Question

A person must pay $ $ 8 to play a certain game at the casino. Each player has a probability of 0.21 of winning $ $ 14, for a net gain of $ $ 6 (the net gain is the amount won 14 minus the cost of playing 8). Each player has a probability of 0.79 of losing the game, for a net loss of $ $ 8 (the net loss is simply the cost of playing since nothing else is lost). What is the Expected Value for the player (that is, the mean of the probabiltiy distribution)? If the Expected Value is negative, be sure to include the "-" sign with the answer. Express the answer with two decimal places.

Answer 1

From the given data, the following Table is calculated:

Event	Value to the player (x)	Probability (p)	xp
Winning	14-8=6	0.21	6 X 0.21 = 1.26
Losing	- 8	0.79	- 8 X 0.79 = - 6.32
Expected Value			1.26 - 6.32 = - 5.06

So,

Expected Value for the player = $ - 5.06

                                              (loss)