Hold'em FAQ

The Hold'em FAQ is brought to you in part by Steven James, author of The Evolution of a Poker Player.

The Evolution of a Poker Player is available from Amazon and Gambler's Book Shop, and you can have your favorite book store order it for you from Baker and Taylor (They'll know who that is.).

Topics:

• What are card odds?
• What are pot odds?
• How can I use odds?

Combinations:

• What are the Sample Spaces?

What are the odds of:

• 1 thru 10 Outs?
• Completing a four flush?
• Catching perfect on 4th and 5th street when I have a three flush?
• Ace, King vs. a pair of Queens on the flop?
• What is expected value?
• Flush Example

What are card odds?

Card Odds are the number of cards that will improve your hand versus the number of cards that will not improve your hand.

By default, any card that is not counted as improving your hand is counted as a card that will not improve your hand. For example, if you have two spades, and the flop has two spades there are nine spades that will improve your hand to a flush. However, let’s consider the case where you have two spades and one of them is the ace. There are three other aces that will improve your hand, but in this example we counted them as cards that would not improve your hand, because we only counted the spades. It's up to you to count them as cards that will improve your hand, or not, depending on what you think you need to win. Card odds can be very simple to calculate. For instance, if you have a four flush on fourth street, the card odds are 4.11 to one against you making the flush on fifth street. Here’s how to get that number. You hold two of the 13 spades and 2 are on the board. That leaves 9 left to make your hand, and 37 that will not make it. Notice that 9 plus 37 is 46, the total number of cards left in the deck for fifth street. Also notice that 37 to 9 reduces to 4.11 to1. Of course there really aren’t 46 cards left in the deck. Your opponents have cards, and there are burn cards. But it’s assumed you don’t know the exact value of those cards. If you do not know the exact value of those cards, it’s customary to count them as being in the deck. If you do know the exact value of those cards, simply adjust your count.

What are pot odds?

Pot Odds are the number of bets the pot offers you for your call, bet, or raise. If you can call for one bet, and the pot has five bets in it, you’re getting 5 to 1. And if you're last to act, it's just that simple. However, if you're not last to act, it's not that simple because you'll need to guess whether or not your opponents will call, or raise, or whatever.

If you're sure they'll call, you can count their call(s) as if they're in the pot. If they raise, you're going to have to consider that as well. It's often correct to call the raise if you're drawing to a good hand. However, the raise has reduced the pot odds you were getting on the next card, and that's usually a bad thing. If they reraise, you're going to have to consider a whole lot more, which is beyond the scope of this pot-odds primer.

How can I use odds?

There are many ways to use card odds and pot odds to help make decisions, but let’s keep it simple for now. (I hate to get confused!) Let’s stay with the flush example we used in the card odds section. Here’s the situation. You have a four flush on fourth street, and you think you must make the flush to win. You know it will cost you one bet to stay for the fifth street card (Because you are last to act.), and you want to know if you should call based on the pot odds compared to the card odds. If the pot has more than 4.11 bets in it, a call would be justified based on the pot odds compared to the card odds. If the pot has less than 4.11 bets in it, a call would not be justified based on the pot odds compared to the card odds. (If you're wondering why we use "4.11," it's because that's the card odds of making the flush. You can review that calculation in the card odds section.) That's the basic concept. However, sometimes there can be a bit more to it than that. Let's consider this contrived, simplified situation. You have a four flush on fourth street against two opponents. Your opponent bets, and it's up to you. The pot contains 6 bets (Let's not be concerned with why there are 6 bets and how they got there.). Your cards odds are 4.11 to 1, and the pot offers you 6 to 1, so you call. Your opponent behind you raises, the other calls, and it's up to you again. Your card odds are still 4.11 to 1, and now the pot offers you 10 to 1 for your call. You liked it when you got 6 to 1, so you've gotta love it now that it offers you 10 to 1, right? RIGHT? And you wish they could raise again, because your next call would be getting 14 to 1! You really love these raises. Right? Not exactly . . . You would certainly be correct to make the call, because of the pot odds compared to the card odds. But, . . . and this is a big, important but . . . you did not really like the raise.

Why not?

Why don't you like the raise when each time you call you're getting higher and higher pot odds? Because in this case you want to get the fifth street card as cheaply as possible, and each raise increases the cost of that card. Even though you got good pot odds on each call, the overall cost of the card was higher, because the percentage of strange money is lower. Without the raise, you got the fifth street card for 6 to 1. But when that opponent raised, and you eventually called, even though you got 10 to 1 for that call, overall you got 5 to 1 for the fifth street card. That's because the pot had 10 strange bets, and it cost you 2 bets. Which brings up a concept that's very confusing to many new players. (Damn! I was afraid this would get confusing. I hate it when that happens!) Money in the pot belongs to the pot, no matter who put it there. That's true, except for the exceptions. As you probably noticed above, at one point we counted all of the money in the pot as the pot's money, and at another point we counted one of the bets in the pot as your money.

