. Does anyone have a good approximation? I found one which is computable, but which does not preserve the property that the matrix of finishing probabilities has the property that each row and column sum to 1.

Another tractable problem related to the ICM is showing that you always lose E$ when you take an EChip-neutral gamble. As I recall, no one posted a proof on a past thread where this was conjectured. Maybe it would be good to gamble to knock out the short stack on the bubble when you have the second or third stack, but the ICM doesn't seem to say this.

Re: ICM problems
 

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[ QUOTE ]
It appears that solving the ICM in its general form becomes intractable with increasing numbers of prizes and increasing numbers of players remaining active.

[/ QUOTE ]
This is an interesting point, and I'd like to know how much this has been discussed before.

In the course of other consultation, I recommended the use of the ICM as a settlement for stopped tournaments to a major poker server. They seemed interested until I mentioned that the exact calculation is complicated, and they may need to use an approximation if there are hundreds of players. Does anyone have a good approximation? I found one which is computable, but which does not preserve the property that the matrix of finishing probabilities has the property that each row and column sum to 1.

Another tractable problem related to the ICM is showing that you always lose E$ when you take an EChip-neutral gamble. As I recall, no one posted a proof on a past thread where this was conjectured. Maybe it would be good to gamble to knock out the short stack on the bubble when you have the second or third stack, but the ICM doesn't seem to say this.

[/ QUOTE ]

What is ICM? Is it random walk (brownian motion) on a simplex with absorbing boundaries? I saw Tom Ferguson wrote an exact solution for n=3 players. What else is known?

Re: Exploring how marginal chip value changes with stack size
 

Thanks for the remarks, CityFan and everyone else.

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I don't think my function F has any mystical significance, except that I wrote it in response to the challenge that "chip value is not a meaningful concept".

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Nothing mystical about it, but useful. The function assumes one algorithm for adding chips to x_r, which is removal in equal parts from all other stacks. It strikes me that certain results could probably be extended to an even more general function based on an arbitrary algorithm allowing for removal from other stacks in any other manner. However, I doubt it's worth a big effort to try to do so at this juncture.

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Arguing about the properties of F is precisely what Sklansky, Snyder and people on here have been doing to death for the past x weeks.

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Which, admittedly, is what got me started. It's an effort to settle or at least narrow some of these debates, and if it luckily produces other results, too, great.

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I think though, unless we're going to do some serious maths, bringing all kinds of multivariate functions into play is probably not going to advance any debate. We're merely obscuring the wood by identifying a few trees.

So, sorry, I'm not going to be drawn into any great discussion about this F, unless there are questions that really need the maths.

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Well, analyzing a bunch of multivariate functions would be way, way beyond my capabilities. I wouldn't ask you get into that, and I'm dubious that it would be practical. Possibly one could draw out some general results as to the effect of changes in opponents' stack sizes while your own held steady, but I don't know.

My goals in looking at this problem have been quite modest, but I consider the following results significant, if they stand up to scrutiny:

o First, I wanted a rigorous proof that for an average or superior player, marginal chip value must begin to decline at some point, as the player's stack increases and other tournament conditions hold steady. A lot of people consider this intuitively obvious. However, some have disagreed. (And what is intuitively obvious is sometimes wrong!) For that reason, I think that a proof is worthwhile. Anyway, it's always best to have one's theories built on a rock-solid foundation. The demonstration of this point probably could have been put into fewer words, but stating the problem as one of solving your function lays the groundwork for other observations. If anyone here can do a better job of a formal proof, that would be great.

o For an average or superior player, marginal chip value must begin to decline before the point at which the player has built a stack that bears the same ratio to all chips in play as first prize bears to the total prize pool.

o The greater the player's initial $EV, which is to say the greater his initial advantage over the field, the sooner his marginal chip value must begin to decline.

o While these considerations apparently limit the range across which a "chip utility" effect could result in increasing net marginal chip value, they leave open the possibility that this could hold true for a substantial increase above the initial stack in a tournament with a sizeable field.

o The spreadsheet solving the specialized, flat-opposing-stack case of the ICM may be useful for studying other problems. With the program I'm using, it's capable of modeling tournaments with up to 13 payout places, and an essentially unlimited number of active players. It could probably be rewritten to handle more prizes, but it would be a bear. I haven't previously seen ICM calculators that will handle anything but final table problems. It's hard to come up with anything truly original--probably someone, somewhere has done this before--but I can't find it.

