Probability in Craps Game

Date: 6/27/96 at 22:44:45
From: Anonymous
Subject: Probability in Craps Game

Dr. Math,

    In the game of craps, the shooter rolls two dice and wins if the 
sum of the dice is 7 or 11 ("natural"); he loses if the sum is 2,3, or 
12 (craps). If the sum is 4,5,6,8,9, or 10, then the result is not yet 
decided.  He must roll the dice again and again, as often as is 
necessary until the initial sum, be it 4,5,6,8,9, or 10  is repeated 
(shooter wins) or until the sum is 7 (shooter loses).  Find the 
probability that the shooter will win in a game of craps.

Richard

Date: 6/28/96 at 18:14:38
From: Doctor Anthony
Subject: Re: Probability in Craps Game

A good way to start is to draw up a square grid with margins 1 to 6 
vertically and horizontally, and the sum of the marginal values 
filling up the grid.  There are 36 grid points, and the probability of 
a sum of 5 (say) is to count the number of 5's in the grid and divide 
this number (=4) by 36, to give a probability of 4/36 = 1/9

We have to enumerate the ways that the shooter could win.

P(7)+P(11) = 1/6 + 1/18 = 2/9

Now consider throwing a total of 4, and needing to repeat this to win. 
Probability of a 4 is 1/12.  Now we have to continue until another 4 
occurs.

This gives rise to an infinite sequence with probability at each throw 
of 1/12 to win, 1/6 to lose (by throwing a 7) and 3/4 to have a 
further go.  We could win after 1, 2, 3, ....... infinity number of 
throws.  The probability sequence looks like this

1/12 + (3/4)(1/12) + (3/4)^2(1/12) + (3/4)^3(1/12) + ....

(1/12){1 + (3/4) + (3/4)^2 + ...} = (1/12)(1/(1-3/4) = (1/12)(1/(1/4))
                                  = 4/12 = 1/3

It's not really necessary to sum the series.  Since rerolls don't affect
whether you win or lose, you can ignore the 3/4 probability and say that
your probability of winning with 4's is 

  P(win) / P(win + lose) = (1/12) / (1/6 + 1/12)

which of course gives the same answer of 1/3 as the series sum.

So complete probability of winning with 4's = (1/12)*(1/3) = 1/36

Since probability of 10 is also 1/12, this too gives an overall 
probability of 1/36

We now consider probability of winning with two 5's or two 9's.  These 
have probability 1/9 of occurring and by a repeat of the argument 
given above, the total probability of winning with 5's is (1/9)*(2/5) 
= 2/45.  The same applies for winning with two 9's.

Finally we consider winning with two 6's or two 8's.  The probability 
of a 6 is 5/36, and a repeat of above argument with infinite series, 
gives a probability of (5/36)*(5/11) = 25/396  and the same applies 
for two 8's.

We now add up all the probabilities of winning

  = (2/9) + 2(1/36) + 2(2/45) + 2(25/396)

  =  (2/9) + (1/18) + (4/45) + (25/198)

  = 244/495   (=0.492929...)
  
-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/