Gilbert Article

[Mason Malmuth]

Hi Everyone:

I just quickly read the article by George Gilbert. While many of you will find it very challenging, if you're playing tournaments, you need to read it.

Best wishes,
Mason

[PairTheBoard]

I don't understand why the GIC numbers are so small. I believe the GIC model assumes you play small pots with edges adequate to produce your overall tourny EV. This would seem ideal to me and would imply a reluctance to play for all your chips when you can grind out your tourny EV with small pots. Yet the chart shows the GIC model gambling All-In First Hand with some of the most minimal edges - even in the 100% Tourny EV case.

PairTheBoard

[NoSelfControl]

That's just the way the numbers came out. I think the GIC probabilities are too small, but don't have the evidence to support my intuition. It would be a mistake to think of the models as suggesting a strategy or style of play. Instead they are an attempt to approximate reality. Look at all the numbers for the type of tournament you play and your ROI. You should shoot for edges somewhere in that range if you are going to risk all your chips "early" in a tournament. In fact, I can't even define what "early" in a tournament means yet.

I think data rather than theory will ultimately elevate one model, perhaps one still to be developed, over another. It might even depend on the individual or particular type of tournament. I've been keeping track of my chips counts since I started thinking about these things last fall. It's a slow process. Anyone willing to collect and donate similar data should contact me.

George Gilbert

P.S. I'm also considering a couple models not mentioned in the article. For instance, just as one can use the power model to estimate the probability of coming in first or the expected value, one could use also (1-b^x)/(1-b) as an estimate of expected value.

[pzhon]

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While many of you will find it very challenging, if you're playing tournaments, you need to read it.

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I disagree. Most tournament players won't get much out of trying to sort through 5 different models (6 if the standard ICM is included). In fact, many players will probably use this article to reinforce their incorrect preconceptions rather than to review their play. Most winning tournament players can wait for the theorists to sort this one out. I think this article is a step in the right direction, but it is not close to being conclusive.

/images/graemlins/diamond.gif I would have liked to see more discussion of actual poker situations involving dead money, and where you cover your opponent.

/images/graemlins/diamond.gif Many players have a ROI much higher than 100% in large MTTs, and the "ability" variable in the article is almost always greater than the ROI. The ability of someone with a 500% ROI should be much higher than 500%. Why were ability values greater than 100% not included? Was it because the edges supposedly required to accept a coin-toss would be (more obviously) implausibly large?

/images/graemlins/diamond.gif There is a fundamental problem with these models. Suppose you are such a great poker player that you can double up with probability 60% in the normal course of play. What edge is needed to accept a coin-toss immediately?

The naive answer is 60%. This is wrong.

Some of the models in the article say you need greater than 60%. By taking the coin-toss, you give up the possibility of stumbling into 2nd place without ever doubling up. Does that seem plausible? In a MTT, it is hard to cash without ever doubling up, much less make the final table and outlast most of the players there. Indeed, it is hard to survive blinds (and antes) larger than your stack without doubling up.

I believe the correct answer is that you need less than 60%. Although you might be 60% to double up through normal play, you expect to miss some opportunities by not having a larger stack now. As Jerrod Ankenman said, "... everyone realizes that you get to apply your super-duper skill to the second $10,000, too, right?"

If you will have a chance to bet everything you have as a 4:1 favorite tomorrow, how much should you bet as a 55:45 favorite today? If the value of money is linear after tommorow, you should bet everything today, to expect to turn each dollar into $1.76 instead of $1.60.

The models don't see how you get your advantage, and how it may be more valuable to accumulate chips earlier. This fundamental modelling problem will be missed by most people who just look for an answer when they hit a few complicated formulas.

[PairTheBoard]

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That's just the way the numbers came out. I think the GIC probabilities are too small, but don't have the evidence to support my intuition.