So, which is right, and which is wrong? Sorry . . . but . . . they're both right.

Each time it's up to you to decide to make the call, you count all of the money in the pot as the pot's money. However, when you calculate your cost of the fifth street card, you count everything you put in the pot on that round of betting, compared to what was in the pot before that round and everything your opponents put in on that round. That's how 10 to 1 turned into 5 to 1. The reason they're both right is because they each answer a different question. One answers the question, "What is the pot offering me now?" The other answers the question, "What was the total cost of the fifth street card?" Another kinda tricky concept for new players is "potential" odds. Some people refer to this as "implied" odds, but they're wrong. Nevertheless, I need to mention it so if you see that term you know it usually refers to the same concept. The concept of potential odds is simple and intuitive. It means that you can count bets that are not in the pot now as being in the pot now if you're sure they'll be there later. For example, if you know opponents behind you will call, you can count their call as being in the pot when deciding what to do. And this concept will come up other places too. If the pot doesn't have enough to justify a call, but you know your opponents will call after you make your hand, you can consider those calls as potentials. That’s the basic concept of using card odds and pot odds. However, some players would argue that there are many other considerations, such as the odds of winning with a bluff, etc. And they'd be right. But, remember, we’re keeping this simple. (So I don’t get any further confused!)

What are the Sample Spaces?

In order to evaluate poker situations we need to know the size of the sample space, and we need to know how many events are in each sample space. An event is one of the possible outcomes of the situation, and all of the outcomes together, comprise the sample space. We also need to know how many of those events are in our favor, and how many of those events are in our opponent’s favor. That’s what combination theory does. Later we’ll use this information to calculate expected values. In a hold’em game we want to know how many events there are for 5th street, 4th street, 4th and 5th street combined, the starting hands, and the flop. Each of these groups is known as a sample space, and each member of the sample space is known as an event. Exploring sample spaces is easy once you get the hang of it, but it can be a challenge to get the hang of it. So, we’ll start of with very simple examples and work our way into the hold’em sample spaces. Let’s start with symbols. Count the symbols: @ @ #. There are 3 symbols. If we think of each symbol as an event, there are 3 events. Therefore, the sample space is 3. What is the ratio of @ symbols, to # symbols? The ratio is 2 to 1. If a symbol were randomly selected, the odds a # symbol would be selected is 2 to 1. The number of negative events, the @ symbols in this case, are usually put first; and the number of positive events, the # symbol in this case, is put second. The odds are read as “2 to 1 against.” The exception to this rule is when the probability of the event is greater than 50%. For example, the odds of an @ symbol being randomly selected are “2 to 1 for.” The larger number stays first and the odds are read as “2 to 1 for,” instead of “1 to 2 against.” This principle is the foundation for calculating card odds for hold’em.

5th Street Sample Space

The 5th street sample space is 46. The sample space is comprised of all unknown cards. We only know the 2 cards we hold and the 4 cards on the board (the 3 on the flop, and the one on 4th street). This leaves 46 unknown cards. So, if we happen to see one or more of the cards remaining in the deck, or our opponent’s cards, or we can tell what cards our opponents hold by some other means, we’ll have to make the necessary adjustments in these calculations. However, for these examples we’ll assume we only know our 2 cards and the 4 on the board.

4th Street Sample Space

The 4th street sample space is 47. That’s the 52 cards minus the 2 cards in our hand and the 3 cards on the board.

4th and 5th Street Sample Space

The 4th and 5th street sample space is 1081. Here’s how we calculate it: 52 cards minus the 2 cards we hold and the 3 cards on the flop, leaves 47 cards that can come on 4th or 5th street. But here the 4th street card and the 5th street card will be counted as a single event. So, we want to know how many 2-card combinations can be made from 47 cards. It’s important at this point to distinguish between combinations and permutations. Combinations are multiple events where the order of each individual event is not important, @# is considered the same as #@. In permutations @# is considered different than #@. For this particular sample space we don’t care about the order, so we use combinations. Our technique counts all of the permutations and then divides out the duplicate events. Here’s how we calculate the permutations. The 47 cards available for 4th street are multiplied by the 46 remaining cards available for 5th street. That’s 2, 162. Now we need to find our factor to divide out the duplications. Since we had 2 events ( 4th street and 5th street) we multiplied 2 by 1, and got 2. Then we divide 2,162 by 2 and got 1,081. That means there are 1,081 2-card combinations from 47 cards.

