Deshaun has a deck of 10 cards numbered 1 through 10. He is playing a game of chance.
This game is this: Deshaun chooses one card from the deck at random. He wins an amount of money equal to the value of the card if an even numbered card is drawn. He loses $3.50 if an odd numbered card is drawn.
A) Find the expected value of playing the game.
(b) What can Deshaun expect in the long run, after playing the game many times?
(He replaces the card in the deck each time.)
A. Deshaun can expect to gain money.
How much per draw?
B. Deshaun can expect to lose money.
How much per draw?
C. Deshaun can expect to break even (neither gain nor lose money).
A) To find the expected value of playing the game, we need to calculate the average amount of money Deshaun can expect to win or lose per draw.
The probability of drawing an even number is 5/10 (there are 5 even numbers out of 10 total numbers), and the amount Deshaun can win in this case is equal to the value of the card drawn.
The probability of drawing an odd number is also 5/10, and the amount Deshaun can lose in this case is $3.50.
So, the expected value (E) can be calculated as follows:
E = (probability of even) * (amount won) + (probability of odd) * (amount lost)
E = (5/10) * (average value of even numbers) + (5/10) * (-$3.50)
The average value of even numbers can be calculated as the sum of even numbers divided by the count of even numbers: (2+4+6+8+10)/5 = 6
E = (5/10) * 6 + (5/10) * (-$3.50)
E = $3 - $1.75
E = $1.25
Therefore, the expected value of playing the game is $1.25.
B) In the long run, after playing the game many times, Deshaun can expect to win an average of $1.25 per draw.
C) Deshaun can expect to gain money, specifically $1.25, per draw.
To find the expected value of playing the game, we need to calculate the average value of the possible outcomes.
A) The probability of drawing an even numbered card is 5/10 (since there are 5 even numbered cards out of 10) and the value of the card is the same as the number on the card. Therefore, the expected value for drawing an even numbered card is:
(1/2) x (2 + 4 + 6 + 8 + 10) = 30/2 = 15
The probability of drawing an odd numbered card is also 5/10, and since Deshaun loses $3.50 in this case, the expected value for drawing an odd numbered card is:
(1/2) x (-3.50) = -1.75
The overall expected value is then:
(5/10) x 15 + (5/10) x (-1.75) = 7.5 - 0.875 = 6.625
Therefore, the expected value of playing the game is $6.625.
B) In the long run, after playing the game many times, Deshaun can expect to gain money. The expected value of $6.625 per draw indicates that, on average, he will make $6.625 for every game played.
C) Deshaun can expect to gain money per draw rather than breaking even or losing money. The positive expected value of $6.625 indicates that, on average, he will make money over the long run.
A) To find the expected value of playing the game, we need to calculate the weighted average of the possible outcomes.
Since there are 10 cards numbered 1 through 10 in the deck, the probability of drawing an even-numbered card is 5/10 = 0.5, and the probability of drawing an odd-numbered card is also 5/10 = 0.5.
The expected value can be calculated as follows:
Expected value = (Probability of winning * Amount won) + (Probability of losing * Amount lost)
Amount won = Value of even-numbered card = sum of even numbers from 1 to 10 = 2 + 4 + 6 + 8 + 10 = 30
Amount lost = $3.50
Probability of winning = Probability of drawing an even-numbered card = 0.5
Probability of losing = Probability of drawing an odd-numbered card = 0.5
Expected value = (0.5 * 30) + (0.5 * (-3.50))
Expected value = 15 - 1.75
Expected value = $13.25
So the expected value of playing the game is $13.25.
(B) In the long run, after playing the game many times, the expected value indicates the average outcome per draw. Since the expected value is positive (+$13.25), Deshaun can expect to gain money in the long run after playing the game many times.
(A) Deshaun can expect to gain an average of $13.25 per draw.
(B) Deshaun can expect to lose money in the long run, as the expected value is positive.
(C) Deshaun can expect to break even (neither gain nor lose money) on average per draw.