How Long is a Roll? - Part 2

6 February 2000

By Donald Catlin

    Last month we faced the question of determining the average number of rolls in a Craps hand.  Recall that the answer was just over eight and a half rolls.  This month I want to address a related question.  How many rolls does it take for a shooter to have an even chance of sevening out?  Experimentally it would go like this.  A billion Crap shooters all roll at the same time.  How many rolls does it take until half of them have sevened out?  This number is called the median number of rolls; the number we computed last month is called the mean.  As we will see, they are different.

    Let me be candid with you; this problem is a bit harder than the typical problem I address in this column.  Nevertheless, we have the wherewithal to solve it so hang in there and I'll try to break it down into digestible pieces.  Here goes.

    To begin with, we have to address a bit of algebra concerning the logical connectives and and or.  (Note: the logical or was first introduced in my June 1999 article Cardano's Gaff Lives On and the logical and was introduced in my September 1999 article An Ear Full of Cider.)  Specifically, I claim that for three events A, B, and C, the event A and (B or C) is exactly the same as the event (A and B) or (A and C).  This is quite easy to see.  If an outcome is in A and also is in B or C, then it is certainly either in A and B or it is in A and C, that is, it is in (A and B) or (A and C).  The converse is just as easy.  If an outcome is in (A and B) or (A and C), then either it is in A and B or it is in A and C or perhaps both.  In either case, it is certainly in A.  It also follows that it is either in B or it is in C, that is, it is in B or C.  Thus it is in A and (B or C).  In other words we have just argued that an outcome is in A and (B or C) if and only if it is in (A and B) or (A and C).

    To save space, I will often write A & B rather than A and B; they mean the same thing.  The idea in the previous paragraph generalizes as follows:

A & (B1 or B2 or ... or Bn) = (A & B1) or (A & B2) or ... or (A & Bn)

(1)

Now I wish to make an additional assumption about the collection of events B₁, B₂,  ..., B_n, namely, I want it to be a partition of the sample space S.  Recall that this means two things.  First every element of the sample space is in at least one of the sets forming the partition.  Another way to say this is simply that S = B₁ or B₂ or ... or B_n.  This means that the left hand side of (1) can be written as A & S which, of course, is just A since every outcome is in S.  In other words we have the result that if B₁, B₂,  ..., B_n is a partition then

A = (A & B1) or (A & B2) or ... or (A & Bn)

(2)

The second feature of a partition is that the events have to be pairwise disjoint, that is, no pair of them can have any points in common.  In this case the events A & B₁,  A & B₂,  ..., A & B_n are pairwise disjoint as well and this should be obvious to you.  So what?  Well, if we take the probability of both sides of (2) as follows

P(A) = P((A & B1) or (A & B2) or ... or (A & Bn))

(3)

then from pairwise disjointness we can rewrite (3) as (see equation (9) in my June 1999 article)

P(A) = P(A & B1) + P(A & B2) + ... + P(A & Bn)

(4)

Recall from my September 1999 article that

P(A & Bi) = P(A | Bi)P(Bi)

(5)

where P(A | B_i) is the conditional probability of A given the event B_i.  Using (5) we can rewrite (4) in the following form

P(A) = P(A | B1)P(B1) + P(A | B2)P(B2) + ... + P(A | Bn)P(Bn)

(6)

Relation (6) is the fundamental tool we will need for our Craps question.

    A Craps hand can be in one of eight states before each roll.  Either the dice are coming out, one of the six points has been established, or the shooter has sevened out and the game is over.  In an actual game, the dice come out again after a seven out.  But for purposes of our calculation, once the seven out occurs, the game is over and the shooter remains in the seven out state forever; the dice never come out again.  Such a state is called an absorbing state.  With this convention, the eight states mentioned above cover all of the possibilities.  In other words they form a partition.  The game cannot be in two different states simultaneously.  For example, it cannot be the case that a Pass Line point is established and the dice are also coming out.  It cannot be the case that a Pass Line point of 4 and a Pass Line point of 8 are both in effect.  And, of course, with our above convention, the game cannot be in any other state once it is in the seven out state.

    Now, as a Craps hand proceeds the game moves from state to state.  Some of these moves are more likely than others; some are impossible.  For example, it is impossible to move from the coming out state to the seven out state on the next roll; a point must first be established.  By convention it is impossible to move from the seven out state to any other state.

    It is time to introduce some notation.  Here are our eight states:

B₁ = coming out state
B₂ = point is 4
B₃ = point is 5
B₄ = point is 6
B₅ = point is 8
B₆ = point is 9
B₇ = point is 10
B₈ = seven out state

On any particular roll, we will have some probability of being in each of these states, that is, we will know P(B₁), P(B₂), ... , P(B_n).  Suppose, for example, that before the fourth roll we know these eight probabilities.  After the fourth roll these probabilities will change.  How?  That is what relation (6) is all about.  Suppose we want to calculate P(B₁) after the fourth roll and we know the eight probabilities before the fourth roll.  Replace the A in (6) by B₁ and obtain

P(B1) = P(B1 | B1)P(B1) + P(B1 | B2)P(B2) + ... + P(B1 | Bn)P(Bn)

