Question 1171872
There is a 26/52 probability of picking a red card.  This can be reduced to 1/2.
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There is a 18/52 chance of picking a black card between 2 and 10.  This can be reduced to 9/26.
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There is a 6/52 chance of picking a black face card.  This can be reduced to 3/26.
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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.
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If you pick a red card you win $1 -- minus $5 for the cost of the game -- leaving your net "winnings" at -$4.
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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.
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If you pick a black face card, you win $10 -- minus $5 for the cost of the game -- leaving your net winnings at $5.
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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:
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<i>Picking a red card:</i> 1/2 * -$4 = -2.00
<i>Picking a black card between 2 and 10:</i> 9/26 * $0 = $0.00
<i>Picking a black face card:</i> 3/26 * $5 = $0.58
<i>Picking a black ace:</i> 1/26 * $95 = $3.65
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Add all of these together, and you have your expected value of the game:
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-2.00 + $0.00 + $0.58 + $3.65 = <b>$2.23</b>