Q.2.26 - Problems - A First Course in Probability

Question

The game of craps is played as follows: A player rolls two dice. If the sum of the dice is either a$2,3,or 12$, the player loses; if the sum is either a $7$ or an $11$, the player wins. If the outcome is anything else, the player continues to roll the dice until she rolls either the initial outcome or a $7$. If the $7$ comes first, the player loses, whereas if the initial outcome reoccurs before the $7$ appears, the player wins. Compute the probability of a player winning at craps.

 Hint: Let $E_i$denote the event that the initial outcome is $i$and the player wins. The desired probability is $\sum _{i=12}^{12}P\left(E_i\right)$. To compute $P\left(E_i\right)$, define the events $E_{i,n}$ to be the event that the initial sum is i and the player wins on the nth roll. Argue that  

$P\left(E_i\right)=\sum _{n=1}^{\infty }P\left(E_{i,n}\right)$

Step-by-Step Solution

Answer

$\frac{244}{495}\approx 0.4929$

Step-1 Given Information

Given In the game of craps a player rolls two dice.

Criteria $1$ : At the initial trial if the sum is $2,3, or 12$ this is a loss if the sum is $7$ or $11$ this is a win if the sum is $4,5,6,8,9$ or $10$go to the second criteria.

Criteria $2$ : Roll repeatedly until the sum is the same as in the initial (first) roll or sum is $7$ . If the sum is $7$ this is a loss, if the sum is as in the initial roll this is a win.

we have to find  the probability of a player winning at craps. 

Step-2 Explanation

Assume that the dice are fair.

Let event $E_i$is that the player wins with initial outcome $i$ and let $E_{i,n}$is the event that the initial outcome is $i$ and the player wins on $n^{th}$ trial.

Therefore,

$P\left(Win\right)=\sum _{i=2}^{12}P\left(E_i\right)$

Because every game can be divided with regard to the outcome of the first throw, and each of those events is mutually exclusive.

The same thing can be applied when calculating the probability of$E_i$ , if a player wins (with an initial outcome $i$ ) in $1,2,3...$ rolls, and those are all possible, and mutually exclusive. Hence,

$P\left(E_i\right)=\sum _{n=1}^{\infty }P\left(E_{i,n}\right)$

From the rules of the game,

$$\begin{gathered} P\left(E_2\right)=P\left(E_3\right)=P\left(E_{12}\right)=0 \\ P\left(E_7\right)=\frac{6}{36}=\frac{1}{6}, \\ P\left(E_{12}\right)=\frac{2}{36}=\frac{1}{18} \end{gathered}$$,

To calculate $P\left(E_{i,n}\right)$ :

The event $E_{i,n}$ means that the $n^{th}$ outcome was $i$, and neither outcome  $i or 7$appears before (except the first throw which is $i$ ) then the player would either win before or lose.

If the probability of the initial outcome is $\frac{x_i}{36}$, there are $x_i$ possible ways of getting the outcome $i$. There are $6$out of $36$ rolls with sum $7$ . Therefore $36-6-x_i=30-x_i$ possibilities where the outcomes are neither $7 or i$

The event $E_{i,n}$ will occur if the first and the last roll should be $i$ and the remaining $n-2$ rolls should be neither $7 or i$. So the possibilities are $x_i .{(30-x_i)}^{n-2}$. The total possible outcomes are ${36}^n$.

hence,

$$\begin{gathered} P\left(E_{i,n}\right)=\frac{x_{i.}{(30-x_i)}^{n-2}.x_i}{{36}^n} \\ ={\left(\frac{x_i}{36}\right)}^2{\left(\frac{30-x_i}{36}\right)}^{n-2} \end{gathered}$$

             

Thus,

$$\begin{gathered} P\left(E_i\right)=\sum _{n=2}^{\infty } {\left(\frac{x_i}{36}\right)}^2{\left(\frac{30-x_i}{36}\right)}^{n-2} \\ =\frac{x_i}{36}.\frac{x_i}{(6+x_i)},i\in \left\{4,5,6,8,9,10\right\} \\ P\left(win\right)=\sum _{I=2}^{12}P\left(E_i\right) \\ \frac{244}{495}\approx 0.4929 \end{gathered}$$