Probability of winning a game of craps

Question

The dice game craps is played as follows: The player throws 2 dice, and if the sum is 7 or 11, he/she wins. If the sum is 2, 3, or 12, he/she loses. If the sum is anything else, then he/she continues throwing until that number appears again, or throws a 7, where the game ends in a loss.

What I do know is that $P(7) = 6/36$, $P(11)=2/36$. So P(winning) in first roll is $8/36$. Furthermore, the probability of having to roll again will be $1-[P($winning 1st roll$) +P($losing 1st roll$)]=24/36$. It's what happens if the game doesn't end in the 1st roll that's got me a bit confused, since it could go on and indeterminate number of rolls. But since it can't be 7,11,2,3,or 12, it depends on if they roll a 4,5,6,8,9, or 10.

Note, this is not a homework problem, but an intriguing one I found in a different book "Probability Models, Sheldon Ross"

Answer 1

The probability of winning directly is, as you calculated, $8/36$, and the probability of losing directly is $(1+2+1)/36=4/36$.

For the remaining cases, you need to sum over all remaining rolls. Let $p$ be the probability of rolling your initial roll, and $q=6/36=1/6$ the probability of rolling a $7$. Then the probability of rolling your initial roll before rolling a $7$ is $p/(p+q)$, and the probability of rolling a $7$ before rolling your initial roll is $q/(p+q)$. Thus, taking into account the probability of initially rolling that roll, each roll that doesn't win or lose directly yields a contribution $p^2/(p+q)$ to your winning probability.

For $p=5/36$, that's

$$ \frac{\left(\frac5{36}\right)^2}{\frac{5+6}{36}}=\frac{25}{11\cdot36}\;, $$

and likewise $16/(10\cdot36)$ and $9/(9\cdot36)$ for $p=4/36$ and $p=3/36$, respectively. Each of those cases occurs twice (once above $7$ and once below), so your overall winning probability is

$$ \frac8{36}+\frac2{36}\left(\frac{25}{11}+\frac{16}{10}+\frac99\right)=\frac{244}{495} = \frac12-\frac7{990}\approx\frac12-0.007\;. $$

Answer 2

Suppose you throw a $4$ and let $p(4)$ your winning probability. At your next roll you have a probability $3/36$ of winning (you throw a 4), a probability $6/36$ of losing (you throw a 7) and a probability $27/36$ of repeating the whole process anew (you throw any other number). Then: $$ p(4)={3\over36}+{27\over36}p(4),\quad \hbox{so that}\quad p(4)={1\over3}. $$ Repeat this reasoning for the other outcomes and then compute the total probability of winning as: $$ p_{tot}={8\over36}+{3\over36}p(4)+{4\over36}p(5)+\ldots $$

Answer 3

Others have covered the same solution, but here's another method using infinite sum.

Preliminaries: \begin{array}{|c|c|c|c|c|c|c|} \hline Sample \in& \{2,12\} & \{3,11\}& \{4,10\}& \{5,9\}& \{6,8\}& \{7\}\\ \hline Probabilities&\frac{1}{36} &\frac{1}{18} &\frac{1}{12}&\frac{1}{9}&\frac{5}{36}&\frac{1}{6}\\ \hline \end{array}

Let the RV indicating the $nth$ throw be $T_n$. Let $W$ be the event of winning \begin{align} \begin{split} P(W) &= \quad P(W|T_1\in \{7,11\})P(T_1\in\{7,11\})\\&\quad+P(W|T_1\in \{4,10\})P(T_1\in\{4,10\})\\&\quad+P(W|T_1\in \{5,9\})P(T_1\in\{5,9\})\\&\quad+P(W|T_1\in \{6,8\})P(T_1\in\{6,8\})\\&\quad+P(W|T_1\in \{2,3,12\})P(T_1\in\{2,3,12\}) \end{split}\qquad\text{(Eqn.1)} \end{align} Calculating the individual conditional probabilities \begin{align} P(W^C|T_1\in \{4\})&=P(T_2=7)+P(T_2\notin\{4,7\})P(T_3=7)\\&\quad+P(T_3\notin\{4,7\},T_2\notin\{4,7\})P(T_4=7)\\&\quad+P(T_4\notin\{4,7\},T_3\notin\{4,7\},T_2\notin\{4,7\})P(T_5=7)\\&\quad+\dots\\\\ \end{align} Since $T_i$ are independent $\forall i$ \begin{align} P(W^C|T_1\in \{4\})&=P(T_i=7)(1+P(T_i\notin\{4,7\})+P(T_i\notin\{4,7\})^2+\dots)\\ &=\dfrac{P(T_i=7)}{1-P(T_i\notin\{4,7\})}\\ &=\dfrac{\frac{1}{6}}{1-\frac{36-9}{36}} = \dfrac{2}{3}\\ \end{align}

Taking the complement, $P(W|T_1\in \{4\})=1-\frac{2}{3} = \frac{1}{3}$ and similarly, $P(W|T_1\in \{10\})=\frac{1}{3}$

We now calculate other quantities \begin{equation} P(W^C|T_1\in \{5\})=\dfrac{P(T_i=7)}{1-P(T_i\notin\{5,7\})}=\dfrac{\frac{1}{6}}{\frac{10}{36}}=\dfrac{3}{5} \end{equation}

Then $P(W|T_1\in \{5\}) = \dfrac{2}{5}$ and $P(W|T_1\in \{9\}) = \dfrac{2}{5}$

And the same way \begin{equation} P(W^C|T_1\in \{6\})=\dfrac{P(T_i=7)}{1-P(T_i\notin\{6,7\})}=\dfrac{\frac{1}{6}}{\frac{11}{36}}=\dfrac{6}{11} \end{equation}

Then $P(W|T_1\in \{6\}) = P(W|T_1\in \{8\}) = \dfrac{5}{11}$. Finally we have $P(W|T_1\in \{2,3,12\})=0$ by the rules of the game.

Substituting all this in Eqn.1 we have \begin{align} P(W) &= 1\times \frac{8}{36} + \frac{1}{3} \times \frac{2}{12}+ \frac{2}{5} \times \frac{2}{9}+ \frac{5}{11}\times \frac{10}{36}+ 0\times \frac{4}{36}\\ &= \frac{976}{1980} \end{align}