George Gilbert


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I think it's too dismisive to say, "that's just the way the numbers came out". When calculations yield results which suprise intuition it deserves a closer look. The result may be trying to tell us something important that we would not have otherwise realized. Or it may indicate either a flaw in the Model or in the calculations which were applied to it - ie. an error in the math.

PairTheBoard

[NoSelfControl]

What I meant was that you singled out the GIC model and not the others. I don't like what I called the PF model, but don't have much of an argument why any of the others should be better than another. If I did, I probably would have only presented the "good" one(s). (I left PF in because learning about it inspired the article.) All of them are only crude first attempts at a model. In that context, in many cases the range of probabilities among the different models is not really that wide. Unfortunately, many common race situtations fall smack in the middle of this range. (Maybe that's confirmation that what seem like tough decisions really are; I don't know.)

I think the last table, comparing my results to the models, along with the comments in the next paragraph, are important. I'd say the agreement was pretty good. HOWEVER, the table was based on one type of tournament (very fast), a single player, and an inadequate sample. It's still suggestive.

[PairTheBoard]

I think you're missing my point. It's more a Math point than a poker one. Why is a model, the GIC, which assumes a player grinds out his Tourny Edge through lots of small favorable gambles, Not insist on Large Edges for Early All-In gambles? How is the math working in this model for it to in fact do just the opposite and actually allow the Smallest Edges for Early All-In gambles of all the models for 50 or more player tournies?

I'm not concerned with whether the GIC result is realistic or not. I'm pointing out that it seems mathematically suprising and thus begs a mathematical explanation.

PairTheBoard

[NoSelfControl]

Ahhhhh! Here's a pure math answer. The probability needed to race is E(x)/E(2x). For simplicity, let x^p and (1-b^x)/(1-b) model the expected value, not the probability of finishing first. (I'm actually looking at the latter model in its own right now.) Since in the models I presented, expected value is "essentially" a function of the probability of finishing first, the mathematical features won't change much. In the first case E(x)/E(2x)=1/2^p, which is very sensitive to changes in p. In the latter case, E(x)/E(2x)=1/(1+b^x). At the beginning of a large tournament, x is very small so that b^x is very close to 1 and E(x)/E(2x) is very close to 1/2.

This brings up another interesting observation. The ever-present caveat is that the models assume we are early in the tournament, so that blinds are negligible and whether one extra hand gets dealt isn't all that important. The above ratio for the Chen-Ankenmann large-pot power model (for expected value) doesn't depend at all on the stack size; i.e. our willingness to race for all our chips doesn't depend on whether we've already managed to quadruple our chips on the first hand or whether we've already lost half our stack on the first hand. On the other hand, for the small-pot model on which the GIC is based, as his/her fraction of the chips in play increases a good player should be less willing to race.

Another comment I want to insert somewhere. I think of poker as a very mathematical game, but one where the math only helps in avoiding fairly major mistakes. Taken collectively, the models are helpful to that extent. Even if someone develops a fantastically accurate model for expected value, it can't do much more than that. There are two other sources of inaccuracy. First, we need to use past tournaments to estimate the parameters for such a model. Just to have a 50-50 chance of estimating your ROI within 5% for a 10-player SnG requires a sample of over 500 tournaments where neither your ability nor your opponents' have changed. (To get within 3% with 95% confidence requires over 10,000.) It would almost certainly take a still larger sample to estimate the parameters for an expected value model to that degree of accuracy. Second, it is unlikely that we can estimate our probability of winning a potential race to within a couple percent. For instance, suppose we have QQ (wish I got it so often). If our opponent could have AA, KK, QQ, JJ, TT, AK, our chip equity is 52.4%. If our opponent could also have 99, it is 56.1%. If instead we drop TT, it is only 47.4%. (Thanks to PokerStove.)

[PairTheBoard]

Thanks George. Good stuff and worth further thought.

PairTheBoard

[SumZero]

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The models don't see how you get your advantage, and how it may be more valuable to accumulate chips earlier. This fundamental modelling problem will be missed by most people who just look for an answer when they hit a few complicated formulas.

