An Answer Page for Exercise 19

Exercise 19

When the shooter makes a pass line bet,   what is the average number of times
that he will roll the dice before his pass line bet is resolved ?

To solve this problem,   I created a formula that uses a power series,   which is
too tedious to evaluate without the help of a program.   Perhaps you can find a better solution.

Using a   java program   to evaluate my formula I found that the average number of rolls is about 3.38.

Here is my solution:

Let the random variable X denote the number of times the dice must be rolled
to resolve the shooter's pass line bet.

Note that X can take on any positive integer value, and
we want to find   E   =   the expected value of   X   =   1*P( X=1) + 2*P(X=2) + 3*P(X=3) + ....

[   The notation   P(X=j)   means "probability that X = j"   ]

In order to evaluate this infinite sum, let's begin by looking at a few special cases.

Case 1   The game ends in just 1 roll.
This can only happen when the come out roll is   2, 3, 12, 7, or 11.
The probability for this event is   12 / 36   =   1 / 3.
So,   E   =   1*(1/3)   +   more_terms

Let's temporarily put aside our interest in E and, instead, focus our attention
on how we will compute the probability that exactly k rolls of the dice will
be needed,   where k = 1 or 2 or 3 or ....

Case 2   The game ends after exactly 2 rolls.
This can only happen when the come out roll establishes a point and
the next roll is either that particular point or a   7.

In other words, we'd have to :

Case 3   The game ends after exactly 3 rolls.
This can only happen when the come out roll establishes a point and
the next roll is neither that point nor a 7 and
the third roll is either the established point or a   7.

That requires that we :

Jumping ahead a little bit,   we consider the requirements for case 6.

Case 6   The game ends after exactly 6 rolls.
This can only happen when the come out roll establishes a point and
the next   6 - 2   rolls yield neither that point nor a 7   and
the sixth roll is either the established point or a   7.

Now look at the most general case for which the game doesn't end on the come out roll.

Case k   The game ends after exactly k rolls,   where k > 1.
This can only happen when the come out roll establishes a point and
the next k-2 rolls are neither that point nor a 7 and
the kth roll is either the established point or a   7.
( Note that since k > 1,   k-2 ≥ 0. )

In other words, for exactly k > 1 rolls,   we need one of these 6 scenarios :

We only need to analyze the first 3 scenarios and then use matching probabilities to handle the last 3.

Since P(8) = P(6)   and   P(9) = P(5)  and   P(10) = P(4),   we can simply multiply
the sum of the results of the above 3 scenarios by 2 to get our final result.

The probability for using exactly k rolls is

2 [ (3/36)*(27/36)^{( k - 2 )} *(9/36) + (4/36)*(26/36) ^{( k - 2 )} *(10/36) + (5/36)*(25/36) ^{( k - 2 )} *(11/36) ]

Which equals
(2/36) * [   (27/36) ^{( k - 1 )}   +   (40/36)*(26/36)^{( k - 2 )}   +   (55/36)*(25/36)^{( k - 2 )}   ]

and is valid for k ≥ 2.

By reducing fractions we can simplify this expression slightly to get
(1/18) * [   (3/4) ^{( k - 1 )}   +   (10/9)*(13/18)^{( k - 2 )}   +   (55/36)*(25/36)^{( k - 2 )}

Define the function h by   h(x) = the probability that the random variable X takes on the value x.

Then   h(x)   =   (1/18) * [   (3/4) ^{( x - 1 )}   +   (10/9)*(13/18)^{( x - 2 )}   +   (55/36)*(25/36)^{( x - 2 )}

Refer back to case 1 and see that
E   =   1*(1/3)   +   the sum of all   x*h(x)   values as x runs from 2 thru infinity.

If you've taken some calculus courses then you might have seen the "ratio test",
which can sometimes tell us whether an infinite series converges or diverges.
The ratio test shows that our expression for E must converge.

Since I don't know what the series converges to,   I wrote a java program which
uses several large values for N to compute the sum of the first N terms in the
series portion of the value for E and then adds 1*(1/3) to that,   giving us an estimate for E.

In order to send the results to a text file instead of to my computer's screen,
I used a modified version of the the java program shown at a link near the top of this page.

It took only a second or two to produce these results: