## Independent Chip Modeling Part 1: Derivation and Sample Analysis

- Independent Chip Modeling Part 1: Derivation and Sample Analysis
- Simulation #1
- Simulation #2
- Simulation #3
- Simulation #4
- Independent Chip Modeling Part 2: Positive Chip EV Doesn’t Mean Positive Monetary EV
- Independent Chip Modeling Part 3: Pushing Decisions from the Small Blind on the STT Bubble As A Function Of Stack Size
- Independent Chip Modeling Part 4: Beyond ICM
- ICM Calculator

**By Tony Guerrera**

## Introduction to Independent Chip Modeling

In cash games, your monetary expectation value (EV) can be directly equated to your chip EV (not accounting for image equity on future hands). Many tournament players don’t go beyond considering chip EV (cEV); however, in a tournament, your monetary EV (mEV) is technically the following:

P(1st)(Payout For First) + P(2nd)(Payout For Second) + … + P(*n*th)(Payout for *n*th)

To accurately assess tournament decisions, we need a way of projecting finishing distributions based on relative stack sizes. Assuming everyone in the tournament is of equal skill, a mathematical model exists that does just that: independent chip modeling (ICM).

## Derivation of ICM

WARNING: This section contains some difficult stuff. If sigma notation and pi notation aren’t exactly your thing, or if you simply care more about applications than derivations, then skip to the next section of this article. If you want to make sure I’m not simply pulling this stuff out of my ass, then read on.

First, we need a way of predicting your probability of finishing in first. One way of doing this is to model a tournament as a series of double-ups. Suppose a tournament contains T chips, and you currently have a stack size of S. The number of double-ups required to win the tournament, *n*, can be derived as follows:

If your probability of doubling up is *D*, then your probability of winning the tournament is . If everyone is of equal skill, then . Substituting in for *D*, we get the following:

is simply the proportion of total chips held. This result agrees with the commonly assumed notion that your probability of finishing in 1st is simply the number of chips you have divided by the total number of chips in the tournament.

Some of you out there might be saying “you can’t just model a tournament as a series of double-ups.” I agree, so as a backup to my math, I had a good friend of mine, Ryan Patterson, run some simulations for me. He ran heads-up and three-way simulations using a few different conditions:

Simulation #1: Heads-up match in which both players went all-in every hand.

Simulation #2: Heads-up match in which the small blind limped and then both players checked the rest of the hand down.

Simulation #3: Three-way match in which each player pushed all-in every hand.

Simulation #4: Three-way match in which each player limped every hand and all three players checked the rest of the hand down.

The results to these simulations are extensive. To preserve the flow of this article, the results are appended to the end. These simulations all support the notion that using is a valid predictor of P(1st). Of course, I haven’t proven that for all possible strategies that the players in a tournament can share, but at least this work suggests that this commonly assumed notion is valid.

Given that , our next task is to figure out P(2nd). Suppose you’re in a tournament consisting of *k* players (including yourself). If you finish in 2nd, one of your opponents finishes in first meaning that, effectively, you win a subtournament consisting of players. The probability of winning one of these subtournaments is , where is the current stack size of the player who is presumed to finish in first.

We can ignore the winning player’s chips in the subtournament since the first place player effectively wins a heads up match that starts with him having chips. For this to happen, the second player needs to accumulate all the other chips in the tournament (i.e. win the subtournament) before going heads-up. Obviously, this isn’t really how tournaments play out, but since we’re assuming that everyone is of equal skill, it doesn’t matter how we model the tournament as long as everyone is playing the same way. A mathematical model doesn’t have to work like the event it’s trying to model. Instead, it simply needs to make accurate predications about the system it’s trying to model, and all the work we’ve done so far ensures that ICM will make accurate predictions provided that everyone is of the same skill.

If we call P(*x*,*y*) the probability that player *x* finishes in *y*th place, your probability of finishing in second is the following (assuming you are player *k)*:

As an example, assume you’re in a tournament with 3 players remaining. You have 5,000 chips, P1 has 3,000 chips, and P2 has 2,000 chips. When you finish in 2nd, either P1 or P2 finishes in first. When P1 finishes in first and you finish in second, you effectively win a heads-up match versus P2 before losing a heads-up match to P1. When P2 finishes in first and you finish in second, you effectively win a heads-up match versus P1 before losing a heads-up match to P2. Your probability of finishing in second is therefore the following:

Assuming you are player *k* in a tournament consisting of *k* players, the general expression for your probability of finishing in *n*th place is the following:

This formula looks daunting, and to be honest, calculating this by hand for anything more than a 4-player tournament becomes a time consuming task; you need to add terms to find your probability of finishing in *n*th place in a tournament consisting of *k* players. Luckily, we live in the computer age, meaning that we can use software to compute our finishing distributions. With that in mind, let’s switch from the theoretical side of poker analysis to the practical side to see how we can use the ICM to make +mEV tournament decisions!

## Using The ICM to Evaluate the mEV of Decisions

To determine whether a decision is +EV, you must apply ICM to each of your possible stacks and then take a weighted average of your equities using the probability of each finish. Suppose you’re in a tournament that’s down to 4 players. 1st pays $500, 2nd pays $300, and 3rd pays $200. Before players post their blinds, players have the following stacks:

Big Blind (You): 1,500

Under The Gun: 4,000

Button: 3,500

Small Blind: 1,000

You have A2o in the big blind and blinds are 100-200. Action folds to the small blind who pushes all-in to 1,000. You put the small blind on a random hand. The possible states of the tournament, with your stack listed first, are given below; the number in parentheses after each set is your mEV found for the respective situation using ICM:

Call and win: {2,500, 4,000, 3,500, 0} ($305.13)

Call and tie: {1,500, 4,000, 3,500, 1,000} ($199.80)

Call and lose: {500, 4,000, 3,500, 2,000} ($74.95)

Fold: {1,300, 4,000, 3,500, 1,200} ($176.78)

Poker Stove, a hand simulation package, finds that your probability of winning is .5295, your probability of tying is .0396, and your probability of losing is .4309. If you call, your mEV is therefore ($305.13)(.5295) + ($199.80)(.0396) + ($74.95)(.4309) = $201.77. This is higher than your monetary EV when you fold; therefore, ICM tells us that this call is +mEV.

ICM is a great tool for evaluating the mEV of decisions in the tournament endgame, in which players can be assumed to be of roughly equal skill. This analysis can be extended to analyze multiple lines of play in order to optimize your mEV. For example, we can look at your mEV if you fold in the big blind with the intention of pushing all-in from the small blind, the button, or under-the-gun. In general, though, ICM is a tool that you should use to analyze the mEV of various moves (blind steals, resteals, calling, etc.) that you can make in your tournaments. By doing your homework with ICM, you should develop an intuitive sense of what to do when you’re actually battling at the tables.

Tony Guerrera is the author of *Killer Poker By The Numbers*.

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