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I think that is an important point. There is an obvious reason why chips early are more important, and that is you want to be in a position to capture the "easier" chips. That is the worse players will, on average, be knocked out before the better ones. So if you are a +20% player against the field as a whole you may be a -5% player against the players remaining when half the field is knocked out. I.e., your ability level against the players remaining should change and decrease as fewer players are remaining.

[NoSelfControl]

Absolutely. Expected value depends not just on stack size, but on stage of the tournament, the particular players who happen to remain (better or worse than normal), their stack sizes, and even things like seating assignments. The models in the article only attempt to provide simple yet plausible estimates of how your expected value might change should you gain or lose a lot of chips early in the tournament. Racing was an easy use for the models. No matter what your style of play, IF your expected value as a function of your stack size at the end of level 1 or 2, say, fits one of the models well, that model will provide an accurate value of the edge needed to race at that stage in the tournament.

[pzhon]

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IF your expected value as a function of your stack size at the end of level 1 or 2, say, fits one of the models well, that model will provide an accurate value of the edge needed to race at that stage in the tournament.

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What are you saying other than, "If it is accurate, then it is accurate?"

As I mentioned earlier in this thread, I don't think these models are accurate when they are based on the ability measure you suggested. I think there are many players with an ability level over 500% in the WSOP main event, and your models grossly overestimate the probability with which they (and others) need to win in order to risk all of their chips. It is unsatisfactory to me if someone with an ability level of 100% needs to plug in an ability level of 10% or -20% to get the right probability.

<ul type="square">Using the same equation as above, it turns out that we would take any edge greater than 48.63 percent. Yes, that’s right. I just made the argument that very good players should actually take slightly negative EV situations early in a tournament, because if they win the hand, they get to use their skill with their new stack.

Matt Matros in To Flip or Not to Flip (http://www.cardplayer.com/poker_magazine/archives/?a_id=15093)[/list]
Now that I've read his article, I'm surprised that you haven't responded to his argument while creating models based on the contrary assumption that you have to be risk-averse in order to have an advantage.

[NoSelfControl]

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IF your expected value as a function of your stack size at the end of level 1 or 2, say, fits one of the models well, that model will provide an accurate value of the edge needed to race at that stage in the tournament.

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What are you saying other than, "If it is accurate, then it is accurate?"

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Yes, in a way. Any model should be tested against real data.

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As I mentioned earlier in this thread, I don't think these models are accurate when they are based on the ability measure you suggested. I think there are many players with an ability level over 500% in the WSOP main event, and your models grossly overestimate the probability with which they (and others) need to win in order to risk all of their chips. It is unsatisfactory to me if someone with an ability level of 100% needs to plug in an ability level of 10% or -20% to get the right probability.

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What do you mean by "the right probability" (and what is the justification that it is right)?
FWIW - For a player whose expected net is 5x the WSOP buy-in, the models give the following probabilities needed to race: PEV 67.5%, PFIC 60.8%, GIC 53.2%
(Note that Matros suggests in the article that he doesn't believe anyone can have such a large edge in expected value. I have no idea.)

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<ul type="square">Using the same equation as above, it turns out that we would take any edge greater than 48.63 percent. Yes, that’s right. I just made the argument that very good players should actually take slightly negative EV situations early in a tournament, because if they win the hand, they get to use their skill with their new stack.

Matt Matros in To Flip or Not to Flip (http://www.cardplayer.com/poker_magazine/archives/?a_id=15093)[/list]
Now that I've read his article, I'm surprised that you haven't responded to his argument while creating models based on the contrary assumption that you have to be risk-averse in order to have an advantage.

