SOLUTION: A person selects a card from a deck. If it is a red card, he wins $1. If it is a black card between or including 2 and 10, he wins $5. if it is a black face card, he wins $10; and
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Question 1171872: A person selects a card from a deck. If it is a red card, he wins $1. If it is a black card between or including 2 and 10, he wins $5. if it is a black face card, he wins $10; and if it is a black ace, he wins $100. It costs $5 to play. What is the expectation of the game?
Answer by VFBundy(438) (Show Source): You can put this solution on YOUR website!
There is a 26/52 probability of picking a red card. This can be reduced to 1/2.
There is a 18/52 chance of picking a black card between 2 and 10. This can be reduced to 9/26.
There is a 6/52 chance of picking a black face card. This can be reduced to 3/26.
There is a 2/52 chance of picking a black ace. This can be reduced to 1/26.
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The cost to play the game is $5. So, this needs to be subtracted off of any winnings.
If you pick a red card you win $1 -- minus $5 for the cost of the game -- leaving your net "winnings" at -$4.
If you pick a black card between 2 and 10, you win $5 -- minus $5 for the cost of the game -- leaving your net winnings at $0.
If you pick a black face card, you win $10 -- minus $5 for the cost of the game -- leaving your net winnings at $5.
If you pick a black ace, you win $100 -- minus $5 for the cost of the game -- leaving your net winnings at $95.
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The expectation is as follows:
Picking a red card: 1/2 * -$4 = -2.00
Picking a black card between 2 and 10: 9/26 * $0 = $0.00
Picking a black face card: 3/26 * $5 = $0.58
Picking a black ace: 1/26 * $95 = $3.65
Add all of these together, and you have your expected value of the game:
-2.00 + $0.00 + $0.58 + $3.65 = $2.23
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