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Backgammon and Probability TheoryBackgammon is a game of chance. For this reason, the optimum strategies depend on probability. The reader has noticed that the entire book is sprinkled with plays whose explanation is based on some probabilistic argument.
In this article we shall present a fundamental discussion of the probability model of backgammon games. The reader is advised to read it carefully, and to try to understand it fully. When two opposing players are equivalent in skill, the player who is most capable of figuring the best play probabilistically, has the best chance of winning the game.
The probability model of a chance phenomenon has the following two components: 1. A list of all the possible outcomes that can take place. These outcomes are called the elementary events (e.e.).
2. The probability of each elementary event as a number between zero and one.
Let us see if we can find these two components for the backgammon games. These games are completely controlled by the throw of two dice. Let us assume that one of the dice is black and the other is white, an assumption that enables us to find the e.e.s a little more easily.
What are all the possible outcomes that can occur when one of the two players throws the dice? We enumerate them by means of an ingenious graph in two dimensions. This graph is shown in Figure below. The horizontal axis in this figure stands for the outcomes of the black die, whereas the vertical axis stands for the outcomes of the white die. There are 36 points in this graph, each of which represents an e.e. for the backgammon model. The notation (W-l,B-2) is used to indicate the e.e. 'The white die is one and the black die is two." Look very carefully at this graph, which is basically a list of all the elementary events. Note, for example, that there is only one e.e. for a specific doublet throw. The throw of (6,6) is represented there by only one point, i.e., when both dice are six. On the other hand, all other nondoublet throws are represented by two e.e.s. For example the throw of (6,2) can take place either as (B-6, W-2) or as (W-2.B-6). This point is essential for our discussion and must be com pletely understood by the reader. Now that we have found all the e.e.s, our next concern is to assign a probability to each one of them. Using the method suggested by the mathematician Laplace, we assign a probability of 1/36 to every point in the graph, i.e., to every elementary event of the model. This is because the chance of throwing any one of them is one out of thirty-six. Naturally, this assignment is made with the assumption that the dice are an honest pair. This completes the probability model of the dice throw. This model can now be used to compute the probabilities of various plays that occur in the four games described in this book. Let us see how this is accomplished.
First we must understand the meaning of the term event. An event is some occurrence of the modeled probabilistic phenomenon that is made up of elementary events. For those who know set theory (now taught in elementary schools), an event is a set. Thus, for example, the throw (2,1) in a backgammon game is an event, since it is made up of the two elementary events (B-2,W-1) and (W-l,B-2). All the situations that arise in a given backgammon game and whose probabilities we desire are events of the probability model. For this reason, the next important item to consider is how to use the probability of the e.e.s to compute the probability of such events. For a model such as ours, the rule for doing this is very simple. It is stated as follows. RULE 1: The probability of an event is the sum of the probabilities of all the e.e.s that make up this event.
This is all that is needed for computing the probabilities of various situations that arise in backgammon games. If the reader has understood the two components of the model, the meaning of the term event, and Rule 1, he should have little difficulty in choosing the best play possible in most common situations. All that he has to do is find out how many e.e.s his event has, and add up their probabilities.
1. Probability of a specific doublet: This is 1/36, since each doublet is an elementary event. 2. Probability of any doublet: The event "any doublet" is made up of the six e.e.s (B-1.W-1), (B-2,W-2), (B-3,W-3), (B-4,W-4), (B-5,W-5), and (B-6,W-6), which are represented by the points on the diagonal in The Figure Using Rule 1 we find the probability of this event to be 1/6. This type of probability is useful in the final stages of a backgammon game. If your opponent is bearing off faster than you are, you will need a throw of doublets to win the game. Naturally, this play would also depend on what type of doublets you might need. But now we are sure that you can figure the appropriate event and its probability yourself. 3. Probability of a sum throw greater than six: Often we need to know the probability that a piece can reach a spot that is greater than six positions away. For example, what is the probability that the dice throw will have a sum total of seven? Well, this event consists of the e.e.s (B-5.W-2), (B-2,W-5), (B-6,W-1), (B-l.W-6), (B-4,W-3), (B-3,W-4). Therefore its probability must be 1/6.
4. Probability of a specific number: This is a little more difficult to compute. The reason is that if the number is greater than six, it can only occur as a sum, whereas if it is less than six, it can occur both as a sum and as a specific number on the face of a die. For example, if you want the probability of a six, you have to find all the e.e.s that will produce it either as a sum or not. Thus, aside from all the e.e.s that have a six in them (eleven of them in Figure 107) you must take into account e.e.s like (B-4.W-2), (B-5,W-1), (B-3,W-3), etc. The probabilities of specific numbers are important in hitting a blot, capturing a prisoner (plakoto), or making a point (moultezim). For this reason we give them all as a table below.
Table A.l Probabilities of Numbers
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