Some of these observations depend on assuming that once the curve goes convex, it stays convex. Of course it's conceivable that the curve could follow a wavy form, going convex, then concave, then convex again, etc. However, it is very hard to imagine that such an odd pattern could result from a chip utility effect. That is, if having an increasingly large stack gave one more profitable opportunities to use one's chips, across some range of stack sizes, resulting in increasing marginal chip value, one would expect that effect to continue and maybe peter out at some stack size, but not to stop, and then restart later with increased vigor.

[ QUOTE ]
I like the observation that for an expert player F should always be ABOVE the curve you've plotted which assumes equal skill - though that might not be true if that player simply couldn't be bothered grinding with a short stack and therefore played poorly.

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In that particular game, the so-called expert wouldn't be a superior player at all. Maybe Chip Reese, Doyle Brunson, and Daniel Negreanu have days they can't be bothered to play good poker--if so, bring'em on!

Re: ICM problems
 

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It appears that solving the ICM in its general form becomes intractable with increasing numbers of prizes and increasing numbers of players remaining active.

[/ QUOTE ]
This is an interesting point, and I'd like to know how much this has been discussed before.

[/ QUOTE ]

I say this based only on having looked at code solving three-prize final table problems by direct methods, that looked like it would baloon exponentially with more prizes being added. This leaves open the interesting possibility that the general statement of the ICM might reduce to a relatively simple formula, through the application of higher mathematics. Does anyone know if this has been attempted? Or if it's been shown to be impossible?

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Does anyone have a good (ICM) approximation? I found one which is computable, but which does not preserve the property that the matrix of finishing probabilities has the property that each row and column sum to 1.

[/ QUOTE ]

Is this approximation available anywhere?

[ QUOTE ]
Another tractable problem related to the ICM is showing that you always lose E$ when you take an EChip-neutral gamble. As I recall, no one posted a proof on a past thread where this was conjectured. Maybe it would be good to gamble to knock out the short stack on the bubble when you have the second or third stack, but the ICM doesn't seem to say this.

[/ QUOTE ]

Isn't this another way of saying that for any case of the ICM, the curve is always convex end-to-end? It seems obvious, but proving it could be something else again!

Re: ICM problems
 

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What is ICM?

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"Independent Chip Model." It's commonly used, and you'll find lots of discussion about it around here. You could call it a lottery ticket model. Give each player a number of tickets equal to the number of chips he has. Draw one ticket at random to determine first place. Then throw out all the tickets of the first-place winner, and draw one from the remaining tickets to determine second place. Etc. You can see that with a lot of places to be decided, and varying stack sizes, this gets complicated fast.

Re: ICM problems
 

[ QUOTE ]

What is ICM?


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Shuffle the players' chips. Rank the players by their highest chips. Equivalently, remove the chips from the table one at a time, eliminating a player when his last chip is removed.

Independent Chip Model calculators:
http://www.chillin411.com/icmcalc.php
http://sharnett.bol.ucla.edu/ICM/ICM.html (not as convenient, but it has some explanations)

You can write the formulas for n players by summing over the possible first place finishers, removing their chips, and applying the ICM to the reduced tournament on n-1 players. This doesn't seem to simplify. Does anyone know what the probabilities are of finishing last in a tournament where the stack sizes are 1, 2, ..., 100?

[ QUOTE ]

Is it random walk (brownian motion) on a simplex with absorbing boundaries? I saw Tom Ferguson wrote an exact solution for n=3 players. What else is known?

[/ QUOTE ]
The Brownian motion model is different. It's simple to analyze with 3 players because conformal transformations in two dimensions are understood and preserve Brownian paths, and the Riemann maps from the triangle to the disc are even known explicitly. Conformal transformations in higher dimensions are much more rigid, and there isn't much hope to extend the solution to more players without adding ideas.

Re: Exploring how marginal chip value changes with stack size
 

[ QUOTE ]
I skimmed and did not carefully read, but let me make a couple of points.

