Using the expected value, it is found that a price of $6 should be charged to make it a fair game.
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- The game is fair if the expected value is 0.
- The expected value is given by each outcome multiplied by it's probability.
- A probability is the number of desired outcomes divided by the number of total outcomes.
In a standard deck, there are 52 cards.
The cost of x.
There are 4 Aces in a standard deck, thus [tex]\frac{4}{52}[/tex] probability of earning 26 - x.
There is 1 King of Hearts in a standard deck, thus [tex]\frac{1}{52}[/tex] probability of earning 204 - x.
There are 4 Jacks and 4 Queens in a standard deck, thus [tex]\frac{8}{52}[/tex] probability of winning 13 - x.
The other 39 cards you lose, thus [tex]\frac{39}{52}[/tex] probability of losing x.
Considering we want the expected value to be 0.
[tex]\frac{4}{52}(26 - x) + \frac{1}{52}(204 - x) + \frac{8}{52}(13 - x) - \frac{39}{52}x = 0[/tex]
[tex]\frac{4(26 - x) + 204 - x + 8(13 - x) - 39x}{52} = 0[/tex]
[tex]104 - 4x + 204 - x + 104 - 8x - 39x = 0[/tex]
[tex]52x = 312[/tex]
[tex]x = \frac{312}{52}[/tex]
[tex]x = 6[/tex]
$6 should be charged.
A similar problem is given at https://brainly.com/question/24905256
