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
$\frac{8}{36}+2(\frac{3}{36}+\frac{2}{45}+\frac{11}{396})=\frac{8}{15}$
.