Question

Suppose you pay $3.00 to roll a fair die with the understanding that you will get back $5.00 for rolling a 1 or a 6, nothing otherwise. What is your expected value?

Answer 1

The expected value of the game where you pay $3.00 to roll a die and win $5.00 for rolling a 1 or a 6 is -$0.33 per roll, indicating a loss over time.

Step-by-step explanation:

The problem is asking us to calculate the expected value of the game in which you pay $3.00 to roll a fair die with the potential of winning $5.00 for rolling a 1 or a 6. To find the expected value, we need to multiply the outcomes by their respective probabilities and then sum these products.

The probability of rolling a 1 or a 6 is 1/6 for each number, so the combined probability for these winning rolls is 1/6 + 1/6 = 1/3. The probability of rolling a 2, 3, 4, or 5 is therefore 2/3 since these are the non-winning rolls.

Let's calculate the expected value (EV):

Winning: (1/3) * $5.00 = $1.67

Losing: (2/3) * -$3.00 = -$2.00

Now we add the two values:

EV = $1.67 - $2.00 = -$0.33

Therefore, the expected value of playing the game is a loss of $0.33 per roll. This means that in the long run, you can expect to lose an average of $0.33 for each time you play this game.

Answer 2

1 down vote accepted The expected value is the sum of the probablity of each outcome multiplied by the "value" of each outcome. Its sort of like a special average. In your case each outcome has probability 16. and you get 5 dollars when you roll 3 or 6 when you roll something else. So the expected value is 06+06+56+06+06+56=106=53. This means loosely that if you played this game a lot you should expect to win on average about 1.66 dollars, while you payed 3 to play it. Clearly this is a great idea if you want to lose money.