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It looks like this refers back to the equation x(20,000) = (.538)(22,000) of his article. This equation seems to be based on flawed assumptions. Suppose you are able to grow your stack from 20,000 to 22,000 over some time period. Meanwhile, if you only had a 10,000 stack, the equation appears to assume that you are just sitting on your hands and are unable to grow it at all. Then at this point you decide to wager all your chips on one bet and it computes the edge you would need. In fact, if you are playing small-to-moderate pots to build this stack, the growth in your stack should be closer to a fixed amount than to a proportion of your stack, so you would expect to have closer to 12,000 chips than 10,000, so the 20,000 on the left side of the equation should be closer to 24,000. (A bit less because you could bust out or fall very low with 10,000 when that wouldn't happen with 20,000, and you've lost the opportunity to rebuild your stack. Or you might not get as much out of a major confrontation with another large stack.) It also implicitly assumes $EV is linear.

[jacksup]

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It looks like this refers back to the equation x(20,000) = (.538)(22,000) of his article. This equation seems to be based on flawed assumptions. Suppose you are able to grow your stack from 20,000 to 22,000 over some time period. Meanwhile, if you only had a 10,000 stack, the equation appears to assume that you are just sitting on your hands and are unable to grow it at all.

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No. If you pass on the coin flip, you're doing so because you want to use your skill on your 10,000 in order to get to 20,000. But that takes time. Meanwhile, if you take the coin flip and win, you get to 20,000 right away. We have to account for this time difference somehow, so I decided to use the 22,000 number to do it. I admit I could've made this more clear in the article.

And I am, of course, assuming the chip EV/$EV relationship is linear--a pretty good assumption for the first hand of the WSOP main event.

Matt

[pzhon]

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If you pass on the coin flip, you're doing so because you want to use your skill on your 10,000 in order to get to 20,000. But that takes time. Meanwhile, if you take the coin flip and win, you get to 20,000 right away. We have to account for this time difference somehow, so I decided to use the 22,000 number to do it. I admit I could've made this more clear in the article.

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Thanks for joining the discussion.

How did you choose the number 22,000?

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And I am, of course, assuming the chip EV/$EV relationship is linear--a pretty good assumption for the first hand of the WSOP main event.

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It looks like you didn't assume the value of chips is linear on the first hand, since that would make the decision easy, and the probability you need would be 50%. Did you assume the value of chips is linear at the end of the first period in which you are assumed to have doubled up or busted after not calling?

[jacksup]

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If you pass on the coin flip, you're doing so because you want to use your skill on your 10,000 in order to get to 20,000. But that takes time. Meanwhile, if you take the coin flip and win, you get to 20,000 right away. We have to account for this time difference somehow, so I decided to use the 22,000 number to do it. I admit I could've made this more clear in the article.

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Thanks for joining the discussion.

How did you choose the number 22,000?

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More or less out of the sky. It's just a guess as to what the EV of the guy who won the coinflip and began the tournament with a 20,000 stack would be at the point in time where his alter ego who passed the coin flip and started with 10,000 chips reached 20,000 chips by using his skill. Feel free to use a smaller number in your own model. Just make sure to account, somewhere, for the idea that 20,000 chips on the first hand is worth more than 20,000 chips three hours later.


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And I am, of course, assuming the chip EV/$EV relationship is linear--a pretty good assumption for the first hand of the WSOP main event.

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It looks like you didn't assume the value of chips is linear on the first hand, since that would make the decision easy, and the probability you need would be 50%.

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Not if you could acquire more chips through your skill and double up at a much greater rate.

[h11]

Ok, your argument seems to be that time has MORE value for a big stack than a smaller one, that 20,000 chips will compound at a faster rate than 10,000, at least for a good player.

That seems logical. I would bet you are right. But is there any data on the extent to which that is true? Is this something experienced players know is true, so there is no real need to see empirical proof?

Anyone out there have such data? Is it true that big stacks compound faster for chumps like me as well as for really good players? That's what I really need to know.

[jacksup]

Sort of. The argument is more that you have to double up a certain number of times to win a tournament, and it's easier to do that if your first double-up comes on the first hand than it is if your first double-up comes at hand 150.

