Simple Arithmetic for Hold'em Players - Part 1 / By: Lou Krieger / Part 2 - Part 3
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Played at its best, hold'em is a wonderful blend of psychology, strategy, luck,
card sense, and self-discipline - where the game's rhythms ebb and flow, and
good players get the money at the end of the day, regardless of fate's
vicissitudes or how strongly the tides periodically surge against them.
While the game's sheen and elegance may seem completely within your control when you're in the zone and playing well, there are, nevertheless, coolly logical, and mathematical underpinnings to the game. Never mind that you sometimes play as though you're in a state of grace where you can seemingly do no wrong; when you're playing hold'em, you're always dealing odds. In lesser moments, don't you recall asking yourself-to- "How many outs do I have?" Haven't you ever wondered, "Is the money I'll win if I call with my straight draw greater than the odds against making my hand?" Remember the last time you called with a pair of eights and were raised? Didn't you wonder whether your opponent had a bigger pair, or was simply raising with big connectors? These questions all have mathematical components at their heart.
In addition to factors such as position, the number of players involved in the hand, and the tendencies of your opponents, some players are also blessed with an intuitive sense about mathematical relationships - although they can't calculate them, or even do a rough approximation in the heat of battle. Most beginners, however, along with others who may have played for years but are not mathematical inclined, simply ignore this aspect of the game. But the price of ignorance is high. These players give up a lot, too much it seems, which may be why they struggle year after year.
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Given the chance, even bright students are prone to avoiding mathematics. People
who would consider themselves egregiously insulted if called "illiterate," readily admit to being innumerate. But mathematics - at least poker math - is not difficult. Follow along and you'll see.
Be forewarned, however, there is some homework involved. Actually, it's more like a classroom assignment. Your assignments throughout this series are indicated by bold type. Whenever you encounter one, don't read any further. Even if you think you don't know the answer, try working out the problem - you might know more than you suspect.
How Many Hands Are Rightfully Mine?
This is not a question that's asked frequently, but when you consider how many hands you should play before the flop, it's in important one. In a nine-handed game with randomly distributed cards, approximately eleven percent of the starting hands should favor you. That's simple, isn't it? Divide one by nine and you get 0.11, or eleven percent. Keep that guideline in mind, but remember-to- It's just a guideline, not an answer.
It doesn't mean you should play eleven percent of your hands. When you're in
late position with lots of callers in an unraised pot, a hand like suited
connectors - that can improve to a very big hand but easily folded if the flop
is no help - certainly should be played, even if you knew it was not the best
hand before the flop. Conversely, if the pot was raised before your turn to act,
some big-card combinations that you might otherwise have raised with - like K-J
or A-T - are now trash hands that should be thrown away.
That's not all there is to it, either. Hold'em, after all, is a game of position, and you'll probably play less than 11 percent of your hands in early position, and far more than eleven percent when you're last to act. Whether calling, folding, raising, or reraising, is your best option will also be influenced by the play of those opponents who act before you.
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Hands like 9♦ 8♦, for example, play better with a full complement of opponents. Others, like K♣ 9♣, are raising hands when only the blinds are active but throw-aways if there's a raise and a couple of calls by the time it's your turn to act.
To produce practical strategies from these general concepts, we need to know just how many starting hands 11 percent represents, as well as when to increase - or reduce - that percentage.
(Exercise 1. Calculate the number of different two-card starting combinations you could be dealt?)
Did you come up with the correct answer? There are 169 unique, two-card starting hands. That assumes, of course, that A♠ A♥ is identical to A♣ A♦ and A♥ Q♥ is equal to A♦ Q♦ - and before the flop they certainly are. Here's how to compute it. You can receive any one of thirteen distinct ranks, from ace up through king, as your first or second card. All you need do, therefore, is multiply thirteen by thirteen to calculate the correct answer.
While that might be the correct answer, it is also not the whole truth. Here's why. A flop that's three-suited can certainly make some holdings much more valuable than others. If the flop, for example, was J♠ 8♠ 2♠ wouldn't you rather have A♠ A♥ than A♣ A♦
Although there are 169 different starting hands, it's also true that there are 1326 different two-card combinations. If you feel lost at this point, stop for a moment. I'll explain. When you're considering 169 different starting hands, you have yet to see the flop, and it's meaningless at this early juncture to distinguish between hands like A♥ Q♥ and A♣ A♦
When you're considering the odds against certain combinations of holdings, it's usually necessary to deal with all of the 1326 different 2-card combinations. Here's why.
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(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-to-
(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-to- 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 4,845. 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.
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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-to-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.
We'll look at after-the-flop calculations in Part 2 of this series.
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