Pre-flop play in PLO 6max. Chapter 1. General

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Pre-flop play in PLO 6max. Chapter 1. General

Here, I continue to publish my research on “Preflop in PLO 6max”.

You will find the foreword here.

Before I get down to reviewing aspects of the game, I will say a few words about the mathematical component of decision-making in poker.

If you can quite easily calculate the probability of at least one of your opponents having a flush draw higher than yours (there is J5s on the flop, and you have 3T of the same suit), then you can just move on to the next chapter. If you feel up to simply believing the author's further mathematical calculations and do not want to overload your brain with even the simplest mathematics, then skip this chapter. You won't lose anything, but it would be useful to return to it once you've completed reading.

 

Contrary to popular belief, success in poker does not necessarily require any mathematical knowledge. A poker player only needs mathematics (or, more precisely, the 'theory of probability' sub-discipline called combinatorics) for the analysis of his own playing, and the theoretical speculation connected with the probabilities formed by finite sets of card combinations.

 

Such a highly-specialized discipline such as combinatorics requires a number of special abilities and tools that significantly exceed the needs of a poker player, as it sets a goal of studying more complicated and deeper issues than simply just calculating hand odds in poker.

 

For this reason I consider even classic combinatorics to be unnecessary in this research. Though players who are familiar with mathematics may dislike both the lack of detailed arithmetic and common terminology, I will sacrifice it for the sake of making this material more accessible to a wider range of poker players.

Instead, I will use a set of simple algorithms that can be used to calculate the necessary probabilities and poker combinatorics. First of all, these are the probabilities of an opponent having a specific card combination. We can limit ourselves to using only simple terminology, and by considering all the necessary problems one by one. The skill of solving these problems will become useful when undertaking further research.

 

Task #1: What is the probability of a certain player's hand having x cards that satisfy a certain parameter (for example, the hand contains two aces, or two cards of a certain suit) ?

 

The solution of this problem (and the base for all subsequent problems) consists of several simple steps.

 

1. We determine the amount of “free” cards in the stack, i.e. the ones that we cannot eliminate the possibility of our opponent having them. If there is no other context to the problem, then we have 52 such cards. If our hand is known, then there are 48 of them. If we also see the flop, there are 45.

 

2. We count how many 'cards of interest' (whose probability of our opponent having them is what we are calculating) are 'free'. If we have one ace in our hand, and we are calculating the probability of our opponent having two aces, then three 'free' aces remain. If we are interested in our opponent's flush draw possibility and there are two diamonds on the flop and one diamond in our hand, then there are ten free diamonds.

These two numbers are the most significant to us.

 

3. Let's number our opponent's cards from 1 to 4. This is done only for the ease of description.

 

4. Let us calculate that the probability that the first card possesses the parameter we are “working out”.

The probability of the first card being of interest to us is calculated by dividing the number of 'cards of interest' by the total number of free cards.

For example, if there are 45 free cards (the flop has been dealt and we know our hand), we are calculating the probability of our opponent's hand containing two aces, and if there is an ace on the flop (or, to put it another way, there are three 'cards of interest' out there in total), then the probability of the first opponent's card being an ace is 3/45, which equals 1/15, or 6.67 percent.

5. Now let us count, using the same method, the probability of the second card possessing that illusive parameter. We have to take into account that there is one free card less than before, as there is also one less 'card of interest', as we are acting with the presumption that the first card is one of the ones we are searching for. In the same way we made the previous calculation, we understand that two free sought cards (aces) remain, and the total number of free cards is 44. So, the conditional probability of the second opponent's card being an ace (conditional meaning that we accept that the player's first card is an ace) is 2/44, or about 4.5 percent.

 

6. Let us now calculate the total probability of the event (that both the first and second cards are the sought ones) by multiplying the probability of the first card being a sought card with the probability of the second card also being a sought card. We use multiplication of the probabilities because the conditional probability reflects that the probability of the second card being a sought card is not true in one hundred percent of cases, but only when the first card of the hand is a sort card as well. In the case of our 'two aces' problem, this probability is [(3*2)/(45*44) =] 6/1980, or 0.3 percent. Please note, that if we are interested in the status of three of the player's cards, we have to repeat steps 5 and 6 again, and if we are calculating the probability of only one card belonging to a certain set for only one card (for example, at least one ace), then these two steps are redundant.

 

7. Now let's pay attention to the idea that the cases that satisfy the statement of the problem are the ones where 'x' player's cards possess the sought parameter, in the case of the current problem where there are two cards:

The two cards could be 1 & 2, or 1 & 3, or 1 & 4, or 2 & 3, or 2 & 4, or 3 & 4. In cases where x=l, then there are four combinations, if x=2 there are six combinations, and if x=3 then there are four combinations. In the context of our problem we can see that we have only analyzed one variant – that cards 1 and 2 are aces. But as there are six possible two-card combinations, we have to multiply the 0.3 percent we calculated in step six, by six. As a result, we end up with 1.8 percent. This then, is the probability of our opponent having two aces if there is an ace on the flop and no ace in our hand.