Starting Hand Sample Space

The starting hand sample space is 1,326. Here’s how we calculate it: 52 cards available for the first card, and 51 cards available for the second card gives us 2,652 permutations. Now we need to divide out the duplications, so we divide by 2 times 1, or 2. Which gives us 1,326.

Flop Sample Space

The flop sample space is 19,600. Here’s how we calculate it: 52 cards minus the 2 cards we hold leave 50 cards for the flop. That’s 50 cards for the first card, 49 cards for the second card, and 48 cards for the third card. Which gives us 50 times 49 times 48, for a total of 117,600 permutations. Now we need to divide out the duplications, so we divide 117,600 by 3 times 2 times 1, or 6. Which gives us19,600.

Card Odds

Sample Space: 47 unseen cards for 4th street, 46 unseen cards for 5th street, 1081 combinations for 4th and 5th street combined.

How to use this information:

If you have a hand that you think will need to improve in order to win, you want to compare the odds of improvement to the pot odds. For example, let’s say you have a four flush on fourth street, and the fifth street card is going to cost you 10 dollars. There are nine cards of your suit remaining, and the chart says the odds for nine outs on fifth street are 4.11 to 1. So, in round numbers, the pot needs to contain at least 40 dollars. That doesn’t mean the pot has to contain at least 40 at the exact moment it’s up to you to act. If the pot contains less, but you are certain there will be callers behind you and their calls combined with the current pot totals more than 40, that’s okay. But, if someone behind you raises, that reduces your pot odds. So, what you’re really looking for at this point is to get at least 4 to 1 for every dollar you put in the pot on this round of betting. For example, if there’s you and 3 opponents, and there’s 30 in the pot and it costs you 10 to call, there isn’t enough in the pot to justify your call. But, if at least one person behind you calls that is enough. But if one of them raises, and everyone else folds, that means you’re getting odds of only 2.5 to 1. That’s because it ended up costing you 20 for that 5th street card, and the pot ended up giving you 50.

Comparing card odds to pot odds needs to be considered as a general guideline, not an absolute rule. Most of the time it’s better to have the pot contain more than the minimum, the more the better. In our flush example, you’d rather have 60 dollars in the pot than 50, even though 50 is enough. That’s because sometimes you’ll make the hand you’re drawing to and still lose. Ouch!

Sometimes you won’t need to make the hand you think you need in order to win, but this usually isn’t the case. So you’ll usually need pot odds better than the card odds in order for the hand to be profitable.

What are the odds of completing a four flush?

Here we’ll take a look at how to calculate the odds of completing a four flush at different phases of the game. We’ll consider any card of our suit a positive event, and any card not of our suit a negative event. The “sample space” is the total number of unseen cards.

Let’s assume we have a heart four-flush on 4th street, and we want to know the odds of making the flush on 5th street. You might want to follow along by getting a deck of cards and laying out the hands as described.

Our sample space is 52 cards minus the 2 cards (both hearts) we hold, and the 4 cards (2 of them hearts) on the board. That’s a sample space of 46 cards. Any card that is not a heart is a negative event, and there are 37 of them. Any heart is a positive event, and there are 9 of them. The odds are 37 to 9 against. This can be reduced to a more useful form by dividing 37 by 9, and 9 by 9, which gives us 4.1 to 1. Reducing all odds to “to 1” makes them easier to use in the game. It’s a bit more difficult to make the flush on 4th street than it is on 5th street. Our sample space this time is 52 cards minus the 2 cards (both hearts) we hold and the 3 cards (2 of them are hearts) on the flop. That leaves us a sample space of 47 cards. Any card that is not a heart is a negative event, and there are 38 of them. Any heart is a positive event, and there are 9 of them. The odds are 38 to 9 against. This can be reduced to a more useful form by dividing 38 by 9, and 9 by 9, which gives us 4.2 to 1. Let’s say you’re sure you can get both the 4th street and 5th street cards for just the bet on the flop. This may be possible because your opponent is all in, or you know your opponent well enough to know they will not bet on 4th street, or whatever. In this situation you need to know the odds of making the flush on 4th street and 5th street combined. Our sample space this time is 1081. Here’s how we calculate it: 52 cards minus the 2 cards we hold and the 3 cards on the flop, leaves 47 cards that can come on 4th or 5th street. But here the 4th street card and the 5th street card will be counted as a single event. So, we want to know how many 2-card combinations can be made from 47 cards.

