Nice -puzzling- and everyone else, this was too much for me, please help?

Question

Suppose that a certain game, that does not cost to play, involves drawing one marble from an urn. Further suppose that the urn contains 4 red marbles, 3 green marbles, 5 white marbles, and 6 black marbles. If you draw a red marble, you win $5. If you draw a green marble, you win $4. However, if you draw a white marble, you lose $3, and if you draw a black marble, you lose $2.

a.) What is the Expected Value of this game.
b.) Is the game fair?

(please explain in detail the answers someone?)

Answer 1

Figure the probability of each outcome first:
P(red marble) = 4/18 = 2/9
P(green marble) = 3/18 = 1/6
P(white marble) = 5/18
P(black marble) = 6/18 = 1/3

The expected winnings would just be:
2/9 * $5 + 1/6 * $4 + 5/18 * (-$3) + 6/18 * (-$2)

This works out to an expected winning of $0.277778.

An alternate way to figure this is:
[(4 * $5) + (3 * $4) + (5 * -$3) + (6 * -$2) ] / 18
= (32 - 27) / 18
= $5/18 = $0.277778

The game is better than fair, since you don't have to pay anything but you are expected to win more than $0. (Technically it is unfair since the expected return is not equivalent to what you pay, but I'm going to keep playing it because it is unfair to the house, not to me!)

Answer 2

Expected value = sum of (probability)*(cost)
=(4/18)(5)+(3/18)(4)+(5/18)(-3)+(6/18)(-2) (there are 18 total marbles)
=5/18

Since this has a positive expected value, this is an unfair game, but it is to the player's advantage (fair games have expected value 0)

Steve

Answer 3

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