(7)

The numbers P(B_i | B_j) are called transition probabilities and they tell us the probability of the game being in the state B_i after the next roll if it is in the state B_j before the roll.  Where do we get these transition probabilities?  Oh, they're easy to calculate. P(B₁ | B₂), for example, is the probability that if a point of 4 is in effect before a particular roll, then on the next roll the dice will be coming out.  This can only happen if the point is made and the probability of that happening is 3/36.  In other words, on any roll P(B₁ | B₂) = 3/36.  What about P(B₈ | B₂)?  Oh, that is the probability of a seven out on a point of 4 and the probability of that is 6/36.  P(B₂ | B₂)?  Well, if neither the 4 nor the 7 happens on the next roll, and these comprise 9 of the 36 possible outcomes, the point of 4 remains in effect.  There are 36 - 9 or 27 ways this can happen so we have P(B₂ | B₂) = 27/36.  Note that P(B₃ | B₂) = 0; do you see why?  Clearly there are 8 x 8 or 64 different transition probabilities.  A nice orderly way to list them is to write them down in an 8 by 8 array,  P(B_i | B_j) will be in the ith row and jth column.  Here it is:

If you look at the second column you will see the numbers that we calculated above.  In other words, the numbers in that column represent the probabilities that starting in the state 2 (point of 4) that you will end up in each of the states 1 through 8 on the next roll.  Here is how we can use this array in an orderly fashion.

    Let me define the state vector to be the set of numbers:

state vector = (P(B1), P(B2), P(B3), P(B4), P(B5), P(B6), P(B7), P(B8))

(9)

Nothing very deep here.  Before each roll the state vector simply lists the probability that the game is in each of the states 1 through 8.  Then according to (6), to find the probability that the game is in the state, let's say 3, after the roll, we multiply each entry of the 3rd row in the array by the corresponding entry in the state vector and add them all up.  Easy!  Let's do an example.

    Suppose that we walk up to the Crap game and see that a point of 4 has been established.  This means that the game is in the state 2 with certainty.  Our state vector is

state vector = (0, 1, 0, 0, 0, 0, 0, 0)

(10)

When we multiply each number in the state vector by the corresponding number in each row of the array, only the numbers in the second column have any effect because everything else is multiplied by zeros.  So, the next state vector is

next state vector = (3/36, 27/36, 0, 0, 0, 0, 0, 6/36)

(11)

The next state vector is more complicated.  The entry in the first position is:

P(B1) = 3/36 x 12/36 + 27/36 x 3/36 + 0 + 0 + 0 + 0 + 0 + 0

(12)

or

P(B1) = 13/144

(13)

The entry in the second position is

P(B2) = 3/36 x 3/36 + 27/36 x 27/36 + 0 + 0 + 0 + 0 + 0 + 0

(14)

or

P(B2) = 46/144

(15)

The third entry will be 1/3, which corresponds to making the 4 and rolling a 5 on the comeout roll.  I trust you get the idea.  If we start out with a point of 4, then the probability that we make the point in exactly two rolls is 13/144 and the probability that we will still have the 4 up after two rolls is 46/144.

    As you can see, the farther we go the worse the arithmetic becomes.  This is clearly a job for a computer.  It is very simple to write a program that will do these computations for us and that is exactly what I did.  The program starts with the comeout roll so the state vector is:

initial state vector = (1, 0, 0, 0, 0, 0, 0, 0)

(16)

The next state vector will just be the numbers in the first column of the array (8); after that it gets complicated but not for my program.  Now remember our original question; we wanted to know on what roll the probability of sevening out becomes 1/2.  Easy.  We simply keep track of the probabilities for state 8, the seven out state.  Because of the way we formulated things, namely that once in the seven out state the game remains there, the number for the eighth state on any roll represents the probability that the shooter will seven out on that roll or before.  Here are the probabilities for state 8 for the first ten rolls.

Roll	Probability of Seven Out
1	0.00000000
2	0.11111111
3	0.22788066
4	0.33264746
5	0.42387109
6	0.50278913
7	0.57095589
8	0.62980865
9	0.68060930
10	0.72445344

Well, there you have it.  By the sixth roll the chances are just a bit better than even that a shooter will have sevened out.

    There are some other interesting things one can conclude from the program's output.  For example, the probability of a seven out on any particular roll can be obtained by subtracting the number for the previous roll from the probability for that one.  The probability that a seven out will occur on exactly the fourth roll, for example, is 0.10476680.   I'll leave you with this interesting bit.  The probability of a seven out on or before the 49th roll is 0.99913902.  Subtracting this number from 1 will give us the probability that the shooter makes it to 50 rolls; the number is 0.00086098.  Not very likely is it?  See you next month with some easier stuff.

For more information about craps, we recommend:

Beat the Craps Out of the Casinos: How to Play Craps and Win! by Frank Scoblete
The Captain's Craps Revolution! by Frank Scoblete
Sharpshooter Craps Audio Cassette Tape (60 minutes) with Frank Scoblete
Winning Strategies at Craps! Video tape hosted by Academy Award Winner James Coburn, Written by Frank Scoblete