Shooting Crap

Crap shooting is a game played with two dice. A first throw of 7 or 11 wins and a first throw of 2, 3, or 12 loses. A first throw of any other number must be repeated to win before a 7 is thrown, else the game is lost. table shows the set of possible scores on the dice, obtained by adding the score on each dice.
        DICE 1      
    1 2 3 4 5 6
  1 2, LOSE 3, LOSE 4 5 6 7, WIN
  2 3, LOSE 4 5 6 7, WIN 8
  3 4 5 6 7, WIN 8 9
DICE 2 4 5 6 7, WIN 8 9 10
  5 6 7, WIN 8 9 10 11, WIN
  6 7, WIN 8 9 10 11, WIN 12, LOSE
The probability of scoring either 7 or 11 is  
\[\frac{6}{36}+\frac{2}{36}=\frac{8}{36}\]
.
Suppose you score 4, then you can keep throwing the dice until you score a 4, unless you score 7, in which case you lose. We can write this as
\[P(4)(P(4)+P(Not \: 4 \: or \: 7 \: Then \: 4)+P(Not \: 4 \: or \: 7 \: Then \: Not \: 4 \: or \: 7 \: Then \: 4)+...)\]
  (1)
\[P(4)=\frac{3}{36}\]
  and  
\[P(7)=\frac{6}{36}\]
  so
\[P(Not \: 4 \: or \: 7)=\frac{9}{36}\]

Hence the probability of winning by first scoring a 4 is
\[\frac{3}{36}(\frac{3}{36}+ \frac{27}{36} \times \frac{3}{36}+ (\frac{27}{36})^2 \times \frac{3}{36})+... )\]

The expression inside the bracket is a geometric series with first term  
\[\frac{3}{36}\]
  and common ratio  
\[\frac{27}{36}\]
, and sum  
\[\frac{3/36}{1-27/36}=\frac{1}{3}\]
, then (1) is  
\[\frac{3}{36} \times \frac{1}{3}=\frac{1}{36}\]
.
The probability of winning by first scoring a 10 is the same.
Suppose you score 5, then you can keep throwing the dice until you score a 5, unless you score 7, in which case you lose. We can write this as
\[P(5)(P(5)+P(Not \: 5 \: or \: 7 \: Then \: 5)+P(Not \: 5 \: or \: 7 \: Then \: Not \: 5 \: or \: 7 \: Then \: 5)+...)\]
  (2)
\[P(5)=\frac{4}{36}\]
  and  
\[P(7)=\frac{6}{36}\]
  so
\[P(Not \: 5 \: or \: 7)=\frac{10}{36}\]

Hence the probability of winning by first scoring a 5 is
\[\frac{4}{36}(\frac{4}{36}+ \frac{26}{36} \times \frac{4}{36}+ (\frac{26}{36})^2 \times \frac{4}{36})+... )\]

The expression inside the bracket is a geometric series with first term  
\[\frac{4}{36}\]
  and common ratio  
\[\frac{26}{36}\]
, and sum  
\[\frac{4/36}{1-26/36}=\frac{2}{5}\]
, then (2) is  
\[\frac{4}{36} \times \frac{2}{5}=\frac{2}{45}\]
.
The probability of winning by first scoring a 9 is the same.
Suppose you score 6, then you can keep throwing the dice until you score a 6, unless you score 7, in which case you lose. We can write this as
\[P(6)(P(6)+P(Not \: 6 \: or \: 7 \: Then \: 6)+P(Not \: 6 \: or \: 7 \: Then \: Not \: 6 \: or \: 7 \: Then \: 6)+...)\]
  (3)
\[P(6)=\frac{5}{36}\]
  and  
\[P(7)=\frac{6}{36}\]
  so
\[P(Not \: 6 \: or \: 7)=\frac{11}{36}\]

Hence the probability of winning by first scoring a 6 is
\[\frac{5}{36}(\frac{5}{36}+ \frac{25}{36} \times \frac{5}{36}+ (\frac{25}{36})^2 \times \frac{5}{36})+... )\]

The expression inside the bracket is a geometric series with first term  
\[\frac{5}{36}\]
  and common ratio  
\[\frac{25}{36}\]
, and sum  
\[\frac{5/36}{1-25/36}=\frac{5}{11}\]
, then (3) is  
\[\frac{5}{36} \times \frac{5}{11}=\frac{25}{396}\]
.
The probability of winning by first scoring an 8 is the same.
Adding up the probabilities of winning gives  
\[\frac{8}{36}+2(\frac{3}{36}+\frac{2}{45}+\frac{11}{396})=\frac{8}{15}\]
.