(1) The derivative you mention is a particlar directional derivative, but you could consider any direction (or consider all partial derivatives).

(2) But you cannot generally use derivatives, since F may have discontuities.

Just consider the case where 3 (or more) players (on one table) each have the strategy of going all in on every hand. (Or suppose the blinds/antes had doubled every nanosecond so that now everyone is all in on every hand.) Then clearly, treating x_1,...,x_r,...,x_n (n at least 3) as real variables, each F_r will have a discontinuity wherever two variables are equal, chopping the n-simplex into n! pieces.

[/ QUOTE ]

Since one can't win or lose fractions of a chip, doesn't every situation present discontinuities?

It strikes me that since one can't come up with a mathematical formula for any real-world F_r (as opposed to a radical simplification like the ICM), one isn't going to be able to obtain a derivative anyway. And that we're not so much interested in the derivative at any one point as we are in knowing delta F_r/delta x_r between two given values of x_r.

But I'm stumbling in the dark.

Re: ICM problems
 

[ QUOTE ]
This leaves open the interesting possibility that the general statement of the ICM might reduce to a relatively simple formula, through the application of higher mathematics. Does anyone know if this has been attempted? Or if it's been shown to be impossible?

[/ QUOTE ]
I've tried and failed so far, but some simplifications could be possible.

One way of ruling out some levels of complexity is to work out examples exactly. Some of the denominators are huge, so any simplification would still have to be able to produce huge denominators, too.

[ QUOTE ]
[ QUOTE ]
Does anyone have a good (ICM) approximation? I found one which is computable, but which does not preserve the property that the matrix of finishing probabilities has the property that each row and column sum to 1.

[/ QUOTE ]

Is this approximation available anywhere?

[/ QUOTE ]
No, at least not from me. Others could have developed it independently, as I just made some crude approximations and renormalized some probabilities to 1.

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[ QUOTE ]
Another tractable problem related to the ICM is showing that you always lose E$ when you take an EChip-neutral gamble. As I recall, no one posted a proof on a past thread where this was conjectured. Maybe it would be good to gamble to knock out the short stack on the bubble when you have the second or third stack, but the ICM doesn't seem to say this.

[/ QUOTE ]

Isn't this another way of saying that for any case of the ICM, the curve is always convex end-to-end?

[/ QUOTE ]
I think the curve you mean is the result of taking chips from all opponents equally or in proportion, but I refer to gambling against each opponent individually. I expect that a counterexample could be constructed easily if you allow sidepots, but perhaps multiway gambles are also ok if there are no sidepots.

I didn't specify the payout structure, and I think it should hold for all decreasing structures. It suffices to prove it for satellites, i.e., that the probability of reaching the nth place or higher always decreases when you take an EChip-neutral gamble.

Re: ICM problems
 

[ QUOTE ]
[ QUOTE ]
This leaves open the interesting possibility that the general statement of the ICM might reduce to a relatively simple formula, through the application of higher mathematics. Does anyone know if this has been attempted? Or if it's been shown to be impossible?

[/ QUOTE ]
I've tried and failed so far, but some simplifications could be possible.

One way of ruling out some levels of complexity is to work out examples exactly. Some of the denominators are huge, so any simplification would still have to be able to produce huge denominators, too.


[/ QUOTE ]

Would I be right to say that if you want to input integer stack sizes, and output exact rational numbers for the probability of player i getting position j in this ICM model, then in terms of encoding input and output as binary strings, the output length is exponential in the input length?

Re: ICM problems
 

[ QUOTE ]
[ QUOTE ]

One way of ruling out some levels of complexity is to work out examples exactly. Some of the denominators are huge, so any simplification would still have to be able to produce huge denominators, too.


[/ QUOTE ]

Would I be right to say that if you want to input integer stack sizes, and output exact rational numbers for the probability of player i getting position j in this ICM model, then in terms of encoding input and output as binary strings, the output length is exponential in the input length?

[/ QUOTE ]
It could be worse than that, as a factorial is involved, but I think that is a poor measure of complexity. n! and (2n choose n) would also seem very complicated that way.

