Simple Arithmetic for Hold'em Players - Part 2 / By: Lou Krieger
/ Part 1 - Part 3 |
In the first of these lessons on hold'em arithmetic you worked your way through
five exercises and learned how to calculate the number of possible two-card
combinations before the flop. There are 1,326 of them, which we arrived at by
determining the number of two-card choices that could be extracted from a
universe of 52 cards. The arithmetic for resolving combinations involves setting
up division problems.
We began by building the top number in our division problem. That top number is
called a numerator - and we derived it by multiplying some of the component
numbers in the universe. We started with the highest number in our 52-card
universe (which, of course, was 52), and continued in descending order until we
had as many numbers as choices.
Next we constructed a denominator, or bottom number in a division problem. It's
the number you divide into the numerator. Our denominator was obtained by
multiplying the numbers of choices we planned to extract from the universe. We
began with the number one, and continued in ascending order. Since we wanted to
choose two cards from the universe of 50, we began with one, and multiplied it
by two (1 x 2), and the result was two.
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That may sound complicated, but it's really quite simple. Stay with me. Here's
an example. To calculate the number of possible two-card choices that can be
extracted from a universe of 52 cards, we multiply 52 x 51 (which equals 2,652)
then divide that by 1 x 2 (which equals 2). The answer is 1,326. Using this
logic, can you solve the following problem? Remember, try solving it yourself.
Once you've finished - or if you simply can't work it out no matter how hard you
try - you have my permission to read the solution.
(Exercise 6. How many different flops are possible?)
This involves nothing more complex than following the procedure described above.
In this case, you want determine how many possible choices of three cards can be
taken from a universe of 50 cards. Why 50, instead of 52? Well, 50 represents
the universe of unknown cards. The two cards in your hand are not included in
this universe because they cannot possibly appear simultaneously in your hand
and on the flop. Your opponents also have two cards each, but since you don't
know the specific cards they're holding, you must assume that any of the 50
unknown cards are equally likely to appear on the flop. Of course, if you
happened to get a peek at one of your opponents cards, then the universe would
be reduced to 49 - since you would also know with complete certainty that the
card you saw in your opponent's hand can't make it to the flop either.
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Setting up the problem involves dividing the product of (50 x 49 x 48) by (1 x 2
x 3). If you don't cancel out to simplify the calculations, you'll wind up
dividing 117,600 by 6. It's that easy. There are 19,600 possible flops - and
it's a handy number to know. In fact it's so handy that we'll use it to solve a
practical hold'em problem - one I'm certain you'll encounter the very next time
you play.
(Exercise 7. If you have a pair in the hole, what are your chances of flopping a
set - or better?)
Here's how to go about it. If you calculate the number of ways to flop a set
when you start with a pair in the hole, you can compare that number with the
19,600 combinations representing the universe of possible flops. It doesn't
sound that tough, does it?
It's actually easier to calculate the number of ways to miss making a set. Let's try that. If, for example, you hold 8♦ 8♣ in your hand, there are two other cards in the deck that will form at least a set from a universe of 50 unknown cards - excluding the rather remote possibility that the flop itself is a set. If the flop was 9♣ 9♠ 9♥, any opponent holding a bigger pocket pair now has a bigger full house, and if someone happened to hold 9♦ he's made quads - although this is an extremely uncommon occurrence.
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So for all practical purposes, if the 8♥ and 8♠ are the only two cards that will make at least a set; the remaining 48 cards will miss. If the first card up on the flop is neither the 8♥ nor the 8♠, then 47 of the remaining 49 cards will also miss, and if the second card brings no help, that third and final flop card will not be the 8♥ or 8♠ 46 times out of the 48 remaining unknown cards.
Your next step is simple. Just multiply the fractions, as follows (48/50 x 47/49
x 46/48). When you multiply the numerators (top numbers) you get 103,776, and
when the denominators are multiplied the answer is 117,600.
What does this all mean? It means you won't flop a set or better 103,776 out of
117,600 times. By subtracting 103,776 misses from the universe of 117,600, you
are left with 13,824 hits. Now you know that flopping a set or better figures to
occur 13,824 times out of 117,600.
But 13,824/117,600 is a large an unwieldy fraction. To reduce it, divide the numerator and denominator by 13,824. When you do, you'll find yourself left with 1/8.5. If you divide 1 by 8 (or 13,824 by 117,600), you get 0.118. Expressed as a percentage, you'll flop a set or better 11.8 percent of the time that you hold a pair in your hand.
(Exercise 8. If you figure to flop a set 11.8% of the time you're holding a pair in your hand, how do you express that in odds?)
Many players are more comfortable talking about odds than percentages, but find it difficult to convert odds to percentages and vice-versa. Actually, it's easy, but before showing you how to make these calculations, it is important to clarify what we mean by odds. Odds are nothing more than a ratio of failures to successes, where the first number represents the predicted failures, and the second number is the expectation of successful events.
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When someone says "What are the odds?" he is really asking: "What are the odds against an event occurring," or, more precisely: "What is the predicted ratio of failures to successes?" If the odds against your horse winning are 8-to-5, it means that if this race were to be run 14 times (8 + 5), your horse figures to win five times and lose eight.
There is a relationship between percentages and odds, and understanding this relationship will help whether you're betting on horses, playing cards, or merely wondering what the chances are that it's going to rain. To change a percentage to odds, subtract the percentage from 100; then divide the result by that same percentage. If something has a 20 percent chance of occurring, this is the calculation-to- (100 - 20) / 20, or 80/20, which is equal to 4. Thus the odds against something with a twenty percent chance of occurring are said to be 4-to-1.
(Exercise 9. Convert odds to a percentage?)
Let's work through the 8-to-5 example. Try it out yourself, but if you're having difficulty, here's how to go about it. If the odds against your horse winning are 8-to-5 you figure to win your bet five times out of fourteen tries. How do you come up with 14? Remember, odds are a ratio of failures to successes. If you add the failures to the successes, you have a universe of 14 events. If this event turns out favorably 5 times out of 14 trials, all that remains is for you to divide the expected wins by the universe of events, or 5/14. The answer is 36 percent. If the odds against your horse winning are 8-to-5, it has a 36% chance of winning.
Now try it yourself, assuming odds of 7-to-2. Add 7 and 2 and you get 9. If you divide 2 by 9, the result is 22 percent. Now you know that a 7-to-2 shot has a 22% chance of winning. If you've been able to follow this so far, take the time to work out some additional problems.
Try your hand at changing odds to percentages, and percentages to odds. With a little practice, you should be able to do this very quickly. When the numbers you're working with are complex, go ahead and approximate, and use rounded numbers that are easier to manipulate. After all, if a 1.86-to-1 shot actually equates to 35%, you won't go too far afield by thinking of it as a 2-to-1 shot, dividing 1 by 3 in your head, and coming up with 33 percent - rather than the more precise figure. In the heat of battle, sometimes that's the best you can hope to do. And since your opponents are probably not doing this at all, you have an edge.
We'll look at more in Part 3 of this series.
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