In the game of craps, a pair of dice are rolled, and people bet on the outcome. For example, you can bet $1 that the sum of the dice will be 12(also called "Boxcars or Midnight"). The probability that you win is 1 over 36 , and if you win, your profit is $30. If you lose, you lose $1.

Question

(a) What is the expected value of your profit? Round the answer to two decimal places.

Answer 1

To calculate the expected value of the profit from the bet in the game of craps, we can use the formula for expected value (EV):

$$EV=(P(\text{win})\times \text{profit if win})+(P(\text{lose})\times \text{profit if lose})$$

From the problem:

• $P(\text{win})=\frac{1}{36}$
• $\text{profit if win}=30$ (you win $30, but your original $1 bet is returned, leading to a total return of $31, but profit is considered as $30)
• $P(\text{lose})=\frac{35}{36}$ (since there are 36 outcomes, and 35 of them are losses)
• $\text{profit if lose}=-1$ (you lose your bet of $1)

Now we can plug these values into the expected value formula:

$$EV=(\frac{1}{36}\times 30)+(\frac{35}{36}\times -1)$$

Calculating each term:

1. For the win: $$\frac{1}{36}\times 30=\frac{30}{36}=\frac{5}{6}\approx 0.8333$$

2. For the loss: $$\frac{35}{36}\times -1=-\frac{35}{36}\approx -0.9722$$

Now we can sum these results:

$$EV=0.8333-0.9722\approx -0.1389$$

Rounding to two decimal places, the expected value of your profit is:

$$\text{EV}\approx -0.14$$

Thus, the expected value of your profit is $-0.14$.

Answer 2

(b) Is it an expected gain or an expected loss? Round the answer to two decimal places.

Answer 3

The expected value we calculated in part (a) was approximately $-0.14$. A negative expected value indicates that, on average, you are expected to lose money when making this bet.

Thus, the answer to part (b) is that it is an expected loss.

To summarize:

• Expected Value: $-0.14$ (rounded to two decimal places)
• Conclusion: This is an expected loss.