Urgent Stats Problem

Question

Alan has a bag with 8

 balls numbered 1

 through 8

. He is playing a game of chance.

This game is this: Alan chooses one ball from the bag at random. He wins 

$1

 if the number 1

 is selected, $2

 if the number 2

 is selected, $5

 if the number 3

 is selected, $6

 if the number 4

 is selected, $8

 if the number 5

 is selected, and $10

 if the number 6

 is selected. He loses $14

 if 7

 or 8

 is selected.

Find the expected value of the playing the game; What can Alan expect in the long run? How much money does he win or lose?

Answer 1

Expected value = (probability of picking #1)(amount won by picking #1) + (probability of picking #2)(amount won by picking #2) + ... + (probability of picking #8)(amount lost by picking #8)

(1/8)(1) + (1/8)(2) + (1/8)(5) + (1/8)(6) + (1/8)(8) + (1/8)(10) + (1/8)(-14) + (1/8)(-14) = 4/8 = $0.50

In the long run, he would win $0.50.

Answer 2

Hi Ishaan,

To find the expected value, you will need to take the probabilities for each possible outcome and multiply it by the payoff for that outcome. After doing so, you then need to add these together.

I'll show you what I mean.

The probability of picking a (1) is 1/8 right? And since the payoff for picking a 1 is 1 dollar, we would do:

1 * 1/8 = 1/8

Similarly, for picking a (2), the probability is also 1/8, but since the payoff is 2 dollars, we would do:

2 * 1/8 = 2/8

Continue doing this until you have all of your values. Then add these together.

The resulting value will be how much Alan could "expect" to win each time he plays the game.

Let me know if you have any questions.