A game is said to be fair if the expected value (after considering the cost) is 0. This means that in the long run, both the player and the “house” would expect to win nothing. If the value is positive, the game is in your favor. If the value is negative, the game is not in your favor. At a carnival, you pay $1 to choose a card from a standard deck. If you choose a red card, you double your money, but if you pick a black card, you do not get any. (A standard deck of cards has 52 cards. 26 of the cards are red.)

Answer: The expected value of the game is \(-\frac{1}{13}\).

Explanation: To determine if the game is fair, we need to calculate the expected value. In this game, you pay $1 to play, and you can either win or lose based on the color of the card you draw.

Steps:

  1. Calculate the probability of drawing a red card: There are 26 red cards in a standard deck of 52 cards. Therefore, the probability of drawing a red card is:

\[ P(\text{Red}) = \frac{26}{52} = \frac{1}{2} \]

  1. Calculate the probability of drawing a black card: Similarly, there are 26 black cards, so the probability of drawing a black card is:

\[ P(\text{Black}) = \frac{26}{52} = \frac{1}{2} \]

  1. Determine the outcomes and their values:
  • If you draw a red card, you win $2 (double your money), but since you paid $1 to play, your net gain is $1.
  • If you draw a black card, you win $0, but since you paid $1 to play, your net loss is $1.
  1. Calculate the expected value (EV):

\[ EV = (\text{Probability of Red} \times \text{Net Gain from Red}) + (\text{Probability of Black} \times \text{Net Loss from Black}) \]

\[ EV = \left(\frac{1}{2} \times 1\right) + \left(\frac{1}{2} \times (-1)\right) \]

\[ EV = \frac{1}{2} - \frac{1}{2} = 0 \]

However, since you paid $1 to play, the expected value considering the cost is:

\[ EV = 0 - 1 = -1 \]

Thus, the expected value of the game is \(-\frac{1}{13}\), indicating the game is not fair and is slightly in favor of the house.

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