Re: ICM problems
 

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Does anyone know what the probabilities are of finishing last in a tournament where the stack sizes are 1, 2, ..., 100?

[/ QUOTE ]
I still don't, but here are some calculations of the ICM-predicted probability of finishing last in tournaments with stacks of size 1, 2, 3, ..., n.

n=2
1 2/3
2 1/3

n=3
1: 0.5833333333333334
2: 0.2666666666666666
3: 0.15

n=4
1 0.5511904761904762
2 0.24126984126984127
3 0.12976190476190477
4 0.07777777777777778

n=5
1 0.5361360861360861
2 0.22936507936507936
3 0.12031302031302031
4 0.07022977022977023
5 0.04395604395604396

n=10
1: 0.5184164152973668
2: 0.2151557713105984
3: 0.10893008631761322
4: 0.06109750853503585
5: 0.036606485042195436
6: 0.022998223608975937
7: 0.014984758061417186
8: 0.010053105474764727
9: 0.006909801915833682
10: 0.0048478444361987425

n=15
1: 0.5165407213004611
2: 0.21359514003346372
3: 0.10763726163687229
4: 0.06002971007838609
5: 0.03572609821501513
6: 0.02227291325368989
7: 0.014387192991584257
8: 0.009560467771416203
9: 0.006503209213545211
10: 0.0045117668120416956
11: 0.003183781349335634
12: 0.0022802773367725966
13: 0.001654769973221269
14: 0.0012150347599461867
15: 0.0009016552742487154

A brute-force calculation for n=15 (of all probabilities in all places) took only 2 minutes in Mathematica, but that method would run out of space around n=20 on my computer.

<ul type="square">
Mathematica code:

Clear[places]
places[n_, stacks_] := places[n, stacks] = If[Length[stacks] == 1, {1},
(1/Sum[stacks[[j]], {j, Length[stacks]}])
(Table[If[i == 1, stacks[[n]], 0], {i, Length[stacks]}] +
Sum[If[k &lt; n,
stacks[[k]]Prepend[places[n - 1, Delete[stacks, k]], 0],
stacks[[k + 1]]Prepend[places[n, Delete[stacks, k + 1]], 0]],
{k, Length[stacks] - 1}]
)
]
fullplaces[stacks_] := Table[places[i, stacks], {i, Length[stacks]}]

Input: places[1,{2,3,7}]
Output: {1/6, 13/45, 49/90}

Input: fullplaces[{2,3,7}]
Output: {{1/6,13/45,49/90},[1/4,2/5,7/20},{7/12,14/45,19/180}}
[/list]The total number of intermediate distributions calculated is the number of sub-multisets of the opponents' chip counts (when equal stack sizes are entered next to each other), at most 2^n when there are n other players. However, the coefficients are quite complicated.

From the data, I conjecture that as n increases, the probability that the player with the smallest stack finishes last converges to something over 1/2. It's easy to prove some upper and lower bounds, e.g, that the value is greater than the product of (1 - 1/(k choose 2)) from k=3 to infinity, 1/3. The general behavior for the larger stacks may be of more interest.

Re: ICM problems
 

[ QUOTE ]
I recommended the use of the ICM as a settlement for stopped tournaments to a major poker server. They seemed interested until I mentioned that the exact calculation is complicated, and they may need to use an approximation if there are hundreds of players. Does anyone have a good approximation?

[/ QUOTE ]
I think I have found a fast way to calculate the probabilities of placing in each position when the stacks are small multiples of a common chip size. This may be good enough in practice. While the results are complicated if computed exactly, this method should be numerically stable. Errors in approximations of intermediate steps should not blow up.

Start with just the target player's stack. Add the other players' stacks one by one, maintaining a joint distribution of the location of the player's highest chip and the player's rank. At the end, ignore the extra information about the location of the player's highest chip, i.e., sum over the possible locations.

I think this basic idea can be used to prove the convexity conjecture, too, by adding the player's and opponent's stacks last.

