Getting Drawn Out in Holdem Vs. Stud
The Author Takes a Controversial Stand Against the Widely Accepted View That it is Easier to Get Drawn Out on in Stud
(The following is an original article from play-video-poker-slots.com)
As you all know, if you've read some of my other articles on play-video-poker-slots.com, I have played a significant amount of Hold Em and Stud over the past couple of years and, as I review my notes, it really seems like I get drawn out on much more often in Hold Em than I do in Stud. I never really dwelt much on it other than to chalk it up to my more than normal distribution of bad luck. Lately, however, I have been focusing more on a possible reason for this and I believe I have come up with one, or at least, a somewhat viable explanation.
Analyzing this apparent dichotomy should also lend itself to answering another more common question that comes up in poker literature and it is one Mason Malmuth has conjectured on before. That is, "What game is more profitable for a professional player, Hold Em or Stud?" I believe this was the title of a Poker Digest article Mason wrote a while back and I believe his conclusion was that Stud is more profitable. Unfortunately I do not have this article handy, nor do I recall his exact reasons, but I believe I have come up with another reason why Stud may be more profitable for the pro (that I do not believe Mason touched upon) and I believe I can back it up with some simple mathematical analysis that will also shed some insight into answering the question I posed in the title of this section.
Drum roll please. No, it does not just seem like we all get drawn out more in Hold Em than Stud because we actually do - for certain types of hands. I realize my views on this matter go against the traditional thinking that it is easier to get drawn out on in Stud because there are more streets, but I stand behind my analysis. Now for an explanation and some simple math - please don't run away. Also, please realize I do NOT plan on attempting to infringe on Mr. Sklansky's role as foremost poker theorist and mathematician. Let's take a ridiculously simplified example to prove my point mathematically.
Let's assume that we are playing Hold Em, we have pocket Aces
and our opponent has pocket deuces
and we are down to the last card to be dealt. Let's also assume that (hypothetically) there are only four cards left to be dealt, two of which are deuces and two of which help neither hand (I realize in most situations there will be many more cards that help neither hand and, thus, the Aces win a much greater percentage of the time).
For illustration purposes, let's suppose there are no chances where both hands could improve, so the only outcomes are: A) either a deuce is dealt out of the remaining four cards and the pocket deuces improve to a set and win or B) nobody improves and our Aces hold up. First, let's analyze (given this scenario) how often the deuces improve and win. With four cards left, two of which are deuces, another deuce appears 2 out of 4 times or this condition has a .5 (50 %) probability of occurring. In other words, the deuces outdraw our Aces, in this Hold Em example, 50% of the time (we are also assuming the remaining Aces are gone). Now I know I am intentionally oversimplifying, but I am just trying to demonstrate something mathematically.
Let's now assume similar conditions, but that we are playing Stud. Again, for illustration purposes, we shall simplify the scenario. Again, we have Aces versus our opponent's deuces and again there are only four cards left to be dealt. We again assume that two of the remaining four cards are deuces and that the single pair of deuces must catch one of the remaining deuces to win the pot. Also, there are no other Aces remaining. No other hands are possible and there are no cards that allow both hands to improve. Just as in the example above, there are only two outcomes: A) either a deuce is dealt out of the remaining four cards and the pocket deuces improve to three-of-a-kind and win or B) nobody improves and our lone pair of Aces holds up and takes down our imaginary pot.
Just as before the chance of a deuce being dealt out of the remaining cards is 2 out of four or there is a probability of .5 (50 %) that this will occur. However, in Stud, unlike in Hold Em, two events must occur (i.e. in mathematical terms, two conditions must be TRUE) in order for our deuces to draw out on our Aces and win. One of the deuces must be dealt out (.5 probability) AND it must land on our opponent's hand. With two separate hands being played there is also ONLY a .5 (50 %) probability of this second condition occurring. When two events must occur to produce a given result, it is said that these events must be ANDed together (e.g. you multiply the individual probabilities of each event occurring, together). In mathematical terms, the probability of an event in which it is necessary for two other conditions to be true is determined by multiplying the probabilities of the individual events together. In our Stud example, one of the remaining deuces must be dealt out of the four remaining cards (.5 probability) AND the deuce must land in our opponent's hand (.5) probability.
In the Stud example, then, the chance of the deuces drawing out on our Aces is ½ times ½ (if we express our probabilities as fractions). One-half times one-half is one-fourth. In Stud, the deuces would only draw out on us one-quarter of the time, or 25% of the time, whereas in Hold Em, the deuces would draw out on us one half of the time, or 50% of the time in our simplified example. So you can see that it not only seems as if we are getting drawn out more in Hold Em than Stud, we actually are and there is a mathematical explanation for it. This is another reason why I believe the professional poker player is better suited to sticking with Stud. The best hand holds up more often and the result is less fluctuation to our bankroll.
I realize that this was a simplified example and did not take into account my opponent's abilities or situations where both/neither hands may improve, etc. However, I do not think the math can be disputed. In conclusion, in Stud hands, cards not only have to "appear", they need to appear in the correct "spot". As such, there are more events which must occur than in a Hold Em hand where the right card just needs to "appear" since both hands share the community cards. In situations where multiple conditions must be true, the likelihood of an event that consists of these multiple conditions is less than the likelihood of an event that is comprised of a single condition. Hope that wasn't too much for you!
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