Expectation, ICM, and Gigabet's "block" theory (long)

[Matt R.]

This post is very long, but I believe it is worth the read. I welcome any questions, comments, or criticism.

After finally making myself get around to looking at ICM, doing some calculations, and figuring out exactly what it tells you, I've come to a few conclusions that I believe link it with chip EV (simple pot odds) and the ideas in Gigabet's stack size theory thread a little while back. My ideas here aren't conclusive or mathematically rigorous, they're just ideas and I think anyone more experienced in math and proofs may be able to rework it into a more complete or usable theory.

Okay, to start I first examined ICM. A lot of this may be obvious to those of you who understand it, but this is where my thought process started. ICM is an equity model used to calculated expected real money return in a tournament, we'll call this E($), by analyzing the distribution of chips among the remaining players. I think ICM works wonderfully for what it is -- an equity model which does not take into consideration blind size, position of the stacks relative to each other, player skill, etc.. These shortcomings have been discussed before, and I think for a lot of situations their relevance should be ignored if ICM is giving you a very large positive E($) for a given play -- again, it appears most people agree on this.

To continue, I first looked at the difference between chip EV for a given play, we'll call this E(C), and the results ICM gives you. E(C) is calculated for a play simply by comparing pot odds to your probability of winning the hand. However, there are a lot of situations where you have a positive E(C), yet ICM tells you that the play is negative E($). Why is this, and where is the E($) going? The simple answer is, it is going to all the players left in the tournament who are not involved in the hand. In close E(C) situations, the player who is an underdog in the hand (including pot odds, so the player who has a -E(C)) is obviously losing E($) because he expects to lose chips in the hand, and chips are directly correlated with expected tournament winnings. However, why is it possible that the player who has a positive E(C) in the hand also has a negative E($). It is because, in certain situations, the probability that he loses the pot times the chip stack he will possess after losing greatly increases his chance of busting out before other players. Until it is heads up, how is $ gained in a tournament? It's not by increasing C -- as long as you "survive", C does not matter until you're heads up. You increase your $ in a typical tourney pay-out structure by other people busting out. It doesn't matter if you sneak into second with 1 chip or 1 million, you still get second place money if you don't win. The only reason the chip difference effects $EV is by greatly altering your chance for first (assuming that you can finish 2nd without accumulating chips. This isn't true in most instances, but we'll make this assumption for simplicity). Here is where this is going -- those who gain in a situation where 2 players are all-in, where even the +E(C) player has a -E($) according to ICM, are those that are sitting out of the hand because of their increased probability of moving up the payout structure due to someone else busting out before them. We'll call this last random variable B, where E(B) is the expectation in tournament winnings from other players busting out before you.

So E($), the number ICM spits out, is equal to E(C)+E(B) (although I don't think it's always accurate; we'll get to this in a second). For a given situation, if you add up your expected chip count return, E(C), for a given play and your expected return for the probability that others will bust out before you when you fold, E(B), you get your ICM number, E($). A couple of examples to illustrate why this is true. You are heads up, so your E(B) must be zero because P(B) is zero -- there's no one else that can bust out to move you up the pay scale. Therefore any +E(C) play will be +E($); this makes sense because you are playing one other player for the remainder of the prize pool above 2nd place. Also note the situation where you are VERY short stacked relative to the blinds, and no one else is. E(B) is very near zero because it's highly unlikely someone will bust before you (this is assuming a single table, we'll leave out MTT's for simplicity). Therefore, your E($) should correspond very closely to your E(C) since you really need to accumulate chips to stay alive. Lastly, look at a situation where you're in the big blind and the small blind has you covered. It is 3 handed, and UTG has one chip left and folds. SB pushes into you. Since P(B) is very close to one, your E(B) will be very high (the exact # will depend on payout structure). You obviously should fold a lot of +E(C) situations since you stand to gain a lot of E($) by shorty busting on the next hand or 2. You'd need to have a very high +E(C) to call.

Something to consider before I move on. I believe that ICM only does an okay job of measuing E(B) in certain situations. This is due to the above shortcomings -- it doesn't take into consideration blind size, position, and player skill. It only takes into account stack sizes relative to each other. Once you factor in very large and rapidly increasing blinds (such as in an STT), your position relative to the other stacks, and skill of the rest of the players, it is obvious that P(B) will change -- leading to differences in E(B). This is why I think there are situations where blinds are huge relative to your stack, you are about to be gobbled up by the blinds, and the other players at your table aren't donks (i.e. they won't do anything stupid to bust out before you) where ICM is inaccurate. When it only uses stack size to measure E(B), a play that it says is -E($) can actually be +E($) when other, non-measurable variables are considered. This is why I think when $EV according to ICM is only slightly negative, it should be ignored in a lot of situations where blinds are high.

A lot of the above may be obvious to some of you. However, I believe there is a link to all of this with Gigabet's stack size theory -- his idea many of you are looking at as "block theory". To see this, let's look at ICM one more time. From before, ICM calculates your expected return by E(C)+E(B). Let's look at a situation where you are the huge stack on the bubble, with one short stack behind you and two medium stacks in front of you. You are on the button. Short stack folds, leaving the 2 medium stacks in the blinds after you. Assuming the 2 blinds aren't crazy and realize that they should fold their way into the money unless they are holding a monster, you can profitably push any two here to pick up the blinds (note that it depends on the assumption they won't call with a lot of hands). Why shouldn't they call? It is per ICM, as stated above. They stand to gain more by folding and waiting for the short stack to bust. Here is the kicker: in order to put yourself in this situation where you can push trash hands profitably to pick up the blinds, you have to possess a large chip stack. Preferably, it should be large enough where you can go all-in, and even if one of the medium stacks decides to call you, you can lose the hand and not be crippled (i.e., you can still wait out the short stack to finish ITM). You can't use this technique to accumulate chips if you're a medium stack pushing into the huge stack -- you have to be the one that covers those who act after you. To put it another way, your large stack is enabling you to COMPLETELY ELIMINATE the E(B) of the medium stacks to act after you -- if they call, their E(B) is zero because they can no longer wait for shorty to bust if they lose. Their E($) is completely dependent upon their E(C), therefore their E(C) is going to have to be very large to profitably call. This means that you can steal the medium stacks' blinds even if they KNOW their hand is a solid favorite over the hand you're pushing.

This leads me to my 4th and last random variable, F. F is your future chip winnings that you stand to gain by using a big stack in the situations outlined in the previous paragraph. I believe this is exactly what Gigabet was talking about in his thread regarding "blocks", although he described it in very different words. So, E(F) is the expectation in chips you will gain if you win a given pot due to the relative stack size situation it puts you in. Put another way, it's your power of wielding a big stack. Note that for E(F) to be large, you have to be a good big stack player, otherwise those chips in the pot are useless when counted towards F.

What does this mean exactly? It means that if you're a good big stack player, and there are enough chips in a given pot to allow you to cover everyone at the table, or even just those to your immediate left (the more you will cover them by, the better obviously), you can profitably call in situations where ICM and pot odds tell you that you shouldn't. Therefore, for someone who is very good at using a big stack:

E($) = E(C) + E(B) + E(F)

If making a -chip EV call only increases your chances of busting out slightly (meaning it is -E(B)), your E($) is still positive if you are a good enough big stack player and the chips in the pot allow you to run over the table (your E(F) is large). So, if a given play doesn't cripple you, the pot is large enough, and you're a good big stack player, you can profitably make these plays that are -chip EV, and you can make calls where pot odds tell you that you shouldn't.