Question 172627This question is from textbook Thinking Mathematically
: Solve the problem that involves computing expected values in a game of chance.
A game is played using one die. If the die is rolled and shows a 2, the player wins $8. If the die shows any number other than 2, the player wins nothing. If there is a charge of $1 to play the game, what is the game's expected value?
Here are some thoughts on how to solve... not sure if I am thinking in the right direction:
Get a 2 the prob. is 1/6 and win $8.00 but pay $1 to play
not a 2 is 5/6 and owe a dollar
1/6*8/1= 8/6...4/3=1.33 - $1.00= $.33
This question is from textbook Thinking Mathematically
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If the die is rolled and shows a 2, the player wins $8.
If the die shows any number other than 2, the player wins nothing.
If there is a charge of $1 to play the game, what is the game's expected value?
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Let the random variable be player gain:
Its values are:
8 with probability (1/6)
-1 with probability (5/6)
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Expected Value = (1/6)8 + (5/6)(-1) = (8/6) - (5/6) = $0.50
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Cheers,
Stan H.
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