Chapter 4-The Science Of Dice

One of the oldest ideas is the belief that there is an element of "chance" in the world, that things do happen which are essentially unpredictable. How did this idea come about? In a roundabout way. Primitive people believed "chance" to be the doings of gods and devils. The Greeks of Homer's day felt that "chance" took one or all of three forms:

At first reading, the following few pages may confuse all those who just can't stomach mathematics. But stay with it. Because no other section in this book will help the crapshooter as much as the theory of chance or probability. The material has been adapted from Dr. Horace C. Levinson's book, "The Science Of Chance," (Rinehart & Company, 1950). And we can't present it any simpler than Dr. Levinson already has. Remember, too, that theories of probability came from gambling houses, and there are no better illustrations of the elementary uses of probability than games of chance — like craps.

So let's begin. Consider one of a pair of dice, an ordinary die with six sides. If we ask what the odds are against throwing any one of the sides, say the six spot, you will reply that they are evidently 5 to 1. And your reply will be not only evident but correct.

If we ask you further why you say that the odds are 5 to 1 against the six spot, you will perhaps answer — The die is symmetrically constructed, and if you throw it a large number of times there is no conceivable reason why one of its sides should turn up more frequently than another. Therefore each side will logically turn up on about one sixth of the total number of throws, so that the odds against any one of them, the six spot, for instance, are very nearly 5 to 1.

Or you may say—At each throw there are six possible results, corresponding to the six faces of the die, and since the latter is symmetrically constructed, they are all equally likely to happen. The odds against the six spot are 5 to 1.

Instead of saying that the odds are 5 to 1 against the six spot, we can as well say that the chance of probability of the six spot is 1 in 6, or 1/6.

So far, that's simple enough.

Now, the first of your replies refers to the result of making a long series of throws. What if defines is therefore called a statistical probability. And notice the words very nearly in your first reply.

The second reply refers to a single throw of the die, and therefore appears to have nothing to do with experience. What it defines is called an a priori probability.

Each of the probabilities just defined is equal to 1/6. This is an example of a general law known as the law of large numbers, which tells us that these two probabilities, when both exist, are equal. We can drop the Latin adjectives now and speak simply of the probability.

Now we have the clue we are after. The chance or probability of throwing the six spot is simply the number of throws that give the six (one), divided by the total number of possible throws (six). Thus, a basic probability principle. And if you think that we've used a lot of words to get here—remember, reasoned in such a way, you'll never forget it

So we come to the fundamental definition of probability. The probability of an event is defined as the number of cases favorable to the event, divided by the total number of possible cases, provided that the latter are equally likely to occur.

Don't fret, we'll explain.

This practical rule tells us how to go about finding the probabilities or chances in a large variety of problems, in particular those relating to games of chance.

To illustrate, let's apply this rule to another problem with one die: What is the chance of throwing either a five spot or a six spot?

First list the possible cases. With one die there are always six, the number of its sides, in other words.

Next pick out the favorable cases, which are the five spot and the six spot there are two favorable cases.

The probability we are looking for is therefore the fraction whose numerator is 2 and whose denominator is 6; it is 2/6 or 1/3. Note: Every probability that can be expressed in numbers takes the form of a fraction.

Now let's get into the game and throw two dice as in our game of craps. What is the chance of throwing a total of 7? Well, first we must know the total number of possible cases. In other words, we must count the number of distinct throws that can be made with two dice. Clearly there are thirty-six.

But before going any further, let's give each of our dice a coat of paint, one blue, the other red. Why? Because unless we have some way of distinguishing the two dice from each other — we'll fall into a stupid error as the novice once did — and some still do.

Now to list the favorable cases, those throws that total 7. If we turn up the ace on the blue die and the six spot on the red die, the total is 7, and we have the first of the favorable cases. But we also obtain a total of 7 if we turn up the ace on the red die and the six spot on the blue die. Thus there are two favorable combinations involving the ace and the six spot. Get the idea?

This simple fact, so obvious when the dice are thought of as painted different colors, was once a real stumbling block to the novice with a slow mind. And believe it or not, it still is. If we continue the listing of the favorable cases, we can write the complete list in the form of a short table

BLUE                 RED
  1                          6
  6                          1
  2                          5
  5                          2
  3                          4
  4                          3

Each line of this table indicates one favorable combination, so that there are six favorable cases. As the total possible cases number thirty-six, our fundamental rule tells us that the probability of throwing 7 with two dice is 6/36 or 1/6.

This result is correct, provided that each of the thirty-six possible cases is equally likely. To see that this is indeed the case, we remember that each of the six faces of a single die is as likely to turn up as another, and that throwing two dice once comes to the same thing as throwing one die twice, since the throws are independent of each other.

Let's be elementary again. Craps is played with two ordinary dice. The man who has possession of them wins immediately if the total of the spots on his first throw is 7 or 11. He loses immediately if this total is 2, 3, or 12 but continues to throw the dice. If the total is any one of the remaining six possible points, he neither wins nor loses on the first throw, but continues to roll the dice until he has either duplicated his own first throw, or has thrown a total of 7. The total shown by the dice on his first throw is called the crapshooter's point. If he throws his point first, he wins. If he throws 7 first, he loses and is required by the rules of the game to give up the dice.

