Casino Dice Game Earnings

Solution & Explanation

Understanding the Problem:

You are playing a dice game at a casino where you win if you roll a sum of 5 with two dice. The payout for winning is $10, and each game costs$5 to play. The task is to determine the expected profit if you continue playing until you win for the first time and then stop.

Step 1: Determine the Probability of Winning

1. Possible outcomes for a sum of 5:

   • (1, 4)
   • (2, 3)
   • (3, 2)
   • (4, 1)

2. Total number of outcomes when rolling two dice:

   • 6 sides on the first die × 6 sides on the second die = 36 possible outcomes.

3. Probability of rolling a sum of 5:

   • There are 4 outcomes that result in a sum of 5.
   • Probability, $P(5) = \frac{4}{36} = \frac{1}{9}$.

Step 2: Expected Number of Trials

1. Geometric Distribution:

   • The geometric distribution models the number of trials needed to get the first success in a series of Bernoulli trials.
   • Expected number of trials, $E(X) = \frac{1}{p}$ where $p$ is the probability of success.

2. Expected number of games to win:

   • $E(X) = \frac{1}{1/9} = 9$.
   • On average, you will need to play 9 games to win once.

Step 3: Calculate Expected Profit

1. Cost of playing 9 games:

   • Cost per game = $5
   • Total cost for 9 games = 9 × $5 =$45

2. Earnings from winning once:

   • Payout for winning = $10

3. Expected Profit Calculation:

   • Expected profit = Earnings from winning - Total cost of playing
   • Expected profit = $10 -$45 = -$35

Conclusion:

The expected profit, if you play until you win once and then stop, is -\35$. This means that, on average, you will lose$35 for each complete cycle of playing until you win. This game, therefore, is not profitable in the long run.