Question

Chance of winning at “craps.” A version of the dice game“craps” is played in the following manner. A player starts by rolling two balanced dice. If the roll (the sum of the two numbers showing on the dice) results in a 7 or 11, the player wins. If the roll results in a 2 or a 3 (called craps), the player loses. For any other roll outcome, the player continues to throw the dice until the original roll outcome recurs (in which case the player wins) or until a 7 occurs

(in which case the player loses).

a. What is the probability that a player wins the game on the first roll of the dice?

b. What is the probability that a player loses the game on the first roll of the dice?

c. If the player throws a total of 4 on the first roll, what is the probability that the game ends (win or lose) on the next roll?

Short answer

1. The probability for wins is 0.222.
2. The probability for loss is 0.08.
3. The probability of wins or loses is 0.25.

Step-by-step solution

Important formula

The formula for probability is$\mathbf{P}\mathbf{=}\frac{\mathrm{favourable}\mathrm{outcomes}}{\mathrm{total}\mathrm{outcomes}}$

The probability that a player wins the game on the first roll of the dice.

The 2 balanced dice are rolled by a player according to information. A player wins if the rolls is 7 or 11, the player loses if the roll is 2 or 3. Other than a 7 or a recurrence of the original outcomes, the player keeps throwing the dice.

The events are

P(2)={1,1}

P(3)={1,2}, {2,1}

P(4)={1,3}, {2,2},{3,1}

P(5)={1,4}, {2,3}, {3,2}, {4,1}

P(6)={1,5}, {2,4},{3,3},{4,2},{5,1}

P(7)={1,6},{2,5},{3,4},{4,3},{5,2},{6,1}

P(8)= {2,6},{3,5},{4,4},{5,3},{6,2}

P(9)={3,6},{4,5},{5,4},{6,3}

P(10)={4,6},{5,5},{6,4}

P(11)={5,6},{6,5}

P(12)={6,6}

$\begin{array}{c}\mathrm{P}\left(\mathrm{WINS}\right)=\mathrm{P}\left(7\right)+\mathrm{P}\left(11\right)\\ =\frac{6}{36}+\frac{2}{36}\\ =0.222\end{array}$

Hence, the probability of wins is 0.222.

The probability that a player loses the game on the first roll of the dice.

$\begin{array}{c}\mathrm{P}\left(\mathrm{loss}\right)=\mathrm{P}\left(2\right)+\mathrm{P}\left(3\right)\\ =\frac{1}{36}+\frac{2}{36}\\ =0.08\end{array}$

Thus, the probability for loss is 0.08.

what is the probability that the game ends (win or lose) on the next roll.

$\begin{array}{c}\text{P(winorloss)}=P\left(4\right)+P\left(7\right)\\ =\frac{3}{36}+\frac{6}{36}\\ =0.25\end{array}$

Therefore, the probability of wins or loses is 0.25.