An Answer Page for Exercise 18

Exercise 18

  How long can a craps game last ?
Fill in the blanks with valid integer values:
    When the come out roll establishes a point of     ____   or   ____  
    there is less than a one percent chance that an additional   ____   or more rolls
    of the dice will be needed to decide the outcome of the shooter's line bet.
 
  Suggestion:     Simply expand the results obtained in solving exercise 17.



Case 1   The shooter's point is 4 or 10

To be specific, assume the shooter's point is 4.

Suppose the shooter rolls the dice N additional times after the come out roll.

Each roll of the dice can be labelled as either a "success" or a "failure",
depending on whether or not the roll determines the outcome of the shooter's line bet.
For this case:

  success     means   rolling either 4 or 7  
  failure     means   rolling any number except 4 or 7  

We can easily compute the probability of getting N consecutive failures.

On each roll we have:
P( failure )   =   1   -   [ P(4) + P(7) ]   =   1   -   [ (3/36) + (6/36) ]   =   3 / 4

Since the dice rolls are independent events, the probability of not getting a 4 or a 7 in N consecutive rolls is ( 3 / 4 )N       for N = 1, 2, 3, ....

Here is a table showing the probability for never getting a 4 or a 7 in N consecutive rolls of the dice.

N consecutive failures when the point is 4
Rolls 1 thru N Probability of not getting   4 or 7   in N consecutive rolls
N = 1 0.751   =   0.75
N = 2 0.752   =   0.5625
N = 3 0.753   ≈   0.4219
N = 4 0.754   ≈   0.3164
N = 5 0.755   ≈   0.2373
N = 6 0.756   ≈   0.1780
N = 7 0.757   ≈   0.1335
N = 8 0.758   ≈   0.1001
N = 9 0.759   ≈   0.0751
N = 10 0.7510   ≈   0.0563
N = 11 0.7511   ≈   0.0422
N = 12 0.7512   ≈   0.0317
N = 13 0.7513   ≈   0.0238
N = 14 0.7514   ≈   0.0178
N = 15 0.7515   ≈   0.0134
N = 16 0.7516   ≈   0.01002
N = 17 0.7517   ≈   0.0075
N = 18 0.7518   ≈   0.0056

Scanning down the 2nd column of this table, we see that the probability for getting consecutive failures doesn't fall below 1% until we reach N = 17.

So, when the shooter's point is 4 or 10,   there is less than a 1% chance that 17 or more additional rolls of the dice will be needed to resolve the shooter's line bet.

 

Case 2   The shooter's point is 5 or 9

To be specific, let's assume that the shooter's point is 5.

Define "success"   and "failure" by:

  success     means   rolling either 5 or 7  
  failure     means   rolling any number except 5 or 7  

On each roll we have:
P( failure )   =   1   -   [ P(5) + P(7) ]   =   1   -   [ (4/36) + (6/36) ]   =   13 / 18   ≈   0.7222

N consecutive failures when the point is 5
Rolls 1 thru N Probability of not getting   5 or 7   in N consecutive rolls
N = 1 0.72221   =   0.7222
N = 2 0.72222   =   0.5216
N = 3 0.72223   ≈   0.3767
N = 4 0.72224   ≈   0.2721
N = 5 0.72225   ≈   0.1965
N = 6 0.72226   ≈   0.1419
N = 7 0.72227   ≈   0.1025
N = 8 0.72228   ≈   0.0740
N = 9 0.72229   ≈   0.0535
N = 10 0.722210   ≈   0.0386
N = 11 0.722211   ≈   0.0279
N = 12 0.722212   ≈   0.0201
N = 13 0.722213   ≈   0.0145
N = 14 0.722214   ≈   0.0105
N = 15 0.722215   ≈   0.0076
N = 16 0.722216   ≈   0.0055

When the shooter's point is 5 or 9,   there is less than a 1% chance that 15 or more additional rolls of the dice will be needed to resolve the shooter's line bet.

 

Case 3   The shooter's point is 6 or 8

To be specific, let's assume that the shooter's point is 6.

Define "success"   and "failure" by:

  success     means   rolling either 6 or 7  
  failure     means   rolling any number except 6 or 7  

On each roll we have:
P( failure )   =   1   -   [ P(6) + P(7) ]   =   1   -   [ (5/36) + (6/36) ]   =   25 / 36   ≈   0.6944

N consecutive failures when the point is 6
Rolls 1 thru N Probability of not getting   6 or 7   in N consecutive rolls
N = 1 0.69441   =   0.6944
N = 2 0.69442   =   0.4823
N = 3 0.69443   =   0.3349
N = 4 0.69444   =   0.2326
N = 5 0.69445   =   0.1615
N = 6 0.69446   =   0.1122
N = 7 0.69447   =   0.0779
N = 8 0.69448   =   0.0541
N = 9 0.69449   =   0.0376
N = 10 0.694410   =   0.0261
N = 11 0.694411   =   0.0181
N = 12 0.694412   =   0.0126
N = 13 0.694413   =   0.0087
N = 14 0.694414   =   0.0061

Scanning down the 2nd column of the above table, we see that the probability for getting consecutive failures doesn't fall below 1% until we reach N = 13.

Summary:

  When the come out roll establishes a point of      4   or   10  
  there is less than a one percent chance that an additional   17   or more rolls
  of the dice will be needed to decide the outcome of the shooter's line bet.
 
  When the come out roll establishes a point of      5   or  9  
  there is less than a one percent chance that an additional   15   or more rolls
  of the dice will be needed to decide the outcome of the shooter's line bet.
 
  When the come out roll establishes a point of      6   or    8
  there is less than a one percent chance that an additional   13   or more rolls
  of the dice will be needed to decide the outcome of the shooter's line bet.
  point is 4 or 10 → rarely [ less than 0.75% of the time ]   need more than 17 additional rolls
  point is 5 or 9   → rarely [ less than 0.76% of the time ]   need more than 15 additional rolls
  point is 6 or 8   → rarely [ less than 0.87% of the time ]   need more than 13 additional rolls