Question

Debra has a deck of 10 cards numbered 1 through 10. She is playing a game of chance. This game is this: Debra chooses one card from the deck at random. She wins an amount of money equal to the value of the card if an even numbered card is drawn. She loses $6 if an odd numbered card is drawn.

a.Find the expected value of playing the game

b.What can Debra expect in the long run, after playing the game many times?

(She replaces the card in the deck each time.) (what will she gain or lose)

Answer 1

We will see that the expected value is $0, which means that after a large number of games, Debra should expect to not win or lose any money.

How to get the expected value?

For an experiment with N outcomes {x₁, x₂, ..., xₙ}, each one with probability {p₁, p₂, ..., pₙ}, the expected value is:

EV = x₁*p₁ + ... + xₙ*pₙ

In this case, the outcomes are:

• x₁ = drawing an odd card = -$6
• On the 10 card deck, there are 5 odd cards, then the probability is: p₁ = 5/10 = 1/2.

• x₂ = drawing a 2 = $2
• There is only one 2 card on the deck, so the probability of drawing it is: p₂ = 1/10

• x₃ = drawing a 4 = $4
• p₃ = 1/10

• x₄ = drawing a 6 = $6
• p₄ = 1/10

• x₅ = drawing a 8 = $8
• p₅ = 1/10

• x₆ = drawing a 10 = $10
• p₆ = 1/10

So the expected value will be:

EV = (-$6)*(0.5) + ($2 + $4 + $6 + $8 + $10)*(1/10)

EV = 0

b) The expected value is 0, this means that over a large number of repetitions, Debra should expect to not win nor lose any money.

If you want to learn more about expected value you can read:

https://brainly.com/question/15858152