The game of craps is played as follows: A player rolls two

Chapter 2, Problem 26P

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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 \(\sum_{i=2}^{12} P\left(E_{i}\right)\). 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 \(P\left(E_{i}\right)=\sum_{n=1}^{\infty} P\left(E_{i, n}\right)\).

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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 \(\sum_{i=2}^{12} P\left(E_{i}\right)\). 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 \(P\left(E_{i}\right)=\sum_{n=1}^{\infty} P\left(E_{i, n}\right)\).

ANSWER:

Step 1 of 3

(a)

We are asked to find the probability of a player winning at craps.

The sum equals each number from two to twelve in the pair of dice, we have

\(\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline x & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline P(X=x) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{1}{6} & \frac{5}{36} & \frac{1}{9} & \frac{1}{12} & \frac{1}{18} & \frac{1}{36} \\
\hline
\end{array}\)

The probability of winning at the first roll, that is obtaining 7 or 11 is,

\(\begin{array}{c}
P_{7}=\frac{1}{6} \\
P_{11}=\frac{1}{18}
\end{array}\)

So the probability of winning on the first roll is,

\(P_{7}+P_{11}=\frac{1}{18}+\frac{1}{6}=\frac{2}{9}\)

The probability of losing at the first roll, that is obtaining 2, 3 or 12 is,

\(\begin{array}{l}
P_{2}=\frac{1}{36} \\
P_{3}=\frac{2}{36} \\
P_{12}=\frac{1}{36}
\end{array}\)

So the probability of losing on the first roll is,

\(P_{2}+P_{3}+P_{12}=\frac{1}{36}+\frac{2}{36}+\frac{1}{36}=\frac{1}{9}\)

Then the probability of making a point will be

\(1-\frac{2}{9}-\frac{1}{9}=\frac{2}{3} \ldots \ldots(1)\)

Now we have already calculated the individual probabilities for the numbers 4, 5, 6, 8, 9 or 10.

Hence the addition of these probabilities is,

\(P(4 \cup 5 \cup 6 \cup 8 \cup 9 \cup 10)=\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{5}{36}+\frac{1}{9}+\frac{1}{12}=\frac{2}{3}\)

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