[pzhon]

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IF your expected value as a function of your stack size at the end of level 1 or 2, say, fits one of the models well, that model will provide an accurate value of the edge needed to race at that stage in the tournament.

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What are you saying other than, "If it is accurate, then it is accurate?"

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Yes, in a way.

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That was not a yes-or-no question. "Yes" was not an appropriate answer. Is English not your first language?

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Any model should be tested against real data.


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Perhaps we have some fundamental disagreements about what is a good model. Yes, the model should predict the data. However, that doesn't make it good when
/images/graemlins/diamond.gif the context is too restricted, such as only applying in the first two levels, unlike the ICM (which is used by many winning players to study the bubble of SNGs),
/images/graemlins/diamond.gif there are free parameters to choose, both explicit (such as your "ability") and implicit (choosing one of your 5 models), and the correct probability may come from choosing a counterintuitive value, such as a negative ability for an expert player, or
/images/graemlins/diamond.gif it fails to be descriptive, or otherwise improve our qualitative understanding of the situation.

You haven't even established accuracy, which leaves your model a complicated mess of questionable value.

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It is unsatisfactory to me if someone with an ability level of 100% needs to plug in an ability level of 10% or -20% to get the right probability.

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What do you mean by "the right probability" ...?

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I mean the probability with which you need to win a coin flip for your whole stack to prefer it to folding when there is no dead money. What alternative could I have meant?

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Note that Matros suggests in the article that he doesn't believe anyone can have such a large edge in expected value. I have no idea.

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"But the best no-limit hold'em players think a $1,000 entry fee is worth $4,000 to $5,000, and in huge events like the World Series of Poker, with many beginners in the field, perhaps as much as $7,000 to $8,000." Harrington on Hold'em, page 12

I'm inclined to believe HOH. That the WSOP main event has continued to grow should increase the ROI of good players.

Matt Matros mentioned his own data about how frequently he doubled up. However, his sample size was small, he might not have been in the level HOH discussed, and when he did double up, on average he probably overshot by a substantial amount.

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Now that I've read his article, I'm surprised that you haven't responded to his argument while creating models based on the contrary assumption that you have to be risk-averse in order to have an advantage.

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It looks like this refers back to the equation x(20,000) = (.538)(22,000) of his article... the equation appears to assume that you are just sitting on your hands and are unable to grow it at all.

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No, you are busy using your skill to double up slowly more than 50% of the time.

Even if you don't like his argument, that doesn't justify introducing an assumption that contradicts his conclusion directly. Of course it meant you came to the opposite conclusion.

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if you are playing small-to-moderate pots to build this stack, the growth in your stack should be closer to a fixed amount than to a proportion of your stack,

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It should be between the two, since you run a risk of busting out during the period (you will bust out almost half of the time in this period), which means you will often not have enough chips to take full advantage of a profitable gamble.

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so you would expect to have closer to 12,000 chips than 10,000

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No. You may argue that the 22k was not plausible, but the 10k was supposed to be turned into 20k or 0. You don't get to double 12k.

I'd like to see generalizations of the ICM chosen carefully and analyzed in context. However, most of this article (and the subsequent discussion on this thread) was a step backwards from the article by Matt Matros.

[NoSelfControl]

My main motivation for writing the article was trying to model expected value in tournaments. Racing is the easiest application, but a model could also apply to rebuy strategy and to analysis of any hand in a tournament - small pot or large pot, two players or several, dead money or no. For hand analysis, even a perfectly accurate $ev model isn’t a cure-all because one has limits on estimating probabilities of the various outcomes. Nevertheless, it would be an improvement over any analysis done using chip ev.