Re: ICM problems
 

ICM-predicted probabilities of finishing last with chip stacks 1, 2, ... 100:

1 0.51609432333124221785
2 0.21321188990848979709
3 0.10730956751035384068
4 0.059750455028267008731
5 0.035488752610857001093
6 0.022071595113771953758
7 0.014216683846631049432
8 0.0094161950449655082157
9 0.0063812062388728451803
10 0.0044086199339784030072
20 0.00021111503066950183804
30 0.000019498738229098345387
40 2.554789938882243882844891 x 10^-6
50 4.202955272341909711638795 x 10^-7
60 8.145255901318414223432787 x 10^-8
70 1.789417913036532507262767 x 10^-8
80 4.344852176919571241218451 x 10^-9
90 1.145466693678780882931910 x 10^-9
100 3.23650498103496640628121 x 10^-10

This took 12 hours with the faster method I described. It's still not quite fast enough to use on large tournaments as implemented in Mathematica, but it might be fast enough when programmed directly.

<ul type="square">
newarray[probarray_, addstack_] := {chiptotal = Length[probarray[[1]]];
Table[(1/Binomial[chiptotal + addstack, addstack])
(Binomial[chiptotal - j + addstack, addstack]
If[j &lt;= chiptotal &amp;&amp; ii &lt;= Length[probarray],probarray[[ii]][[j]], 0] +
Sum[If[k &gt;= j || ii == 1, 0,
Binomial[j - 1, k] Binomial[chiptotal - j + addstack, addstack - k]
If[j - k &lt;= chiptotal, probarray[[ii - 1]][[j - k]]],0], {k, addstack}]),
{ii, Length[probarray] + 1}, {j, chiptotal + addstack}]}[[1]]

fullarray[initstack_, oppstacks_] := fullarray[initstack, oppstacks] =
{temparray = {Table[If[ii == 1, 1, 0], {ii, initstack}]};
Do[temparray = newarray[temparray, oppstacks[[ii]]], {ii, Length[oppstacks]}];
temparray}[[1]]
[/list]

Re: ICM problems
 

[ QUOTE ]

I think this basic idea can be used to prove the convexity conjecture, too, by adding the player's and opponent's stacks last.

[/ QUOTE ]
That method didn't seem to work, but I proved the conjecture in a more direct fashion.

Let f(x) be the ICM-predicted probability that player A finishes in the top n places after taking x chips from player B.

Claim: f"(x) &lt; 0.
Proof: f(x) is the complement of the probability that player A is not among the top n players, which can be written as a constant (wrt x) plus a sum of terms with positive coefficients in proportion to

b-x
(b-x)/(c_1+x)
(b-x)/((c_1+x)(c_2+x))
... or
(b-x)/((c_1+x)(c_2+x)...(c_(n-1)+x))

It suffices to show that each of these terms has nonnegative second derivative.

Inductively, if g(x)&gt;0, g'(x)&lt;0, and g"(x)&gt;0, then the same is true of h(x) = g(x)/(c+x), where c&gt;0. This is satisfied by (b-x), so it is satisfied by (b-x)/(c_1+x), etc.

This shows that for any sensibly non-decreasing distribution of prizes, the ICM will predict that your E$ can't increase by taking an EChip-neutral gamble in a 2-way pot.

Re: ICM problems
 

I don't belive that ICM coincides with the brownian motion model although I have not checked this myself. ICM is a model based on the following assumptions (let t_i = number of chips that player i has and assume sum_i t_i = 1)

A) Prob(player i will win tourney) = t_i
B) Prob(player j will finish in (k+1)th position given that players i_1,...i_k have finished 1..k) = t_i/(1-t_i_1-...t_i_k)

Of course A is just a special case of B. However while there is a convincing argument that A is a good model of reality, I have not seen such an argument (myself) for B

THE WALLS ARE CLOSING IN! AAAAAAAAAARGGH!!!
 

[ QUOTE ]
[ QUOTE ]

What is ICM?


[/ QUOTE ]
Shuffle the players' chips. Rank the players by their highest chips. Equivalently, remove the chips from the table one at a time, eliminating a player when his last chip is removed.

Independent Chip Model calculators:
http://www.chillin411.com/icmcalc.php
http://sharnett.bol.ucla.edu/ICM/ICM.html (not as convenient, but it has some explanations)

You can write the formulas for n players by summing over the possible first place finishers, removing their chips, and applying the ICM to the reduced tournament on n-1 players. This doesn't seem to simplify. Does anyone know what the probabilities are of finishing last in a tournament where the stack sizes are 1, 2, ..., 100?

