Poker Articles

Simple Arithmetic for Hold'em Players (Part 2)
By Lou Krieger

This is the second in a multi-part series appearing on Poker Pages:

(Exercise 2. Suppose I said that the guy sitting next to you will raise only with aces, kings, or A-K. If you're holding a pair of queens, what are the chances you're already trailing?)

Although the arithmetic isn't difficult, there are a number of components to this problem. Here's how to solve it. The only hands better than a pair of queens at this point are a pair of aces or a pair of kings. Queens, after all, are stronger than A-K right now, since A-K has to improve on or after the flop in order to beat you.

To solve Exercise 2, there are two additional problems you have to work out before you continue:

(Exercise 3. How many different ways can you make a pair of aces or a pair of kings? How many ways can you make A-K?)

There are only six ways to make aces - or any pair for that matter - but 16 ways to make A-K. Follow these examples and you'll see how easy it is to calculate. Here are all the possible ways to make a pair of aces: A A", A A , A A , A"A , A"A , and A A . Just take the aces from any deck of cards and you'll be able to figure this out even if you don't know how to calculate the answer arithmetically.

In mathematics, what you've just done by laying out cards is called a combination. It answers this question: "How many ways can you choose two items (in this case, each possible pair of aces) from a universe comprised of four aces?"

The mathematical process involves multiplying components of the universe in descending, but sequential order. How many components? Select as many as there are choices. That's step one. Then multiply each component of your choices, in ascending order. That's step two.

In this case, you'd multiply 4 x 3, or 12, and 2 x 1, or 2. Then set up a division problem, with the product of the universe calculation on top (the numerator, in case you've forgotten) and the product of the choice calculation on the bottom (the denominator). Then divide the numerator by the denominator. The answer, of course, is 12/2, or six.

Here's another example. If you wanted to calculate how many ways you can choose four items out of a universe of 20 items, you'd multiply 20 x 19 x 18 x 17, and divide the result by 4 x 3 x 2 x 1. The answer works out to be 4845. Of course, like any other problem, you can cancel out numbers to simplify the calculation. Just as long as you treat both the numerator and denominator the same, you can't go wrong. In this problem, you can divide the 4 in the denominator by 4, yielding 1, and 20 in the numerator by 4, which yields 5. You can also divide 3 by three, yielding 1, and 18 by 3, which yields 6. Then you can divide 2 by itself, and 6 by 2. This simplifies the problem to (5 x 19 x 3 x 17) divided by one. Canceling out can simplify problems, and if you are working with big numbers, it can often keep your pocket calculator from giving you an error message,

It's even easier determining how many ways you can make A-K. There are four aces, and four kings - and since any of the aces can combine with any of the kings, the answer is 16. It's a simple case of multiplying four times four. The reason you can't simply multiply when calculating how many pairs of aces can be extracted from a four-ace universe, is that aces can't combine with themselves. If you've been dealt a hand like A A , you're playing with a bad deck!

Since there are only six ways to make a pair of aces, there are obviously only six ways to make a pair of kings. If your opponent will raise only with A-K, a pair of aces, or a pair of kings, chances are greater that he raised with A-K, since there are 16 ways to make A-K but only 12 ways to make a pair of aces or a pair of kings!

You can do some other useful calculations with what you've learned thus far. For example, since there are 1326 possible two card holdings, but only six ways to make aces, try your hand at this problem.

(Exercise 4. What are the odds against being dealt a pair of aces before the flop?)

Since there are six ways to make a pair of aces, all you have to do is divide 1,326 by 6 to figure this out. The answer, of course, is that you'll receive aces, on average, once every 221 hands. Expressed in odds, you are a 220:1 underdog. Since you now know how rare aces are, as well as how to calculate it, try not to get even more upset than you usually do when you that rare hand gets cracked!

(Exercise 5. If you raise with any pair of 10s or higher, as well as A-K, A-Q, A-J, K-Q, or K-J, what percentage of your starting hands does that represent?)

There are five pairs (T-T, J-J, Q-Q, K-K, A-A, each of which can be made six ways) or 30 paired hands you'll raise with, as well as five big-card combinations, (each of which can be made in 16 ways) or 80 big cards. The total (80 + 30) equals 110 raising hands. Since there are 1,326 possibilities, you'll probably raise about eight percent of the time (110/1326) x 100 = 8.3 percent.

Next time we'll look at after-the-flop calculations.