Basic Poker Math
I know that a lot of you are less than thrilled by mathematical dissertations, but it’s all a part of the game and you must have a grasp on at least a few basic principles in order to be successful at Hold ‘em Poker, so please bear with me. I’ll try to make this as simple, easy to understand and brief as possible. If you’re a student of my Blackjack School, you’re hopefully already familiar with the term, “expected value” (EV), but it’s not something you hear about a lot in the poker world. For whatever reason, most poker players, authors, commentators and so forth seem to prefer using “odds” to describe a situation. For example, a particular play may have odds of “4 to 1 against”, which basically means it has a 20% probability of happening.
The terminology of odds have always confused me and because of that, I wanted to teach myself, and you, a quick way of doing calculations in your head, so I’ve decided to go more with probability when calculating EV, rather than odds. I mean, does 5 to 1 odds mean a 16.67% probability or a 20% probability? While there’s not a huge difference between the two, being consistently wrong about how you figure your chances in a given situation will eventually cost you some hard-earned $$$. But for those of you who’d rather deal with odds, let me show you the easiest way to convert probability to odds. Any probability that’s expressed as a percentage can be converted to odds by first subtracting the probability from 100, then dividing the result by the probability. For example, in the case of a 16.66% probability it’ll look like this: 100 minus 16.66 = 83.34 divided by 16.66 = 5.00 or 5 to 1 “against”. In the case of a 20% probabilty, it’ll look like this: 100 minus 20 = 80 divided by 20 = 4.00 or 4 to 1 odds “against”.
What do 5 to 1 odds “against” mean in the real world? Well, it means that for every 6 times you try the whatever you’re talking about, it’ll work once. More confusion, right? The clue for getting a good grasp on this is to add the 5 to the 1 to get 6. Out of 6 attempts, 1 will work, so the odds are 5 to 1 “against.” Isn’t it really just more simple to say you have a 16.66% chance of success? That’s what I’m going to do as I take you through this course, use probability in conjunction with bet size to arrive at EV (expected value, remember?). For example, if your $10 bet has a probability of success of 20%, your EV is $10 x 20% (or 0.20) = $2.00. It’s what we do in Blackjack all the time; a hand of 6,4 versus the dealer’s 7 has an EV of -.476 if you stand (!!!), an EV of +.293 if you hit and an EV of +.406 if you double. It’s just a matter of choosing the highest EV in the play of your hand, so you should double 6,4 vs. 7.
Unfortunately, it doesn’t work exactly that way in Hold ‘em Poker, because your hand is always being compared to the other players’ hands and, as the old saying goes, “Any hand can be a winner in poker”. Rather than measuring the value of a given hand, I’m going to show you how to evaluate the expected value of your bets with the idea that if you make all (or almost all) of your bets in situations where you have a “positive” EV, you can’t help but make a profit. This doesn’t mean you’re going to win every hand, just like there’s no guarantee you’re going to win every time you double 10 vs. a dealer’s 7 in Blackjack. But, if you do it often enough, in the long run you’ll make a profit.
Let me give you a quick example of what I mean. Let’s say that you hold a hand of 10, J offsuit in the “pocket” in a $10/$20 limit game and the flop comes 10, J, 6 (I’m ignoring suits here). You now have Two-pair and, if you choose to play this hand through to its conclusion – two more cards – there is a 16.5% chance that you’ll catch another 10 or Jack, thus ending up with a Full House. Now remember that the math can’t tell you if the Full House you make is a guaranteed winner because another player may have a higher Full House or Four-of-a-kind, etc. when all the cards have been dealt. But, the math can tell you if betting on your Two-pair makes sense. Let’s say all of the pre-flop betting has resulted in a pot total of $60, the bet after the flop comes to you and the pot is now worth a total of $90. Should you make a bet on this hand?
First of all, you have Two-pair, regardless of what happens and that alone may be enough to eventually win, so it has a value of its own, but let’s ignore that for the moment. However, let’s assume that a Full House has nearly a 100% probability of winning the pot, as most Full Houses do. With a 16.5% probability of making a Full House from your hand, the EV of your bet is 16.5% of $90 = $14.85. If the bet you have to make is $10, then you have a definite positive EV and should make the bet. If the bet you must make is $40, it’s not as clear-cut a choice. That’s because players betting after you may or may not add more to the pot’s value, plus you’ll undoubtedly have to make additional bets after the “turn” and “river” cards are dealt. But all we can really do is play our hand one bet at a time, while taking into consideration what other hands are being formed by the other players; don’t forget that the flop, turn and river cards belong to them, too. As we get further into the lessons, I’ll show you how to “read” other players’ hands by how they bet or don’t bet and that will help you in your decision-making process for situations like this where a hand with a positive EV can be suddenly transformed to one with a negative expectation.
Whether or not you make a $40 bet for the hand shown above is immaterial to this situation. What really matters is that you know the probability of making the hand from the flop, forward and you use that to guide your betting. But, and it’s a big “but”, if you choose to make the $40 bet, be aware that it’s probably a negative EV bet at the moment and, if you make them often enough, you’ll eventually lose all of your $$$. I say “probably” because at this point I cannot precisely quantify the value of your Two-pair other than to say that the only hand it beats is a Pair, but that’s often enough to win a pot in Hold ‘em. If we somewhat arbitrarily assign a probability of 20% to the Two-pair winning the pot, then the total EV for that hand is about $33 (20% x $90 = $18 + $14.85), so a $40 bet is a borderline decision at best and a $30 bet seems reasonable. However, a $60 bet would be a real “gamble” and you should know that before you make the call.
