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