Chapter 2.10, Problem 138E

To determine

Find the probability that the player wins a game of craps.

Answer to Problem 138E

The probability that the player wins a game of craps is 0.4929.

Explanation of Solution

Calculation:

Let A denote the event that the player wins the game, B denote a sum of i on first toss and C_k denote a sum of k before obtaining a 7.

The possible outcomes when rolling two dice are shown below:

$S=\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

The formula for finding the probability that the player wins a game of craps is shown below:

$P\left(A\right)={\displaystyle \sum _{i=1}^{12}P\left(A\cap B_i\right)}$

From the given information, $P\left(A\cap B_2\right)=P\left(A\cap B_3\right)=P\left(A\cap B_{12}\right)=0$.

Also, $P\left(A\cap B_7\right)=P\left(B_7\right)=\frac{6}{36}\text{ and }P\left(A\cap B_{11}\right)=P\left(B_{11}\right)=\frac{2}{36}$.

For k = 4, 5, 6, 8, 9 and 10, $P\left(A\cap B_k\right)=P\left(B_k\cap C_k\right)=P\left(B_k\right)P\left(C_k\right)$

Here, the tosses are independent.

Let F_k denote the event that player do not throw a k or a 7, for k =4, 5, 6, 8, 9, 10.

The Event C_k is obtained as follows:

$\begin{array}{c}P\left(C_k\right)=P\left(B_k\right)+P\left(F_k\cap B_k\right)+P\left(F_k\cap F_k\cap B_k\right)+\mathrm{..}\\ =P\left(B_k\right)\left(1+P\left(F_k\right)+P{\left(F_k\right)}^2+\mathrm{...}+P{\left(F_k\right)}^n+\mathrm{...}\right)\\ =\frac{P\left(B_k\right)}{1-P\left(F_k\right)}\end{array}$

The probability of C₄ is obtained as given below:

$\begin{array}{c}P\left(C_4\right)=\frac{P\left(B_4\right)}{1-P\left(F_4\right)}\\ =\frac{\frac{3}{36}}{1-\frac{27}{36}}\\ =\frac{\frac{3}{36}}{\frac{9}{36}}\\ =\frac{3}{9}\\ =\frac{1}{3}\end{array}$

Similarly, the probabilities of $C_5$, $C_6$, $C_8$, $C_9$, and $C_{10}$ can be obtained.

From the sample outcomes, it can be observed that,

$\begin{array}{l}P\left(B_4\right)=P\left(B_{10}\right)=\frac{3}{36},P\left(C_4\right)=P\left(C_{10}\right)=\frac{1}{3},\\ P\left(B_5\right)=P\left(B_9\right)=\frac{4}{36},P\left(C_5\right)=P\left(C_9\right)=\frac{2}{5},\\ P\left(B_6\right)=P\left(B_8\right)=\frac{5}{36},P\left(C_6\right)=P\left(C_8\right)=\frac{5}{11}\end{array}$

Therefore,

$\begin{array}{l}P\left(A\cap B_4\right)=P\left(C_4\cap B_4\right)=P\left(C_4\right)P\left(B_4\right)=\frac{1}{3}\times \frac{3}{36}=\frac{1}{36}\\ P\left(A\cap B_5\right)=P\left(C_5\cap B_5\right)=P\left(C_5\right)P\left(B_5\right)=\frac{2}{5}\times \frac{4}{36}=\frac{2}{45}\\ P\left(A\cap B_6\right)=P\left(C_6\cap B_6\right)=P\left(C_6\right)P\left(B_6\right)=\frac{5}{11}\times \frac{5}{36}=\frac{25}{396}\\ \text{Similarly }P\left(A\cap B_9\right)=\frac{2}{45},P\left(A\cap B_8\right)=\frac{25}{396},P\left(A\cap B_{10}\right)=\frac{1}{36}\end{array}$

The probability that the player wins a game of craps is obtained as follows:

$\begin{array}{c}P\left(A\right)={\displaystyle \sum _{i=2}^{12}P\left(A\cap B_i\right)}\\ =0+0+\frac{1}{36}+\frac{2}{45}+\frac{25}{396}+\frac{6}{36}+\frac{25}{396}+\frac{2}{45}+\frac{1}{36}+\frac{2}{36}+0\\ =0.4929\end{array}$

Thus, the probability that the player wins a game of craps is 0.4929.