Some poker experts like to use “implied odds” when making a decision like this and they want you to figure out how many players will call your bet so the total pot before the next bet comes due can be used to calculate your EV, which they call “pot odds”. Well, that sounds good and is certainly valid if you’re able to predict just who is going to bet and how much they’ll bet. My problems with that concept are many, not the least of which is that it encourages a certain amount of wishful thinking on your part, plus it’s yet another layer of calculation that’s being added to what is already a fairly complex equation. Just as in Blackjack, I prefer to err on the side of conservatism when $$$ are involved, so rather than use implied odds, I prefer to use the odds presented to me as the hand progresses. Let’s continue along and play out the Two-pair we have by making a $30 call after the flop. Now comes the “turn” card and it may well give us our Full House. But, if your luck is like mine, it won’t so we’ll have to face more decisions in betting. (If we made the Full House with the turn card, I’m assuming we’ll welcome and call any bet or more likely, raise the pot for the balance of play.) With the turn card out, we now have to re-evaluate if our hoped-for Full House can still win the pot. Don’t forget that Four-of-a-kind beats a Full House, as does a Straight Flush, so we have to evaluate the impact of the turn card on other players’ hands. It didn’t help us, but it might have helped them.
If you remember, we had a hand of 10, J and the flop came 10, J, 6. Because I’m ignoring suits in this example, let’s rule out the possibility of a Straight Flush, but even if the flopped 10, J were suited, the best anyone could have is a 4-card Straight Flush (called a S.F. “draw”) and the odds are greatly in favor of them making either a Straight or Flush, both of which lose to a Full House, so we can’t spend our time worrying about losing to a Straight Flush. I’ve played thousands of hands and have lost to a Straight Flush only one time. But that little, lonely 6 that came on the flop could be a problem. It’s not inconceivable that some other player has 6,6 “in the hole” and s/he is going to be thrilled to see it, because those Trip 6s will beat our Two-pair if we don’t improve.
But we have set our course and will go forward, although not blindly. By calculating our EV after the flop, we are not done with all of the calculations for this hand, as we would be in a no-limit game where we went “all in”. If a player who has just been passively checking or calling now comes out with a bet or raise after the turn card is dealt, we must take that into consideration when the bet comes to us. In a Limit game where we cannot go “all-in”, which guarantees us to see the last two cards without further betting, we have to – once more – calculate our pot odds to see if it’s a positive EV. Let’s say the the turn is the 5 of spades, a card that probably helped no one, but a player acting before us now bets $20 and the pot is offering us $110 for a $20 call. We still have two-pair, which might be good enough to win the hand, but now – with only one card to come – the probability of making a Full House has dropped to 4 chances of 46 or 8.7%. (See that? We can make our FH by catching one of the two remaining Jacks or one of the two remaining 10s, thus 4 “outs” among the 46 cards we haven’t seen). For a pot at $110, our EV is 8.7% x $110 = $9.57, but we must call with a $20 bet.
But, you may ask, what about the bets we already have in the pot; don’t they have a place in our calculations? The short answer is “no”. Those $$$ are gone, so to speak and we’ll only get them back if we win the hand. Think about it: If we don’t call, they’re lost anyway, so I don’t count our previous bets when calculating EV, only the full value of the pot, thus an EV of $9.57 with only 4 “outs”. You’ll hear that a lot in the poker world; the number of “outs”, so let me take a minute to explain it.
Up to this point in our play, we’ve seen 6 cards; our two “hole” or “pocket” cards and the four community cards on the “board”, three from the flop and the one turn card. That leaves 46 cards unseen and we can only assume, at least for mathematical purposes, that the two Jacks and two 10s that will help us remain in the deck. That, indeed, may not be the case, but we have no way of knowing otherwise unless someone shows us their hand. So it’s just like in Blackjack; if we don’t see it, we don’t count it. Of course, we’re not counting the cards here, so the math is now very simple. Four cards of 46 help us so we have 4 “outs”, or a 2/23 probability of making our hand at this point. Does this mean that the pot now has to be 11.5 times the size of our bet in order for us to call a bet? Not really, because we could have the best hand with two-pair; after all, someone may be bluffing or has a lower two-pair such as 10s and 6s, etc. If this were a no-limit game where we could go “all-in” after the flop, then 6 to 1 pot odds would be satisfactory because no more bets can be made, plus we’re guaranteed to see both the turn and river cards. But in a limit game, we should calculate the pot odds after every card is dealt.
I’m going to give you the percentages of success for making various hands that you may encounter after the flop (5 cards seen), then those same numbers based upon staying with the hand until the end (7 cards seen), but first I want to show you an easy way to check the validity of your bet in the heat of battle, so to speak. If you have a probability of 16.5% in making your Two-pair into a Full House, that means the pot should be at least six times the value of your bet for it to carry a positive EV. Why six? Multiply 16.5% by 6 and you get 99%. A figure of 100% is the threshold of positive expectation, but for me, 99% is close enough because we have some extra EV built into the play due to the possibility of the two-pair winning on their own. Knowing this little trick will allow you to quickly calculate the pot odds in the manner I’ve described above by multiplying the bet times 6 and then comparing that figure to the pot total at the time it’s your turn to bet. That’s very easy to do in a limit Hold ‘em game because of the uniform bet size and not so easy in a pot limit or no-limit game. But for now, we’re discussing limit Hold ‘em, so I won’t confuse the issue.
Let me give you an example of how this works. Let’s say the pot is $90 and you must bet $10, minimum. Well, six times $10 is $60 and the pot is “paying” you $90, so make the bet. Were the pot only $40, you’d be facing a negative expectation of $20 if you make the bet. Conversely, if the pot is, say, $300, you could bet $40 and still have a positive EV. If nothing else, this method of play removes a lot of anxiety from the game; should I call, bet, fold or raise… oh, what to do?
(Source: GameMaster)
