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Best of Donald Catlin  # Why Dealer Qualification Works

2 May 2004

Many of the newer casino games have a feature called dealer qualification. The notion of qualification per se has been around for a long time, for example, Jacks or better in Poker or the dealer in Blackjack requiring a hand of 17 or better. As far as I know, dealer qualification as implemented in the way we describe below, was first seen in Caribbean Stud and Three Card Poker. Since the advent of these games, other games have come along with this same feature. In this article I will explain how dealer qualification works and construct a simple mathematical model to illustrate why it works. Those of you who have wanted to see more mathematics in this column should be happy. The rest of you should not despair though; a quick review of your ninth grade algebra book (quadratic expressions) will be all that you'll need.

Let us suppose that we have some casino table game in which the basic structure of the game gives the player a fair shake or even a slight edge. The game with the player having a slight edge is the more interesting of the two, so we will assume that. Since we are going to assume that all hands are equally likely, this advantage will have to come about because of some condition such as the player taking ties or some other player advantage in the rules. If the player's probability of beating the dealer is p and the probability of getting beaten by the dealer is q then the player's expected return per unit wagered is just

 e = p - q (1)

So, we are assuming that e is positive. Naturally, any casino would be crazy to offer the game in that form so here is how they can modify it.

The player makes a one-unit ante. The player then receives his hand. At this point the player often has the option of folding the hand. The issue of when the player should fold is a future topic. Right now we are interested in the situation in which the player doesn't fold. If the player decides to play the hand, then he is required to bet an additional unit, sometimes more than one unit, to stay in the game. We will look at the case where the additional bet is equal to the ante.

The dealer's hand is then dealt. If the dealer's hand is less than a previously specified hand, called the dealer's qualifying hand, the player's additional bet is returned and he is paid even money on the ante. Otherwise, the dealer is said to qualify and the player and dealer hands are compared. The settled bet is either a two-unit win for the player or a two-unit loss for the player.

I am going to make a simple mathematical model of the above game in order to see how this qualifying condition affects things. Will the model be exactly right? No, it won't be. For one thing I am going to make a continuous model rather than a discrete model. Also, I am going to make the simplest assumptions I can that are, nonetheless, plausible and produce the kind of features that an exact model would have. I'll be careful to point out these assumptions to you as we proceed.

We start by assuming that the "hands" in this game are represented by real numbers on the unit interval [0,1]; 0 is the worst hand one can get and 1 is the best hand that one can get. In Three Card Poker, 0 would be the unsuited 2,3,5 hand and 1 would be the suited Ace, King, Queen. We further assume that each point on the interval [0,1] is as likely as the next. Notice that this is not exactly the way a real game would work. In Three Card Poker there are 60 ways of choosing an unsuited 2,3,5 hand and only 4 ways of choosing a suited Ace, King, Queen hand. But my purpose here is not to study Three Card Poker or any other specific game with dealer qualification. Rather, I want to see the mathematical underpinnings of dealer qualification to see why it works the way it does. The best chance of seeing the dynamic here is to consider a game in which all of the outcomes are equally likely because that is the simplest situation one can have.

Very well, let us suppose that x represents the qualifying hand. If the dealer's hand is less than x then the dealer doesn't qualify. Notice that under our assumptions the probability of this happening is just x since that is the length of the interval from 0 to x. Similarly, the probability of the dealer qualifying is 1 - x.

