The game of chuck-a-luck consists of rolling 3 dice. The player picks a number, 1, 2, 3, 4, 5, or 6. Winning and losing depends on the number of times the number appears on the top of each die. If the number appears once then the winnings are equal to the bet. If the number appears twice the winnings are twice the bet. If the number appears 3 times the winnings are 5 times the bet. If the number doesn't appear then the bet is lost. A $5.00 bet is made on the number 3. what is the expected outcome of the bet?
0.840
1.354
0.354
-0.354
I was thinking D..
To determine the expected outcome of the bet, we need to calculate the probability of each possible outcome and multiply it by the corresponding winnings or losses.
In this case, the player bets $5 on the number 3. Now let's calculate the probability of each outcome:
1. The number 3 appears once: To calculate this probability, we need to determine the probability of the number 3 appearing on one of the three dice, and the probability of the other two dice not showing a 3. The probability of a specific die showing a 3 is 1/6, and the probability of it not showing a 3 is 5/6. Since there are 3 dice, we multiply these probabilities together: (1/6) * (5/6) * (5/6) = 25/216.
2. The number 3 appears twice: Similarly, we need to calculate the probability of the number 3 appearing on two of the three dice, and the probability of the third die not showing a 3. The probability of two specific dice showing a 3 is (1/6) * (1/6), and the probability of the other die not showing a 3 is 5/6. There are three possible combinations for the two dice showing a 3: (1/6) * (1/6) * (5/6) = 5/216. Since there are three possible combinations, we multiply this probability by 3: (5/216) * 3 = 15/216.
3. The number 3 appears three times: This is the simplest case since we only need to calculate the probability of all three dice showing a 3. The probability of a specific die showing a 3 is 1/6. Since there are three dice, we raise this probability to the power of 3: (1/6)^3 = 1/216.
4. The number 3 doesn't appear: To calculate this probability, we need to determine the probability of each die not showing a 3. The probability of a specific die not showing a 3 is 5/6. Since there are three dice, we multiply these probabilities together: (5/6)^3 = 125/216.
Now that we have the probabilities, we can calculate the expected outcome:
Expected outcome = (Probability of outcome 1 * Winnings for outcome 1) + (Probability of outcome 2 * Winnings for outcome 2) + (Probability of outcome 3 * Winnings for outcome 3) + (Probability of outcome 4 * Winnings for outcome 4)
Expected outcome = (25/216 * $5) + (15/216 * $10) + (1/216 * $25) + (125/216 * -$5)
Simplifying this equation, the expected outcome is approximately -$0.354.
So, the correct answer is C: -0.354.