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# An Introduction to the [0, 1] Model for Ranges

One of the problems with studying poker as a whole is that different poker games have different ways of dealing the cards across multiple streets. If you want to study poker on its own, then you need to have some sort of standardized model for dealing with ranges. It doesn’t have to be a perfect model, but it should make calculations relatively easy and be relatively easy to understand. The [0, 1] model is pretty good for this, and here we’re going to introduce you to what this model is and how it works.

The notation [0, 1] means that a player’s range is determined by randomly picking a real number between zero and one, including both zero and one. In this model, the lowest number wins, so if one player is dealt a 0.3 and another player is dealt a 0.6, the player with the 0.3 wins. This can be a little confusing at first, but after you look at some examples, it starts to make more sense.

How to Break Up Ranges Based on Strategic Options

Let’s suppose that you’re facing an all-in bet from a single opponent and you want to use the [0, 1] model to represent your range. You’re going to have a calling range and a folding range, and the calling range should be stronger than the folding range. If you decided to call 30 percent of the time, then your calling range would be [0, 0.3] and your folding range would be [0.3, 1]. You can find the size or frequency of your calling range by doing 0.3 – 0 = 30%, and you can find the size of your folding range by doing 1 – 0.3 = 70% as a quick shortcut.

How This Ties Into the EV Calculations Tutorials

Over the past few months, I have done a series of tutorials showing how EV calculations can be performed using simple addition, subtraction and multiplication. You can think of this as an extension to those tutorials that will give you a way to study poker on a more theoretical level. The [0, 1] model is used a lot when studying game theory as relates to poker, and being able to use it for those applications will help your poker game in a major way.

With an understanding of what this model can do for your game out of the way, let’s look at some simple applications.

Non-Polarized Betting and Calling Scenario

Suppose we’re on the river heads-up with a \$25 pot and \$20 behind. Both players have the same [0, 1] range after the OOP player checks. The IP player shoves with [0, 0.40] and checks with [0.40, 1]. This basically means he’s shoving a non-polarized range of the top 40 percent of his hands. When the OOP player faces a bet, he calls with [0, 0.60] and folds with [0.60, 1]. We want to know each player’s EV in this situation. Let’s start by working out all of the possible outcomes like we normally would from the EV calculations tutorial series:

1. IP checks, IP wins
2. IP checks, OOP wins
3. IP bets, OOP folds
4. IP bets, OOP calls, IP wins
5. IP bets, OOP calls, OOP wins

Now we need to know the chances of each of these happening. For scenarios 1 and 2, there’s a 60 percent chance that he checks. To figure out how often IP wins, remember that OOP will always win with a hand from 0 to 0.4, and then they will split the other 60 percent of hands. OOP will then win 70 percent of the time with IP winning 30 percent of the time after a check.

For the last three scenarios, realize that IP bets 40 percent of the time. OOP will fold 40 percent of the time as well. When OOP calls, IP’s range will be [0, 0.40] and OOP’s range will be [0, 0.60]. If you look at OOP’s range, from 0 to 0.4, he and IP will split those evenly. Then from 0.4 to 0.6, OOP will lose every time. This means that OOP’s going to win 1/3 of the time and IP’s going to win 2/3 of the time.

The last bits of information that we need are the profit levels for each of the possible outcomes. Let’s do this from the perspective of the IP player. In scenario 1, IP takes down the pot of \$25. In scenario 2, IP has a profit of \$0. With scenario 3, IP takes down the pot again for a \$25 profit, and scenario 4 has IP taking down the \$25 profit and the \$20 call for \$45 total. Finally, the fifth scenario has IP losing his bet for a profit of -\$20.

Putting It All Together

If you have followed through my EV Calculations Tutorial series, then you know what comes next. We have to find the EV of each possible outcome, and we do that by multiplying the chance of each outcome by the profit for each outcome. Here is a quick summary:

EV of Outcome 1 = (0.60)(0.30)(\$25) = \$4.50
EV of Outcome 2 = (0.60)(0.70)(\$0) = \$0.00
EV of Outcome 3 = (0.40)(0.40)(\$25) = \$4.00
EV of Outcome 4 = (0.40)(0.60)(2/3)(\$45) = \$7.20
EV of Outcome 5 = (0.40)(0.60)(1/3)(-\$20) = -\$1.60

Total EV for IP = \$14.10

We can do the same calculation for the OOP player and see who is getting the advantage here. The following is a summary of that:

EV of Outcome 1 = (0.60)(0.30)(\$0) = \$0.00
EV of Outcome 2 = (0.60)(0.70)(\$25) = \$10.50
EV of Outcome 3 = (0.40)(0.40)(\$0) = \$0.00
EV of Outcome 4 = (0.40)(0.60)(2/3)(-\$20) = -\$3.20
EV of Outcome 5 = (0.40)(0.60)(1/3)(\$45) = \$3.60

Total EV for OOP = \$10.90

As we can see, it’s the IP player who is getting the advantage with a \$3.20 difference in EV when the two players follow these strategies. Next week, we’ll look at more ways to use the [0, 1] model for ranges.