Now, if the dealer qualifies, let us let px represent the conditional probability that the player beats the dealer and qx represent the conditional probability that the dealer beats the player. Certainly px and qx ought to have something to do with p and q respectively but are not necessarily equal to them. If x = 0 then certainly they are equal, but what if x were very close to 1? This would mean that the player would be playing against a very good dealer hand. As a result, one would expect that px would be very close to 0 and qx would be very close to 1. A simple model that would give us this feature is as follows:

 px = p(1 - x) (2)

Since the condition qx + px = 1 must be satisfied, it follows from (2) and the fact that p + q = 1 that

 qx = q + px = q + (1 - q)x (3)

With the above assumptions we can now calculate the player's expected return in this game. If the dealer fails to qualify (probability x) then the player receives one unit. If the dealer does qualify (probability 1 - x) then the player risks two units with the probabilities of winning and losing of px and qx respectively. Hence

 exp = x + (1 - x)(2)( px - qx) (4)

If we substitute (2) and (3) into expression (4) and rearrange the resulting terms we obtain

 exp = x + (1 - x)(2p - 2q - 2x(p - q) -2x) (5)

Noting from (1) that e = p - q the expression in (5) is simply

 exp = x + (1 - x)(2e - 2ex - 2x) (6)

If we carry out the indicated multiplication in (6) and collect like terms we have

 exp = 2(e + 1)x2 - (4e + 1)x + 2e (7)

There are several things we should note about the expression in (7). First of all, if we substitute x = 0 into it, meaning the dealer always qualifies, then the player's expectation is just 2e. That is what we would expect. If we substitute x = 1 into this expression, meaning that the dealer never qualifies, then the player's expectation is 1. Again, this is what we should expect. Finally, we note that the expression in (7) is a quadratic. If we were to set y equal to the expression on the right of (7) and plot the graph it would be a parabola that opens upward. If you are unfamiliar with that term, let me just say that it is a U-shaped curve with the sides of the U pointing outward rather than just straight up. The major question for us, however, is where is the bottom of the curve? Is it above or below the x - axis? Why do we care? We care because if it is above the x - axis, then the house can never obtain an edge; the player's expected return would always be positive.

The following reconfiguration of (7) takes a bit of algebra, so I'm just going to carry it out for you and report the result. If you would like to try it yourself the important buzz words are 'completing the square'. You can check to see that it is correct, if you wish, by carrying out the indicated squares and multiplications and recovering (7); otherwise you'll have to take my word for it. Here is the result:

 exp = 2(e+1)[x-(4e+1)/(4e+4)]2 - (1-8e)/(8e+8) (8)

Notice that if we set x = (4e + 1)/(4e + 4) then the first (squared) term is zero and we are just left with an expected return of -(1-8e)/(8e + 8). This number is negative as long as 8e is less than 1 or, in other words, as long as e is less than 0.125. If e = 0.02, for example, then for x = 0.265 we would have a house edge of around 10.3%. Of course, optimal folding could cut this number some and we'll look at this in a later article.

The model shows us that as we increase x from zero the player's edge drops from e and eventually becomes zero. Past this point, as we keep increasing x the player's edge reaches a low point and then begins increasing. There comes a point where the player's edge becomes positive and eventually when x is 1 the player's expected return is 1 (100%).

Actual games behave in this fashion. Although the graph of a player's expected return in an actual game would not be a parabola, it would look a lot like one. I can tell you from experience that as one raises the qualifying criterion from the lowest hand, any advantage the player may have had decreases until the house has an advantage. As the qualifying hand is raised the house edge increases - up to a point. Then the house edge starts to drop. If one raises it too much the game becomes favorable to the player. Just as in our model, if the player's edge is initially too high the house can never obtain an edge. I recently worked on developing a game that, for awhile, had a player edge in the 12% range. Try as I could I couldn't obtain an edge for the house using dealer qualifying; now I see why.

See you next month and I promise - no algebra!

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Best of Donald Catlin
Donald Catlin Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers
Donald Catlin
Don Catlin is a retired professor of mathematics and statistics from the University of Massachusetts. His original research area was in Stochastic Estimation applied to submarine navigation problems but has spent the last several years doing gaming analysis for gaming developers and writing about gaming. He is the author of The Lottery Book, The Truth Behind the Numbers published by Bonus books.

#### Books by Donald Catlin:

Lottery Book: The Truth Behind the Numbers