In poker tournaments, when your stack gets shorter, the value of each remaining chip increases. A stack with 100 chips is worth more than a tenth of a stack with 1,000 chips. This is caused by the distributed prize structure of the tournament.
One basic assumption
To calculate the actual value of a player's tournament chips, one important assumption is made:
|Assumption: A player's chances of winning the tournament equal his share of the total chips.|
If you have half of the chips, your winning chances are 50%. With 10% of the total chips, you're 10% to win it all. If you have all the chips (100%), you're the winner per definition.
This assumption is based on another assumption: that you're as good as the average player in the tournament. This assumption can be debated, but for now it makes the calculations manageable.
Effect of the prize distribution
Due to the distributed prize structure in poker tournaments, first prize isn't the only value for the players. The second-place finisher also wins money, and you have a certain probability of coming in second, which adds to your equity. The same goes for all paid places.
To calculate the expected value in the tournament based on your present number of chips, we need to find the probabilities for your finishing in each one of the paid places, multiply it by the corresponding prize money, and add up all these pieces of equity.
And that's what ICM is all about. The concept as such isn't complicated. However, if there are a lot of players and a lot of money places, the calculations quickly become terrible. There are also a number of more or less questionable assumptions that underlie the calculations.
Three-player ICM example
Let's look at a small-scale example to show how the calculations are done. Three players remain in the tournament and there are two prizes:
Aaron has half the chips, and if there were only a first prize of $400, his expectation would be 0.5 * 400 = 200 dollars, according to the basic assumption above. Now, however, if he doesn't take first place he still has a chance to finish second.
The magic touch of ICM
Here comes the trick of the whole shebang, listen up!
If Aaron doesn't win, there are two possible cases: Bea wins or Caleb wins. The probability for each of those cases equals the player's share of the total chips, as we saw above. What we need to do is calculate Aaron's chances of finishing second in each of the two cases.
If Bea wins (a 40% chance), Aaron needs only defeat Caleb to come in second. They have 600 chips between them and Aaron has 83% of them. Analogously to the basic assumption we made in the beginning of this article, his chances of defeating Caleb is 83% (if Bea wins).
It should be noted that this part is debated. Several models to estimate second place probabilities have been proposed. But this is the calculation that the online IDM calculators are using and we will motivate it below.
In the same way, if Caleb wins (10% chance), the probability that Aaron will beat Bea to the second place is 500/900, that is, 56%.
So, the total probability for Aaron to finish second is:
P = 0.40 * 0.83 + 0.10 * 0.56 = 0.39
All in all, Aaron's equity in the tournament right now is 239 dollars. It's calculated like this:
E = 0.50 * 400 + 0.39 * 100 = 239
Short stacks have more equity than they deserve
Now we see the interesting effect of the distributed prize structure: despite having half the chips, Aaron's equity is less than half the total prize pool. He's not 50% to win second place, which makes him lose some equity.
If we do the same calculations for Caleb, his equity in the tournament turns out to be $57.
Caleb's probability to finish second:
P = 0.50 * 0.20 + 0.40 * 0.17 = 0.17
And his total equity in the tournament:
E = 0.10 * 400 + 0.17 * 100 = 57
This is more than 10% of the total prize pool. Caleb's 10% share of the chips is worth more than 10% of the prize pool. The equity that Aaron loses has to go somewhere; it goes to Bea and Caleb.
If we divide the players' equities with the number of chips they have, we see how much each of their chips is worth:
Equity per chip ($)
Short stacks pay more than big stacks
It's like we said in the introduction: the shorter your stack, the more valuable each of your chips.
From this fact, smart people have drawn a host of conclusions about correct tournament strategy. For example, you can study a tournament situation and use card odds in connection with ICM rather than the usual pot odds to find your optimal decision. More about this in another article.
For now, let's just notice that ICM explains much of the logic behind playing a big stack: The short stacks' bets are worth more than the big stack's. Each time a big stack makes a bet against a short stack, the short stack must pay a higher price to call. No surprise then that tall stacks can bully shorter stacks.
More details about ICM
For those of you who want more, we'll flesh it out a bit. We'll stay with the three-player example above.
Let's use the following notation for statements about the tournament outcome:
A1 = "Aaron places first"
A2 = "Aaron places second"
B1 = "Bea places first"
C2 = "Caleb places second", etc
We use the following notation for the players' chip stacks:
A = Aaron's chips
B = Bea's chips
C = Caleb's chips
At the start of the article, we made the basic assumption that a player's chances of winning the tournament equals his or her share of the total chips, which we call a, b and c.