 

Please note that this probability is only purely mathematical. The way your opponent plays can tell you much more about your opponent's hand than mathematics. Calculations do contain a certain measure of inaccuracy which is caused by the fact that hands of the type AAAx are taken into account several times, but due to the rarity of such hands (you can work this probability out for yourself!) this inaccuracy is relatively insignificant.

In addition, the results arrived at could have been calculated a little easier, but I decided to fully analyze the problem in a way that makes it clear to players who have never dealt with the mathematics before.

 

Working this out is much easier in practice. Here are a couple of examples.

 

Example 1:  How often are hands dealt of the type AAxx, i.e, that contain two aces?

Solution: The probability of the first card in a hand being a certain value is 4/52, or 1/13. The probability of the second card of being of the same type is 3/51. Having multiplied the fractions (1*3/13*51) we get the total of the probability of both the first and second cards as being aces as 3/663, or 0.45 percent. Multiply this number by six (as there are six possible two-card combinations from the four cards in the hand) and we arrive at 2.7 percent when rounded to one decimal.

 

Example 2: This problem is more advanced. The flop is dealt: 3h Jh and a card of another suit. We have 5h and Th in our hand, which is not a bad flush draw. What is the probability that our opponent also has a flush draw, and a higher-ranked one at that?

Solution: The probability of our opponent's first card being a heart is 9/45. The probability of our opponent's second card being a heart as well is 8/44. This makes the total probability as (9*8/45*44), or 3.64 percent. If we multiply this by six (six possible two-card combinations), we arrive at the probability that our opponent also has a flush draw as being 21.8 percent, when rounded to one decimal. This is quite a number! Now let's calculate the probability that one of our opponent's hearts is of a higher value than ours. To beat our ten, our opponent must have an ace, king or queen (the jack is out on the flop). Aside from the ace, king and queen, our opponent could have a 9, 8, 7, 6, 4, or 2 of hearts. To arrive at the probability that our opponent's flush draw is higher than ours, first let us calculate the probability of his cards not being an ace, king or queen. The probability of his first card being one of the small cards listed is 6/9 (there being nine cards not accounted for – two in our hand and two in the flop). The probability of his second card also being one of the small cards listed is 5/8. Therefore, the total probability of both his cards being small is (6*5/9*8) 30/72 or 41.7 percent. The probability that at least one of our opponent's flush cards is an ace, king or queen is therefore (100–41.7=) 58.3 percent. If we multiply this number by the probability of our opponent's flush draw, we arrive at the solution to our problem as being (21.8 percent of 58.3 percent) = 12.7 percent.

 

Task #2. What is the probability of at least one player's hand meeting the requirements of not less than x cards possessing a certain parameter (for example, the hand contains two aces or two cards of a certain suit)?

 

Problems such as this are solved by narrowing down the previous one. First, we calculate the required probability for each opponent separately, as if he is the only one playing against us. After that, we calculate the probability of the required condition not being met. Multiplying this probability by itself several times – as many times as the number of opponents – we get the probability of the requirement not being met by any of our opponents. Finally, if we subtract the number we arrive at from 1 (if we are using fractions) or 100 (if using percentages), we arrive at the probability of at least one opponent meeting the required condition.

 

Example 3: What is the probability of at least one of six players being dealt a hand that contains two aces.

Solution: Based on our results from Example 1, the probability of at least one of our opponents not having an ace is (100–2.7 =) 97.3 percent, or 0.973 in decimals. If we multiply this number six times by itself (i.e. 0.9736 = 0.973 *  0.973 * 0.973 * 0.973 * 0.973 * 0.973) we arrive at 85 percent (rounding up to the nearest whole number). This means that the probability of at least one player having two aces will be equal to (100–85 =) 15 percent.

 

Example 4: The conditions are the same as in Example 2, but there are three opponents. 

Solution: The probability of one of our opponents not having a flush draw or having a lower one is (100–21.8 percent, from Example 2) 87.2 percent. The probability of none of the players having a higher flush draw is (0.8723 = 0.872 *  0.872 * 0.872) = 66.5 percent. The answer therefore is that the probability of one of our opponents having a higher flush draw is 33.5 percent.

 

Task #3. Calculate the probability that the three cards dealt on the flop will meet required conditions.

 

As this problem is very similar to those given previously (the flop substitutes the role of one player), we will only work through one basic example.

 

Example 5: We were dealt T♠T♦3♣3♥ as our hand. We are seeking the probabilities of completing a set on the flop.

Solution: We need to calculate the probability of at least one card on the flop of being either a three or a ten. There are therefore, four sought cards. The total amount of free cards is 48. Let us calculate the probability of a ten or a three not arriving on the flop. The number is equal to (44/48) * (43/47) * (42/46), or 79,464/103,776, or 76.6 percent. Subtracting that from 100, we arrive at 23.4 percent.

This ends our brief review of the type of problem-solving necessary for further learning of pre-flop play. Later on we will learn to solve more complicated problems, but as the same principles are used for solving them, they will not seem so difficult and deserve a separate chapter of reviews.

 

Next article: 

Chapter 2 (Introduction + UTG-AAxx hands)