Doing Arithmetic With Odds

Mathematics, Strategy

September 6, 2007

Previously, I’ve tried to convince you that doing your poker math in odds notation is far easier, once you get used to it, than using percentages. Now, I want to show you how to do some of the arithmetic usually associated with percentages faster and easier using odds. But first we need to lay some groundwork.

Base & Acuuracy

In percentage arithmetic, you represent additional accuracy by increasing the number of digits, or significant figures you retian in your calculations. In odds calculations, you retain additional accuracy by doing the calculation in a higher base. The base of a set of odds is simply the sum of the two (or more) numbers. That is to say, it’s the total number of outcomes. So odds of 7:1 against would have a base of 8. If the rules change so that an additional outcome is now a win for you, you would have odds of 6:2 against, still with a base of 8.

Now, here’s the deal: in the previous article I stated that odds are simply ratios, and therefore that you can legally multiply or divide each side by the same number, and retain the same odds. Odds of 3:1 against are exactly the same as odds of 6:2 against. However, it’s more complicated than that. Suppose I say the odds are “about 3:1” against or “about 6:2 against”. Those may appear to be identical statements, but they’re not. Assuming that in each case I chose the most accurate ratio in that base to make my statement, the first statement says the odds are between 2.5:1.5 against and 3.5:0.5 against while the second statement says that the odds are between 5.5:2.5 and 6.5:1.5 against. Clearly the second statement is a much stronger statement than the first. In other words, when odds are stated in a higher base, it implies greater accuracy.

Since it’s easy to switch between bases, this brings up the question: what base should we use in our calculations. And this question has two answers. First, it’s always OK to use the “true” base that represents the true number of events. You may later need to increase the base for other calculations, but the true base is always an acceptable place to start. Beyond that, you want to use a base that’s sufficiently large you get “enough” accuracy but not so large that doing arithmetic in your head with the numbers involved becomes too difficult. As I said previously, I believe that most poker calculations can be done with an error of a couple of % without harming your decisions. This means that a base in the realm of 50 or so is about right, and that should leave numbers small enough that doing math on them isn’t too hard.

Basic Arithmetic Concept

The basic concept for doing arithmetic with odds is simple: most operations involve moving numbers from one side of the ratio to the other without changing the base. However, to preserve accuracy, a sufficiently high base must be used. So the first step is always to convert to an appropriate higher base.

A doesn’t happen and B doesn’t happen (A or B happens AKA the logical OR)

Give two events A and B, and the odds against each, the question is what are the odds against one or the other happening. Or, given -A:A, -B:B, compute (-A and -B):(A or B). In layman’s terms, a situation where the odds get better when you combine the two options. This is easy to do:

  • Covert the odds against A into a sufficiently high base
  • Determine the base of the odds against B, which I’ll call B’
  • Divide -A by B’, multiply that result by B, and move that many elements from the -A bin to the A bin
  • Handle the remainder – split the remainder from that division in the same ratio as -B:B, and move those corresponding to B from the -A to the A bin.

Now, that probably sounds incredibly complicated, but it’s actually very easy to do. Suppose you know the odds against the Colts winning the Superbowl and the odds against the Patriots winning, and you want to know the odds against either of them winning.

Colts: 11:2 against
Patriots: 5:1 against

So here’s what we do: convert the Colts odds to 44:8 against. Divide 6 (the base of 5:1) into 44, and get 7. Move 7*1=7 units from the loss side to the win side, giving you odds of 37:15 against. Then note the remainder is 2, which not enough to move any more over.

Final result: 37:15 against either one winning. Note that this result is accurate to within a a percent ( 28.8% vs. 29.5%) even though no fractional or decimal numbers were used and we had to fudge a bit since 6 doesn’t evenly divide into 44. This pleasantly small amount of error stems from choosing the right base. While that computation wasn’t exactly easy to do, I’d say it’s much easier than doing 15.4% + (84.6% * 16.7% ) = 29.5% which I know I couldn’t do in my head.

The Logical AND

As you might guess, the logical “and” operation looks much like the logical “or” operation only you’re moving numbers the opposite direction. Given -A:A and -B:B, you want to compute (-A or -B):(A and B). So you do the following steps:

  • Covert the odds against A into a sufficiently high base
  • Determine the base of the odds against B, which I’ll call B’
  • Divide A by B’, multiply that result by -B, and move that many elements from the A bin to the -A bin
  • Handle the remainder – split the remainder from that division in the same ratio as -B:B, and move those corresponding to -B from the A to the -A bin.

Almost the exact same instructions, just with the directions flipped. For example, suppose I want to know the odds against a name pro wining the next WSOP main event and a name pro (not necessarily the same one) winning next year’s WPT Championship. Let’s say I believe the odds are

7:1 against a name pro winning the WSOP ME
2:3 against a name pro winning the WPT championship (ie. 3:2 in favor).

Then I convert the base on 7:1 to 49:7 and divide 7 by 5, getting 1 and a remainder of 2. So I move 1*2=2 cases over, getting 51:5. The remainder is 2, which is enough to move one more over, so we get final results of 52:4 – accurate to within 0.4%

Relative vs. Absolute Accuracy

Note that all the accuracy figures presented here are absolute in terms of percentage. Since that’s the main type of error that concerns poker players. If you want low relative error, you need a higher base. But the results here are perfectly good for poker purposes.

The Perfect Base

Notice the following: for all the computations above, if you use the base that is the least common multiple of A’ and B’, you get perfect accuracy.  So in the second example, if I had used a base of 40 instead of 56 I would have gotten odds of 37:3 which is an exact result.  Depending on what A’ and B’ are, this may not be feasible (since the resulting base may be too big) but in that example it would have been both easier and more accurate.  When you choose the perfect base, you never have to fudge at the end.

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