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How Long is a Craps Roll?5 January 2000
In this article and next month's article I want to look at a question that gets kicked around the Craps table now and then. How long is a Craps hand? That is to say, how many rolls does it take until the shooter sevens out? There are a couple of ways to look at this question and they are distinctly different. One way is to ask how many rolls it will take to have a better than even chance of a seven out. The other is to ask how many rolls, on average, a Craps hand will last. The answers to these two questions are quite different and Craps players sometimes get them confused with each other. This month I'll address the latter question, which is the easier of the two questions. Here we go.
What is the average number of rolls that a Craps hand will last? This is the same as asking for the expected number of rolls in a Craps hand (see the discussion of the exp function in my July 1999 article Right question, Right Answer, Hopefully). To be precise, the sample space here consists of all possible Craps hands, an infinite set. If h represents one particular Craps hand in this set, we could denote by n(h) the number of rolls that particular Craps hand lasts. In other words, n is what we called a random variable back in the July article. To calculate exp(n) one would ostensibly have to find the probability prob(h) of each craps hand, form the product prob(h)n(h) for each hand h, and add up all of these numbers for each and every hand h. Whew! An infinite sum, no less. Although this is conceptually the right idea, we cannot possibly carry out such a calculation. What can we do? Here's the plan.
Suppose we could figure out the average number of rolls per Pass Line decision and also the average number of Pass Line decisions per seven out. The number we want, the average number of rolls per seven out, would simply be the product of these two numbers. Fine, but how do we get hold of these numbers? Here's how.
I want to stop for a moment and look at a more general question and one that will be central to our solution of the Craps problem. Suppose that I have a series of independent trials such that each trial results in one of two outcomes; I'll call them success and failure. Remember independence means that no set of outcomes has any influence on any other set of outcomes. Suppose that p represents the probability of success and q represents the probability of failure at each trial. I keep repeating the trials until I get a failure. What is the expected number of trials in this situation? In other words, how many trials can I expect on average until a failure occurs?
Let me give you a simple example. Suppose that the trial consists of rolling a single die. Success will mean that a non-five is rolled and failure will be the occurrence of the five. In this case p = 5/6 and q = 1/6. How many rolls can I expect on average until the five occurs?
Getting back to the general question, let us let e represent the expected number of trials until failure. Note that on the first roll I could get a failure and the number of trials would be 1. This would happen with frequency q. On the other hand, I could have success on the first trial and this happens with frequency p. In this case I would do a second trial. Notice, however, that because of the independence of the trials, when I face this second trial the number of trials until failure I expect from this point is the same as the number I faced at the first trial, namely e. Thus, counting the fact that one trial has already occurred, if I get a success on the first trial the expected number of trials until failure, counting from the first trial, is 1 + e. Remember this happens with frequency p. Weighting 1 (failure on the first) by q and 1 + e (success on the first) by p we get the equation
Equality (4) is the answer to our question. In the die rolling example above where q = 1/6, (4) would tell us that, on average, it will take 6 rolls until a 5 occurs. In other words, if we did this die rolling experiment over and over many times, each time keeping track of how many rolls it took until the 5 appeared, then took the average of all of these numbers, the result would be a number near 6.
Okay, we're ready to roll. First let us calculate the expected number of rolls for a Pass Line decision. To do this we will calculate the expected number of rolls for each of the possible things that will lead to a Pass Line decision and weight each one by the probability that it occurs. If a Natural or Craps occurs on the comeout the, number of rolls is 1; this happens with probability 12/36. Suppose, on the other hand that a 4 is rolled; this happens with probability 3/36. In this case the probability that the bet will be settled on any particular succeeding roll is 9/36 (3 fours and 6 sevens) or 1/4. According to (4) above, interpreting a settlement as failure, it will take an average of 4 rolls to settle the bet. Well, not quite. Remember the comeout roll has to be counted, so the figure is really 4 + 1 or 5. This figure will be weighted by 3/36 in order to calculate the expected number of rolls for a Pass Line decision. The complete set of calculations is given in the table below.
Since 1 = 165/165, the total is (165 + 392)/165 or 557/165. This fraction is approximately 3.376. In other words, if you played many, many Crap games (each decision is a game), kept track of the total number of rolls, and divided the number of rolls by the number of games, the result would be approximately 3.376. According to Stewart Ethier from the University of Utah as quoted in Peter Griffin's book (Extra Stuff: Gambling Ramblings, Huntington Press, Las Vegas, 1991, p. 168), this result was first published in the American Mathematical Monthly in 1909.
Next we want to determine the number of Pass Line decisions per seven out. Think of this as follows. We bet the Pass Line at Craps over and over. When a Pass Line decision is reached, that is considered as one game. We want to keep playing and add the total number of games played until a seven out is reached. In terms of relation (4), A Natural, Craps, or a Point made is a success; a seven out is a failure. So we need to calculate q, the probability that a seven out occurs given that a Pass Line decision is reached. Easy. Just refer to my November article Independence Day where I give the probabilities for Craps. All you have to do is add up all of the probabilities for a seven out on each of the six possible points. The result is 784/1980 or 196/495. According to (4), it follows that the expected number of Pass Line decisions per seven out is 495/196 or approximately 2.5255.
Well now, we have our problem solved. The expected number of rolls, call it enr, in a Craps hand is:
enr = 557/165 rolls/Pass Line decision
Next month we'll look at the harder problem of determining how many rolls it takes to have an even chance of losing the dice. See you then.
This article is provided by the Frank Scoblete Network. Melissa A. Kaplan is the network's managing editor. If you would like to use this article on your website, please contact Casino City Press, the exclusive web syndication outlet for the Frank Scoblete Network. To contact Frank, please e-mail him at firstname.lastname@example.org.
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