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The game of craps is a very popular casino game, and relatively easy to play.  This makes it amenable to analysis using the laws of probability, to determine the likelihood that a player wins the game.  One version of the game of craps is played as follows:

A player rolls two dice, and the sum of the numbers rolled is obtained.  Now:

1.  If the sum is 7 or 11, the player wins immediately.

2.  If the sum is 2, 3, or 12, the player loses immediately.

If any other sum is obtained, the original sum is noted and the following procedure is followed to finish the game.  (Of course, in an actual casino game, bets on the outcome may be placed before the initial rolls of the dice, and possibly again at this stage and in subsequent stages).

The two dice are rolled repeatedly until either the original sum is achieved again, or the sum of the two dice is 7.  Now:

3.  If the original sum is achieved before 7 is achieved, the player wins.

4.  If 7 is achieved before the original sum, the player loses.

So, the question is, what is the probability that a player will win the game?

To solve this problem, we proceed in stages.  First, let us sort out and write down the number of ways that the player can win the game.  We will define one event for each of these ways of winning.  From the specification of the way this game of craps is played, we define.

Let A be the event that the player wins by throwing a score of 7 or 11 (step 1)

Let B be the event that the player wins by initially throwing a score of 4, and then wins by throwing a score of 4 once again before a 7 is scored.

Let C be the event that the player wins by initially throwing a score of 5, and then wins by throwing a score of 5 once again before a 7 is scored.

Let D be the event that the player wins by initially throwing a score of 6, and then wins by throwing a score of 6 once again before a 7 is scored.

Let E be the event that the player wins by initially throwing a score of 8, and then wins by throwing a score of 8 once again before a 7 is scored.

Let F be the event that the player wins by initially throwing a score of 9, and then wins by throwing a score of 9 once again before a 7 is scored.

Let G be the event that the player wins by initially throwing a score of 10, and then wins by throwing a score of 10 once again before a 7 is scored.

Events B to G follow from steps 3 to 4 of how to play the game.  Of course, the throwing of a score of 7 or 11 are not included in these events, since throwing a 7 or 11 wins immediately on the first throw.  Finally, let us define the event W that the player wins by any of the means available to him.

Now, The events A to G each consist of a single winning outcome, and each of these winning outcomes are disjoint (i.e. there are no winning outcomes that belong to more than one of the events A to G).  Therefore the intersections of any two or more of the events A to G is the empty set, and so

We can evaluate each of these probabilities in turn.

The event A is that a 7 or 11 is thrown on the first throw of the dice.  To calculate the probability of event A, we note that of the 36 possible ways that two dice can land, 6 of them result in a sum of 7, and 2 of them result in a throw of 11.  Thus there are 8 outcomes that result in a win on the first throw of the dice, and so

The event B is that a score of 4 is thrown on the first throw of the dice, and then that a score of 4 is obtained before a 7 on subsequent throws of the dice.  This can be considered to be the intersection of two events X and Y, X being the probability that a 4 is thrown on the first throw, and Y being the event that a 4 is thrown before a 7 on subsequent throws.

In the article on Bayes’ theorem, it was shown that for independent events

So, we need to evaluate the probabilities of X and Y.

Since there are three ways of throwing a total score of 4 from two dice, we have

To evaluate the probability of Y, we note that the problem can be restated in the following terms:  what is the probability of throwing a score of 4, given that either a 4 or a 7 is achieved on the roll of the dice.  No other scores are of interest.  We are only interested in those throws of the dice that result in a score of 4 or 7.  If a score of 4 is obtained, then the criterion of throwing a 4 before a 7 is achieved.

Hence, in the notation of conditional probabilities, we can define three more events (!) P, Q and R, such that P is the event that a score of 4 is obtained, Q is the event that a score of 7 is obtained, and R is the event that a score of 4 or 7 is obtained.  Then, from Bayes’ theorem,

where the last step follow because the conditional probabilities on the right hand side of this expression are clearly unity.  Thus, since there are three ways of scoring a 4 and six ways of scoring a 7, we have

We can now finally evaluate the probability of the event B:

That is, the probability of scoring 4 on the first roll of the dice, and then a score of 4 before a 7 on subsequent rolls, is 1/36.

The probabilities of the events C, D, E, F and G can be evaluated in a similar manner.  It is not hard to show that:

Thus the total probability of winning the game is

That is, the probability of winning is just less than 50%.  The odds just favour the house when this version of craps is being played.

An alternative version of this game is that steps 3 and 4 at the beginning of the article are changed such that the player must achieve the original throw before a 7 or 11 is achieved.  In this case, following the same logic as above and noting that there are two ways of throwing a total score of 11 when rolling two dice, the probabilities of winning via events B through to G become:

Therefore the total probability of winning is

Therefore, in this version of the game, the odds favour the house to a greater extent.

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