P(A1) ≡ a = A / (A+B+C)
P(B1) ≡ b = B / (A+B+C)
P(C1) ≡ c = C / (A+B+C)
When it comes to the probability of finishing second, there are two ways for a player to do this - one with each of the other players as winner.
These probabilities are given by the following expressions:
P(A2) = P(A2|B1)*P(B1) + P(A2|C1)*P(C1)
P(B2) = P(A2|B1)*P(B1) + P(A2|C1)*P(C1)
P(C2) = P(A2|B1)*P(B1) + P(A2|C1)*P(C1)
where P(A2|B1) means the probability for A2 given B1 - the probability that Aaron finishes second given that Bea wins. This conditional probability is given by the Aarons share of the total chips when Bea's have been deduced, the total chips held by Aaron and Caleb:
P(A2|B1) = A / (A + C)
With this, the first of the three expressions resolves as follows:
P(A2) = P(A2|B1)*P(B1) + P(A2|C1)*P(C1) = A/(A+C)*B/(A+B+C) + A/(A+B)*C/(A+B+C) = A/(A+B+C)*[B/(A+C) + C/(A+B)] =
a*[ b/(1-b) + c/(1-c) ]
The other probabilities follow by symmetry:
P(B2) = b*[ a/(1-a) + c/(1-c) ]
P(C2) = c*[ a/(1-a) + b/(1-b) ]
Second place probabilities
We saw that the probability for a player to finish second if a certain other player wins was taken as his or her share of the chips among the remaining players - after deducting the present chip count of the future winner.
People have questioned this assumption and proposed other solutions, such as random walk simulations from the present chip stands. But if you look at it the other way around it seems quite reasonable.
If Bea is to win the tournament, then first one of the other players must be eliminated, and then the other. Aaron has 500 chips and Caleb 100. It might seem reasonable that Caleb will be first out in about 5/6 of the cases if you repeat this situation often?
Or put differently, that Caleb's 83% to finish in third place? Which is the same as saying that Aaron finishes second, of course.
Right or wrong, this is the assumption used in the ICM calculators that are available on the internet.
More than three players
So far we've only looked at the case with three players. With more players, the principles are the same but the calculations become more involved.
To outline this, let's look shortly at the case with four players and three paid places.
Chips: A, B, C, D
Finishing probabilities: A1, A2, A3, A4, B1, B2 etc.
For the probabilities of a player to finish in first, second and third place, we continue to split events into the possible cases (which are exclusive and exhaustive):
The same old assumption about first place probabilities:
P(A1) = A/(A+B+C+D)
Three possible cases for finishing second:
P(A2) = P(A2|B1)*P(B1) + P(A2|C1)*P(C1) + P(A2|D1)*P(D1)
A more involved expression for finishing third:
P(A3) = P(A3|B2)*P(B2) + P(A3|C2)*P(C2) + P(A3|D2)*P(D2) = P(A3|B2)*[ P(B2|C1)*P(C1) + P(B2|D1)*P(D1) ] +
P(A3|C2)*[ P(C2|B1)*P(B1) + P(C2|D1)*P(D1) ] +
P(A3|D2)*[ P(D2|B1)*P(B1) + P(D2|C1)*P(C1) ] =
[ P(A3|B2,C1) + P(A3|B2,D1) ] *
[ P(B2|C1)*P(C1) + P(B2|D1)*P(D1) ] +
[ P(A3|C2,B1) + P(A3|C2,D1) ] *
[ P(C2|B1)*P(B1) + P(C2|D1)*P(D1) ] +
[ P(A3|D2,B1) + P(A3|D2,C1) ]*
[ P(D2|B1)*P(B1) + P(D2|C1)*P(C1) ]
As usual, for second place probabilities we use the player's share of the remaining chips when the winner's chips have been deduced.
For example: P(B2|C1) = B/(A+B+D)
If player A finishes third while player B finishes second, either C or D wins, which again gives us two excluding and exhaustive cases where we use the player's share of the remaining chips when the winner's AND the second place finisher's chips have been deduced:
P(A3|B2) = P(A3|B2,C1) + P(A3|B2,D1) = A/A+D + A/A+C
As you can see, the P(A3) is a bit on the ugly side, and then you can imagine what P(120) would look like if you move into a realistic multi table situation. We won't do that here =)