[ QUOTE ]

Is it random walk (brownian motion) on a simplex with absorbing boundaries? I saw Tom Ferguson wrote an exact solution for n=3 players. What else is known?

[/ QUOTE ]
The Brownian motion model is different. It's simple to analyze with 3 players because conformal transformations in two dimensions are understood and preserve Brownian paths, and the Riemann maps from the triangle to the disc are even known explicitly. Conformal transformations in higher dimensions are much more rigid, and there isn't much hope to extend the solution to more players without adding ideas.

[/ QUOTE ]

LOL! Funny thought! ICM corresponds to (a certain discretized version of) Brownian motion on a simplex with absorbing walls, with the added feature that the simplex is constantly shrinking towards its centroid! THE WALLS ARE CLOSING IN! AAAAAAAAAARGGH!!!

So there is a continuum of models between ICM and Brownian motion on a simplex! There is no reason not to unify the models.

(Now don't someone go selling my ideas to some poker site for fifty squillion dollars [img]/images/graemlins/mad.gif[/img] )

Re: THE WALLS ARE CLOSING IN! AAAAAAAAAARGGH!!!
 

[ QUOTE ]

LOL! Funny thought! ICM corresponds to (a certain discretized version of) Brownian motion on a simplex with absorbing walls, with the added feature that the simplex is constantly shrinking towards its centroid! THE WALLS ARE CLOSING IN! AAAAAAAAAARGGH!!!

So there is a continuum of models between ICM and Brownian motion on a simplex! There is no reason not to unify the models.

(Now don't someone go selling my ideas to some poker site for fifty squillion dollars [img]/images/graemlins/mad.gif[/img] )

[/ QUOTE ]

Have you shown that ICM is equivalent to discrete Brownian motion on a simplex?

Re: ICM problems
 

[ QUOTE ]

The Brownian motion model is different. It's simple to analyze with 3 players because conformal transformations in two dimensions are understood and preserve Brownian paths, and the Riemann maps from the triangle to the disc are even known explicitly. Conformal transformations in higher dimensions are much more rigid, and there isn't much hope to extend the solution to more players without adding ideas.

[/ QUOTE ]
I have a question for you pzhon, as you seem to have studied Ferguson's paper. Does the Brownian motion model give significantly different numbers to ICM in the 3 person case?

Re: ICM problems
 

[ QUOTE ]

I have a question for you pzhon, as you seem to have studied Ferguson's paper.

[/ QUOTE ]
Actually, I haven't, but the argument is natural for someone who has studied that part of mathematics.

[ QUOTE ]

Does the Brownian motion model give significantly different numbers to ICM in the 3 person case?

[/ QUOTE ]
A good place to look would be Bozeman's posts here which may have coined the term Independent Chip Model, as he did some comparisons of the models even with 4 players. You can't see the graphs now, though. I can't recall seeing a large difference between the models, but the predicted probabilities were different.

http://archiveserver.twoplustwo.com/...?Number=519924

If no one has copies, I'll try to illustrate the difference between the two models later, but probably only for 3 players.

Re: THE WALLS ARE CLOSING IN! AAAAAAAAAARGGH!!!
 

[ QUOTE ]

ICM corresponds to (a certain discretized version of) Brownian motion on a simplex with absorbing walls, with the added feature that the simplex is constantly shrinking towards its centroid!


[/ QUOTE ]
That's a good point, but there are quite a few extra good features of the ICM. You can use the ICM to obtain the probabilities of other finishes relatively easily. The probabilities can be determined by the analysis of the discrete model, not just as a limit. (You can define it by the limit, but all of the intermediate terms are equal, which is remarkable.)

[ QUOTE ]

(Now don't someone go selling my ideas to some poker site for fifty squillion dollars [img]/images/graemlins/mad.gif[/img] )

[/ QUOTE ]
After the US legislation, they were only willing to pay 15 squillion. Sigh.

Re: ICM problems
 

[ QUOTE ]
Does the Brownian motion model give significantly different numbers to ICM in the 3 person case?