You might ask first whether the player throwing the dice has the odds with him or against him. We gave you the answer before, but we'll answer again in another way. As it is well known that when craps is played in gambling houses, the house never throws the dice. And as we know that the odds are always with the gambling house, we may feel reasonably certain in advance — that our calculations will show that the odds are against the player with the dice. Other questions, indeed, can be asked concerning odds in favor of or against making various throws. They will be answered as we go along

3 or 11                                     2/36 or 1/18
4 or 10                                     3/36 or 1/12
5 or 9                                       4/36 or 1/9
6 or 8                                       5/36
7                                              6/36 or 1/6

By this table and the rules of the game, we know that the probability that a player throwing the dice will win on his first throw is 8/36 or 1/9. The chance that he will win or lose on the first throw is therefore 3/9 or 1/3. The probability that the first throw will not be decisive is 2/3.

We have now taken care of the cases where the first throw yields one of the following points—2, 3, 7,11,12. Suppose now that some other total results from the first throw, a 6 for instance. Six, then, becomes the player's point. He must throw a 6 before a 7 in order to win. What are the chances? The chance of throwing 7 is 6/36, and the chance of throwing 6 is 5/36.

Before we answer, let's put in right here an example of reasoning, which well applies in this case and others.

One might be tempted to reason as follows: The ratio of the chances of throwing a 7 to those of throwing a 6 is 6 to 5. Therefore the probability that a 6 will appear before a 7 is 5/11. The probability that 7 will appear first is 6/11.

This is an example of bad reasoning which gives the correct result. It is bad reasoning because, as we have stated it, the conclusion does not follow from the principles of probability. Also, we have assumed that either a 6 or a 7 is certain to appear if we roll the dice long enough.

So let's make this reasoning correct. The player's point is 6. The probability that he will throw a neutral point (all throws except 6 and 7) is 25/36. The chance that he will make two consecutive neutral throws is (25/36)2, and the probability that he will make this throw x times in succession is (25/36) x.

For the game to continue indefinitely, it would be necessary for the player to throw an indefinitely large number of neutral throws in succession. But the probability of doing so is (25/36) x, which becomes smaller and smaller as x increases. Since we can make it as small as we please by taking x large enough, we can legitimately consider the probability that the game will not end as 0.

Read that over again. It's not as complicated as it might sound.

Now, with all neutral throws thus eliminated, there remains to be considered only the throws of 6 and 7. We can now conclude that out of the eleven possible cases that give 6 or 7, five favor the 6 and six favor the 7. Therefore, the probability of throwing a 6 before throwing a 7 is 5/11. Correct result with correct reasoning.

It is easy to make the corresponding calculation for each of the six possible points. The probability is the same for the point 6 as it is for 8, the same for 5 as for 9, and so on, just as in the preceding table. The calculation for each of the possible points gives the following results:

PROBABILITY OF THROWING INDICATED POINT BEFORE THROWING A SEVEN

POINTS                      PROBABILITY
4 (or 10)                      3/9
5 (or 9)                        4/10
6 (or 8)                        5/11

We wish to know the probability, before the first throw in the game is made, that the crapshooter will win on each of the points just listed. In order to win in this manner he must of course, neither win nor lose on his first throw.

We can find what we wish by combining the two preceding tables as follows:

POINTS                                  PROBABILITY
4(or 10)                                   3/36 X 3/9   = 1/36
5(or 9)                                     4/36 X 4/10 = 2/45
6(or 8)                                     5/36 X 5/11 = 25/396

This simply means that the probability that the crapshooter will win on a point specified in advance, say point 5, is 2/45, and the probability that he will win on point 9 is also 2/45.

To find the total probability that the crapshooter will win, we add to his probability of winning on the first throw, which is 2/9, the sum of the three probabilities shown in this table, each multiplied by 2. This gives a probability of 244/495, or 0.49293.

This is the sort of result we expected to begin with. The odds are against the crapshooter, although by only a very small margin. In fact, in the long run, the crapshooter loses (or the gambling house wins, if there is one) only 1.41 per cent of the amounts staked. Compare this to the loss of 2.7 per cent on the numbers in roulette.

A great deal of the interest in the usual crap game comes from the side bets of various sorts. And many of these bets consist in giving odds that one total will appear before another.

The fair odds on all such bets (regardless of what's marked on the average crap table layout), including those totals that cannot be a player's point, by the rules of the game, are shown in the following table.

The table gives the odds against throwing one or the other of the totals given in the left-hand column before throwing one or other of the totals in the top line.

The listing together of two numbers, such as 4 and 10, is done only to abbreviate. The odds as given apply to either 4 or 10, not to both on the same series of throws.

Odds Against Throwing                        Before Throwing
                                3 (or 11)  4 (or 10)  5 (or 9)  6 (or 8)     7
2 (or 12)                  2 to 1       3 to 1        4 to 1      5 to 1     6 to 1
3 (or 11)                                  3 to 2        4 to 2      5 to 2     6 to 2
4 (or 10)                                                   4 to 3      5 to 3     6 to 3
5 (or   9)                                                                  5 to 4     6 to 4
6 (or   8)                                                                                6 to 5

Thus, the correct odds against throwing a 4 before throwing a 6, for instance, are 5 to 3. And so on. So if a player ever says to the shooter, "Two to one you get a 6 before a 4"—and this and like bets are often foolishly made—it's a bad bet. In the long run he will lose 1/24 of the amount staked, as he will in varying degrees with all other such odds given—odds that are out-of-line with the facts above.

As we said at the start, this section on craps is the most important in the book. It will profit you to reread it—maybe many times. In other words, know your odds in the game of craps and expect returns accordingly. Of course, you'll continue tempting "chance" and fooling with "luck"—as many crap-shooters do and always will. But it's the wise gambler, and most often the winning gambler, who keeps such hunches to a minimum. In the long run, the theory of probability runs the game.