It’s important at this point to distinguish between combinations and permutations. Combinations are multiple events where the order of each individual event is not important, @# is considered the same as #@. In permutations @# is considered different than #@. For this particular sample space we don’t care about the order, so we use combinations. Our technique counts all of the permutations and then divides out the duplicate events.

Here’s how we calculate the permutations. The 47 cards available for 4th street are multiplied by the 46 remaining cards available for 5th street. That’s 2, 162. Now we need to find our divisor to divide out the duplications. Since we had 2 events ( 4th street and 5th street) we multiplied 2 by 1, and got 2. Then we divide 2,162 by 2 and got 1,081. That means there are 1,081 2-card combinations from 47 cards. To find out how many of those events are negative, and how many are positive, we take the 9 hearts and multiply them by the 38 non-hearts. This gives us 342 events where one of the two cards on 4th or 5th street is a heart, or 342 positive events. But, there are others. It’s also possible that both 4th street and 5th street will be a heart, so we need to include those events also. We need to know how many 2-card combinations can be made from the 9 hearts. There are 9 hearts available for 4th street, and then 8 hearts available for 5th street. That’s 9 multiplied by 8 for a total of 72. We factor out the duplications by dividing by 2 for a total of 36. This gives us an additional 36 positive events. Now we add our 342 positive events to our 36 positive events, for a total of 378. That’s 378 ways to make our flush. Since our sample space is 1081, we subtract our 378 positive events from it to find the number of negative events, 703. So, that gives us 703 negative events, and 378 positive events. The odds are 703 to 378, or about 1.86 to 1.

What are the odds of making a three-flush?

23 to 1

However, we need to put this question into its proper perspective. Players don't usually want to know this because they plan to play it, well, some do, but usually it's because they suffered a bad beat and want to know just how bad it was. Nevertheless, here's how to calculate it. Let's say we have two spades, and one spade comes on the flop. Actually, the odds are the same if we have one spade, and two come on the flop. So it doesn't matter which way we define it. And, there are two simple ways to solve this question.

First, we'll use combinations.

Our sample space is 47 choose 2. Why 47 choose 2? Because the deck originally had 52 cards. We have two cards in our hand, and there are three on the flop. That leaves 47 cards unknown, and there are two of them to come. One on 4th street and one on 5th street. (47 x 46) / (2 x 1) = 1,081 So, our sample space is 1,081. How many of those events are two spades? 10 choose 2. Why 10 choose 2? Because there are 10 spades left, and we need two of them. [(10 x 9) / (2 x 1) = 45] So, there are 45 combinations of two spades. Therefore, the odds are 1,036 to 45, or about 23 to 1. 23 to 1 expressed as a percentage is (45 / 1,081 = .0416) 4%. Or, 1 / 24 = about 4% also.

Second, we'll use probabilities.

If my memory serves me correctly, (Don't bet on it.) we'll use the multiplicative property of dependent probabilities. Since we need 4th street to be a spade, and we need 5th street to be a spade, we take the probability of each and multiply them. That's 10 / 47 x 9 / 46 = .0416 Why 10 / 47 and 9 / 46? Because there are 47 cards available for 4th street, and 10 of them are spades, and there are 46 cards available for 5th street and 9 of them are spades. So, the probability is about 4%. Amazing! Isn't it? The same answer both ways. Well, you may not be impressed, but I am. I usually can't get the right answer just one way, more less two ways ;-) Good luck.

Ace, King vs. a pair of Queens on the flop?

Here’s the situation: If we're heads up and hold ace, king, and our opponent holds a pair of queens, what are the chances of us catching at least a pair on the flop without any queens on the flop? If you’re not interested in the answer, stop here. If you are, read on . . . Okay then. Let’s get to it. This is a basic combination problem. The first thing we do is establish the sample space, and then count the positive and negative events. Our sample space for the flop is 48 choose 3,

[(48 x 47 x 46) / (3 x 2 x 1)]

which is 17,296. That’s all of the possible flops. The positive events are all flops with at least one ace or king, without any queens. The negative events are all flops without an ace or king, and all flops with at least one ace and/or king with a queen. Let’s count the negative events first. You might ask, why not be positive and count the positive events first? I’d just roll my eyes and say, let’s get on with the negativity, Miss Sunshine. We’ll be positive later. We'll count the negative events, and then subtract them from the sample space to get the positive events. If we count the number of flops that don't have an ace or king, we have 42 choose 3,

[(42 x 41 x 40) / (3 x 2 x 1)]

which is 11,480. If we subtract the flops without an ace or king from all flops

(17,296 – 11,480 = 5,816)

we have 5,816 flops with an ace or king. Some of these flops will have a queen also, so let’s count them and subtract them out. Let’s see, we could have one ace or king with two queens.

