Question

a card is drawn from a standard 52 - card deck. calculate the expected value for the game. a player must pay 6 dollars to play the game, which must be subtracted from the winnings. if a diamond is drawn, the player wins 24 dollars, otherwise, they lose their 6 dollars. calculate the price that would make the game fair. the expected value of this game is square dollars. (round to the nearest cent as needed.) the fair price of this game is square dollars. (round to the nearest cent as needed.)

Explanation:

Step1: Calculate the probability of drawing a diamond

In a standard 52 - card deck, there are 13 diamonds. So the probability \(P(\text{diamond})=\frac{13}{52}=\frac{1}{4}\), and the probability of not drawing a diamond \(P(\text{not diamond}) = 1-\frac{1}{4}=\frac{3}{4}\).

Step2: Calculate the expected value

The formula for expected value \(E(X)=\sum_{i}x_{i}P(x_{i})\).
If a diamond is drawn (\(x_1 = 24 - 6=18\) dollars, because the player paid 6 dollars to play), and \(P(x_1)=\frac{1}{4}\).
If a non - diamond is drawn (\(x_2=- 6\) dollars), and \(P(x_2)=\frac{3}{4}\).

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Wait, no. Let's correct. The formula for expected value when the cost to play is \(C\) and the winning amount is \(W\) (if event \(A\) occurs) and loss amount is \(-C\) (if event \(A\) does not occur).
The expected value \(E(X)=(W - C)\times P(A)+(-C)\times(1 - P(A))\)
Here \(W = 24\), \(C = 6\), \(P(A)=\frac{13}{52}=\frac{1}{4}\)

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For a fair game, the expected value is \(0\). Let the price to play be \(x\).
If a diamond is drawn, the net winning is \(24 - x\), and if a non - diamond is drawn, the net winning is \(-x\)

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Set \(E(X) = 0\)

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Answer:

The expected value of this game is \(0\) dollars.
The fair price of this game is \(6\) dollars.