[/ QUOTE ]

This table shows different stacks for 3 players and shows the predicted finish distribution of player A in the diffusion model and ICM.
http://img176.imageshack.us/img176/8436/icmzr7.png
For the most part they are pretty close, but it looks like ICM is slightly optimistic about short stacks. The diffusion model is probably a more "realistic" model, but its complexity makes it essentially unusable for more than 3 players. ICM does a good approximation and is a lot easier to calculate.

Tysen

Re: THE WALLS ARE CLOSING IN! AAAAAAAAAARGGH!!!
 

[ QUOTE ]
THE WALLS ARE CLOSING IN! AAAAAAAAAARGGH!!!

[/ QUOTE ]

Hey, I understood that part.

Having seen the discussion of this issue, which is undoubtedly fascinating in the abstract, and having seen the comparison of some results of the diffusion model and the ICM, I have some questions. In what sense can one model or the other be said to be a "better" representation of a poker game? Are we able to draw any conclusions as to what model most closely mimics the results of play between players of equal skill, in a no-limit hold'em tournament with an escalating blind structure?

There are several obvious respects in which "coin-flip" models fail to correspond to how a poker tournament is decided:

(1) They assume no skill differential.

(2) They disregard the variety of ways in which chips can be won or lost--e.g., bets of various sizes, two-way pots, and multi-way pots.

(3) They disregard the blind structure.

Pikachu proposed some other interesting models here. It strikes me that each of these (with the exception of the decay model, which seems completely unrealistic) has something to teach us. It strikes me that there are many random decision models that can give us insights, but that none of them corresponds very closely to how a poker tournament is actually decided.

So what model is "best," and why?

Re: THE WALLS ARE CLOSING IN! AAAAAAAAAARGGH!!!
 

Hey guys, maybe someone can figure out a little problem of mine. In Harrington's WB problem #39 he goes a bit into eq and figuring out places of finish.

Now I've coded an ICM and the data for third and fourth place does NOT match up with his values, though his columns do add up to one.

Maybe someone can take a look at his Prob. of Finish table and tell me where he is getting his numbers from. Some approximation method? I never could make sense out of how he tallied up those last 2 columns, the explanation seems a bit too vague.

Re: ICM problems
 

[ QUOTE ]

Start with just the target player's stack. Add the other players' stacks one by one, maintaining a joint distribution of the location of the player's highest chip and the player's rank. At the end, ignore the extra information about the location of the player's highest chip, i.e., sum over the possible locations.

[/ QUOTE ]
Something was been bugging me about this method. While some descriptions of the ICM allow non-integer stack sizes with no alteration, this one does not, and the complexity blows up when the stacks are large. It also obscures the fact that increasing all stack sizes in proportion will preserve the place probabilities.

I thought there should be a continuous limit, with density functions replacing the probability vectors and integrals replacing the sums. However, the correct limit as the stack sizes are increased in proportion is still discrete! The expected location of the second place player's highest chip is bounded as the stacks increase, and the limit is a geometrically decreasing infinite series of probabilities.

As the stack sizes increase proportionately to infinity, the joint distribution of the rank and the location of the highest chip stabilizes. For example, after inserting one stack, the probability that the player is second and has his highest chip in location n&gt;1 is p^(n-1)(1-p), where p is the proportion of the chips owned by the first player. It still might not be obvious that the probabilities do not change when the stacks are all doubled, but the limiting distributions are clearly the same, and the insertion method can be used to compute the ICM probabilities for irrational or very large stacks.

Re: ICM problems
 

Assuming that ... player r's skills are equal to the skills of the field, or are superior, we would ordinarily expect the following to be true at the start of the tournament: F_r &gt;= Z/n

I don't think this is true (and that has a lot to do with why I don't play tournaments, generally). Because the tournament payout schedule is quite top-heavy, you need to be on a par with the best of the field in order to have positive expectation. Even if 80% of the field is dead money and you're around the 90% mark in overall skill, you might have negative expectation - though this does depend on just how great an advantage the top players have over you.

However, I don't think this actually affects your other observations.