(A or K, Q, Q) There are 3 aces and 3 kings, and 2 queens remaining in the deck. So, it’s 6 choose 1 (6 / 1 = 6), times 2 choose 2.

[(2 x 1) / (2 x 1) = 1]

That’s (6 x 1 = 6), so there are 6 of them. And we could have two of our cards with one queen. That's (6 choose 2) times 2.

[(6 x 5) / (2 x 1) = 30]

There are 30 of them. And we could have one ace or king with one queen and any other non-ace, -king, or -queen (6 x 2 x 40 = 480), there are 480 of them. So, we subtract the flops with a queen or queens from the flops that have an ace or king.

5,816 – 6 – 30 – 480 = 5,300

That’s 5,312 flops out of 17,296 that have at least one ace or king without a queen. So, the odds are 11, 996 to 5,300 or about 2.26 to 1 against. Okay, okay! I know . . . we originally asked for the chances, and we got odds. Wow, this is a tough crowd! So, the chances are 30.6 percent. Now, this is for Miss Suzy Sunshine. We’ll count the positive events, and get the same result. Our sample space is still 17,296. We have 4,680 flops that have one ace or king and any other two cards that are not a queen, 6 times (40 choose 2).

[6 x [(40 x 39) / (2 x 1)] = 4,680]

There are 4, 680 of them. We have 600 flops that have two of our cards and any other card except a queen, which is (6 choose 2) times 40.

[[(6 x 5) / (2 x 1)] x 40 = 600].

And there are 20 flops that have only our cards, which is 6 choose 3, [(6 x 5 x 4) / (3 x 2 x 1) = 20] Add them all together 4,680 + 600 + 20 = 5,300. Does that number look familiar?

WHAT THE HELL DO YOU MEAN, NO!!!

Okay, I get it . . . you were just kidding. Very funny. Well, there you have it! Two different ways to answer the same question. And, amazingly enough, we got the same answer both ways!

Wasn’t that fun!?!?!

What is Expected Value?

Expected value is the theoretical worth of an event that is repeated an infinite number of times, and it’s the foundation of gambling. The Law of Large Numbers says that if an event is repeated many times the outcome of that event will be about the same as the theoretical probability of that outcome. This allows us to use expected value to evaluate specific options in the play of poker hands, and use it to determine the most profitable option. Many people think of the Law of Large Numbers as the “law of averages.” Although the “law of averages” has become widely accepted outside of academia and the serious study of probability, within academia and the serious study of probability it does not exist. In order to calculate expected value we must be able to establish our sample space, and we must be able to break that sample space down into its relevant events. Once we’ve identified each event, we determine whether it’s a positive event, or a negative event.

Then we establish the value of each event. With that complete, we just simply add together all of the values for every event. If we get a positive number we’re earning money on that bet. If we get a negative value we’re losing money. By dividing the total amount won or lost by the sample space, we get our expected value, or expectation, for each event. That’s the amount we can expect to earn every time we make that particular bet, regardless of whether or not we collect the chips.

This will always be the way to calculate expected value. In poker, as with all other games of chance, we want to have a positive expectation as often as possible.

Expected Value - Flush Example

Here’s an example to introduce you to expected value calculations. The situation is deliberately simplified to make the calculations simple, in order to make the concept easy to understand.

Here’s the situation. You have a four flush on 4th street against one opponent. You each put in one bet before the flop, on the flop, and on 4th street. If you make the flush your opponent will call your 5th street bet, and you will win. If you do not make the flush you will fold. You and your opponent are in the blinds, that way there is no other money in the pot to consider. The question is, what is the value of your hand under these circumstances? For this example your opponent’s cards are not defined, so the sample space is 46. If your opponent’s cards were defined, the sample space would be 44. Of the 46 remaining cards 9 make your flush, and 37 don’t. Let’s use 2-4 limit for this example. The 37 times you miss the flush and lose the pot, you lose 8 dollars. That’s 2 dollars before the flop, 2 dollars on the flop, and 4 dollars on 4th street. For a total of (37 x 8 = 296) 296 dollars lost. The nine times you make the flush and win the pot, you win 12 dollars. That’s 2 dollars before the flop, 2 dollars on the flop, 4 dollars on 4th street, and 4 dollars on 5th street. For a total of (9 x 12 = 108) 108 dollars won. Your expected value is the sum of your 296 dollars lost and your 108 dollars won (296 - 108 = 188), which is a loss of 188 dollars. The 188-dollar loss divided by the sample space of 46 reveals that on average you would lose about 4 dollars per hand.