Question 382989: In a carnival game, a person wagers $2 on the roll of two dice. If the total of the tow dice is 2,3,4,5, or 6 then the person gets $4 (the $2 wager and $2 winnings). If the total of the two dice is 8,9,10,11, or 12 then teh person gets nothing (loses $2). If the total of the two dice is 7, the person gets $0.75 back (loses $0.25). What is the expected value of playing the game once?
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! You made a mistake writing the problem. I am assuming you meant "If the total of the two dice is 7, the person gets $1.75 back (loses $0.25)."
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There are 6 ways out of 36 to roll a 7: 6-1 1-6 5-2 2-5 4-3 3-4.
There are 36/2 - 6=18-3=15 ways to roll 2,3,4,5,6 and 15 ways to roll 8,9,10,11,12.
On 15 roll he loses 30 and on 15 he wins 30 and on 6 rolls he loses 1.50
On 36 rolls he has a net loss of $1.50
1.50/36=-$0.04 the expected value of playing the game once.
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Ed
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