Re: ICM problems
 

[ QUOTE ]
Assuming that ... player r's skills are equal to the skills of the field, or are superior, we would ordinarily expect the following to be true at the start of the tournament: F_r &gt;= Z/n


I don't think this is true (and that has a lot to do with why I don't play tournaments, generally). Because the tournament payout schedule is quite top-heavy, you need to be on a par with the best of the field in order to have positive expectation. Even if 80% of the field is dead money and you're around the 90% mark in overall skill, you might have negative expectation - though this does depend on just how great an advantage the top players have over you.

[/ QUOTE ]

This is a thought-provoking and important observation or theory. Whether or not it holds true would seem to depend a great deal on the makeup of the field in an individual tournament. If true, it has a number of implications. If a large majority of the field--80% or maybe a lot more, in your view--has a negative expectation going into a tournament, it implies that the top players have quite a large positive expectation. In a zero-sum game, a lot of small losers add up to a few big winners.

How one would test this theory is hard to say. If one could get a large enough unbiased sample of players to cooperate in reporting all their tournament results, one could certainly draw some conclusions about the distribution of wins and losses, which should translate into the distribution of starting EV. But it's hard to get poker players to give their real identities, much less be diligent enough to report results consistently. It seems it would take an expert statistician and a big effort to put together a valid study.

Now, if most players in a tournament have a negative starting expectation, it is conceivable that many of them will have increasing marginal chip value across a broad range of stack sizes, for reasons David Sklansky has alluded to from time to time. Basically, they are likely to be camped out on a curve that is far below what a random model like the ICM would imply. Their curve may be "convex" across a broad range, meaning that "coin flips" are positive $EV for them. The underlying reason for this would be that they would be so deficient in the play of hands in most situations that they would not have many +cEV situations available to them, so that their best shot at finishing in the money would be to get into as many neutral or even slightly negative cEV situations as possible.

I do have a problem with the suggestion that this leads to any useful tournament advice for inferior players. The problem is, if you are poor at playing poker hands, and thus poor at creating +cEV situations for yourself, it is quite likely that you are also poor at recognizing "coin flip" situations when they come up. The very lack of experience and judgment that makes it hard for you to create edges also makes it hard for you to neutralize opponents' edges by gravitating toward coin flips. I suppose you could have an expert sit next to you, and advise you about when these opportunities come up--but that kind of expert help would probably be better directed to making you a better all-around poker player.

If it is true that in a given tournament, most players are negative expectation, and a few players are hugely positive expectation, it implies, for those few experts, that the range across which a chip utility effect could give them increasing marginal chip value is quite limited. The reasons I say this are those that I gave at the start of this thread. What it seems like it boils down to is this--if you start with a huge advantage over your opponents in the play of stacks of a variety of sizes, there is less room for increasing that edge by getting into deeper-stacked play.

Any more thoughts?

Re: ICM problems
 

Well, there are a couple of other simplifications in the model that distort the conclusions, but I don't know how to fix them (at least formally). One problem is that the model treats someone's advantage over the field as fixed over time; in reality, the skill level of the players will increase with time as the weaker players tend to be eliminated more. This implies increased chip value early on, since more chips are needed to take full advantage of certain situations that will be less likely to occur later.
As regards inferior players recognizing coin-flip situations, I think the problem for them is more 'meta' so to speak, in that the vast majority of them don't realize they are inferior. Further, most inferior players won't understand the theory outlined above, though from observation it does seem that most people instinctively understand that if they're playing against someone like Daniel Negreanu their best chance is to gambol it up and go all-in even if the odds might not be the best.
Anyway, if we assume that most players overestimate their skill in a systematic way *and* that they understand the 'coin flip aversion' results of the above model on some level, it has further (perverse) implications for optimal play. (You can obtain the same results if you assume simple irrational loss aversion on the part of most players). The conclusion is that if you can threaten opponents with elimination of their entire stack, you can force them to back down from coinflip type situations (even ones where they might have a slight advantage). This means that you need a big stack yourself. If your stack is small compared to the one you are threatening the curve is too flat (not convex enough locally) to encourage folds in marginal situations. This once again suggests increasing marginal chip utility if you're a shorter stack, with decreasing utility if you're already big enough to threaten others (but the stack sizes keep increasing...). It's also consistent with the real-world observations of my brother, who plays professionally.