Question

Suppose that you are offered the following "deal." You roll a six sided die. If you roll a 6, you win $12. If you roll a 2, 3, 4 or 5, you win $1. Otherwise, you pay $10.

a. Complete the PDF Table. List the X values, where X is the profit, from smallest to largest. Round to 4 decimal places where appropriate.

Probability Distribution Table

X	P(X)

b. Find the expected profit. $ (Round to the nearest cent)

c. Interpret the expected value.

• If you play many games you will likely win on average very close to $1.00 per game.
• You will win this much if you play a game.
• This is the most likely amount of money you will win.

d. Based on the expected value, should you play this game?

• No, since the expected value is negative, you would be very likely to come home with less money if you played many games.
• Yes, since the expected value is 0, you would be very likely to come very close to breaking even if you played many games, so you might as well have fun at no cost.
• No, this is a gambling game and it is always a bad idea to gamble.
• Yes, because you can win $12.00 which is greater than the $10.00 that you can lose.
• Yes, since the expected value is positive, you would be very likely to come home with more money if you played many games.

Answer 1

X values = -10,+1,+12

P(X) probability of winning x

P( -10 ) = 1/6 . { only if we roll 1 so one favourable out of 6)

P(1) = 4/6   { if we roll 2,3,4,5 so 4 favourable out of 6)

P( 12 ) = 1/6 . { only if we roll 6 so one favourable out of 6)

a)

X	P(X)
-10	0.1667
1	0.6667
12	0.1667

b) Expected Profit . = ∑X*P(X) = -10*0.1667 + 1*0.6667 + 12*0.1667 = 1.0001 = 1

Answer is $1

c) The interpretation will be if you play many games you will likely win on average very close to $1.00 per game.

d) Answer => Yes, since the expected value is positive, you would be very likely to come home with more money if you played many games.

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