If one is solely interested in the probability needed to race early in a tournament, the end of Matt Matros’ article contains a gem of an idea. It is reasonably accurate, completely customized to the individual, and takes minimal bookkeeping and no math. The tournament should really be large enough that you will usually either double up or bust out early in the tournament. I wouldn’t let this stop me from using it even for single-table SnGs, but I would be more cautious in relying on the answer. Simply compute the fraction of the time you double up before busting out of a tournament. This fraction is approximately the edge you need to race. Race with less of an edge or fail to race with a greater edge and you are costing yourself $ev. Estimating this fraction to within 0.03 with a 90% success rate would take about 750 tournaments. However, using a smaller sample is better than speculation based on no data.

Given that you can only estimate your edge in any particular hand to within a few percent at best, it would be fine to stop here. Of course, I won’t (see moniker). If time is truly not too much of a factor, this estimate is actually somewhat of an underestimate of the edge needed to race. At the instant you double up, you generally overshoot twice your initial stack by varying amounts. Let x be your stack and E(x) your expected value (early in the tournament, which I define to be early enough that we can ignore any time dependence). Choose units so your initial stack size is 1. Let’s use Matt’s data of doubling up 67 times in 127 tournaments. Then Matt’s total prize money is
127 E(1). Because, presumably, all of the prize money is compressed into the 67 tournaments where he doubled up, we have 127 E(1) = 67 E(2) and the probability needed for Matt to race is E(1)/E(2)=67/127=52.8%. However, Matt didn’t exactly double up, he actually crossed the doubling threshold with multiples of his initial stack x1, … , x67, each at least 2. Therefore, the real equation is more like
127 E(1)= E(x1)+ … +E(x67)
and the edge needed to race is
E(1)/E(2)=(E(x1)/E(2)+ … +E(x67)/E(2))/127 &gt;= 52.8%.
One can’t go much further without knowing the rough distribution of the xi and, ideally, the $ev. After tracking one tournament and just playing around with several cases and estimates, I’d say it appears that Matt’s edge needed to race should be in the 54%-58% range, i.e. 1-4% higher than the base estimate. A quick and dirty estimate for E(xi)/E(2) that requires only keeping track of one’s chip count at the moment of doubling and a little more calculation, yet seems roughly ballpark, is 0.5+ xi/4. (Again xi is the ratio of one’s chip count just after doubling to the initial stack size.)

You could do similar calculations for the middle stages of a tournament at whatever point you decide starts the middle stage, but it would require a lot more data because you should partition the data based on the size of your stack. You should probably also introduce a time factor here, perhaps something in the spirit of counting the times you double up after some deadline as one-half of a double, I don’t know. (Here, my definition of middle stage is more or less that lots of people still have at least borderline-deep stacks, but lots of the dead money is gone.)

One last remark. Suppose one views expected value E(x,t) as a function of stack size x and time in the tournament t. It’s fairly well-behaved as a function of x: increasing and, for a good player, almost certainly concave down nearly everywhere. On the other hand, the behavior of E with respect to t is pretty tricky to get any handle on, even qualitatively. Consider how the expected value of the initial stack changes over time. At the start, it is simply your expected value for the tournament. In the middle stages of the tournament, you would be extremely short-stacked with low expected value. However, near the end of a tournament, say the final table, the expected value of the initial stack has risen from the ashes and is quite high.

George

[pzhon]

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Simply compute the fraction of the time you double up before busting out of a tournament. This fraction is approximately the edge you need to race.

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No, it isn't, as was discussed repeatedly on this thread and in his article.

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Consider how the expected value of the initial stack changes over time. At the start, it is simply your expected value for the tournament. In the middle stages of the tournament, you would be extremely short-stacked with low expected value. However, near the end of a tournament, say the final table, the expected value of the initial stack has risen from the ashes and is quite high.

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The idea that an initial stack in the middle of a tournament would have a low expected value seems like a common misconception of casual players. It should be contradicted by any reasonable equity model. It certainly is by the ICM. (It is also contradicted by your other statements on how the equity varies with the chip stack.)

This discussion is going backwards. I'm really disappointed.