Statistical Calculations





Introducing Statistical Calculations

2006 Randy Saylor

In an introductory article, Simple Probability Calculations, I explained how to determine some starting hand possibilities and apply them to a game of poker. This article takes the calculations a step further.

The more accurate way of determining the number of combinations that are possible in card-dealing situations is to use combinatorial mathematics. This is a faster way to come up correct answers in complex situations. The calculation (52x51)/2 is simple enough for determining the number of possible two-card starting hands in Texas Holdem, but arithmetic is insufficient for more difficult calculations.

The proper way to ascertain the number of two-card starting hands using combinatorial mathematics is to use the combination function: C(52,2). This is read as how many two-item combinations can be selected from 52 possible items when order of selection makes no difference?

The method of calculating C(x,y) is:

C(x,y) = ____x!_____
                  (x-y)! y!

where ! stands for the factorial function. The factorial of a number is the multiplication product of that number multiplied by each number one below it, down to one. For example, 5! = 5x4x3x2x1 = 120.

To calculate C(52,2), we plug the numbers into the formula:

C(52,2) = ______52!______
                         50! * 2!

= (52x51x50x49xx4x3x2x1)
(50x49xx4x3x2x1) * (2x1)
removing common factors from the top and bottom of that fraction yields

= (52x51)/(2x1) = 2652/2 = 1326

which is the same result we obtained when considering the number of possible starting two-card hands in Texas Holdem.

Complex Combinatorial Mathematics

Since both methods can be used to arrive at the same result, why would we ever choose the more complicated one? This is because arithmetic breaks down when confronted with more complex probability calculations. The example C(52,2) used above was to demonstrate the method.

Say we need to answer a question that is slightly more complex, like the probability of a single opponent holding two suited hole cards when a flush is made possible by the flop. You are dealt J J in early position. It is folded to you, and you raise. All fold except the big blind, who calls. The flop is T 5 3. You have an overpair, but no straight or flush possibility. The only real way for your hand to improve is by catching another jack. Your opponent may or may not have flopped a flush. What is the likelihood that you are already beat by a flopped flush?

In this situation, there are five known cards, T 5 3 J J. There are forty-seven unknown cards, two of which are held by your opponent. What are the chances of your opponent having two hearts in the hole?

The combination C(47,2) gives us the total possible starting hands held by your opponent out of the 47 unknown cards. This number is 1081. The combination (10,2) gives us the total of the ways that the ten unseen hearts can be dealt in two-card combinations. This result is 45. So of the 1081 possible holdings for your opponent, 45 of them include two hearts. Dividing 45/1081 yields 0.0416, or a 4.16% chance.

If you bet out on this flop and are raised, should you fold? Not necessarily, unless you have a specific read on this opponent as playing super-tight. The only hands that beat you here are AA, KK, QQ, TT, 55, 33, T5, T3, and 53, or any two hearts. Many players will also use a semi-bluff raise here with only a single heart, hoping to win if 1) you fold; or 2) they hit their draw and improve their hand.

So what are the possibilities that your opponent holds a single high heart? We will assume that a player holding only a single heart lower than ten will not draw to the flush. This requires simple multiplication. There are four high hearts (ace, king, queen, jack) among the forty-seven unknown cards. After our opponent gets one of them, there are forty-six other cards that could complete his starting hand, but we cannot count the other nine unseen hearts, since we are considering the possibility that your opponent is drawing to a high flush. 4 * 35 = 140. Dividing this into 1081 shows that there is a 12.95% chance that your opponent has only a high heart and is still drawing.

If you bet the flop, and your opponent raises, they might still be drawing. If the turn card is not a heart, you might try betting out again, since they will no longer have the odds to draw to their flush. If they raise, you can probably lay it down, confident that you are beaten. If they simply call, and the river doesnt help the flush draw, you can probably safely bet. A realistic hand for your opponent in that case might be A T.

Can accurately determining the probabilities associated with specific situations give you the right answers in the heat of battle? Of course not. Tricky opponents and clever plays can always trip you up, but getting a thorough understanding of the possibilities when you are offline (or away from the casino) can help you make better choices when you are in the heat of battle. Learning how to calculate the odds when you analyze your play will help take you a lot further in your advancement as a poker player.