Answer to Question #124503 in Statistics and Probability for Dan Degrossa

Question

A person must pay 9$ to play a certain game at the casino. Each player has a probability of 0.11 of winning 15$, for a net gain of 6 (the net gain is the amount won 15$ minus the cost of playing 9$).

Each player has a probability of 0.89 of losing the game, for a net loss of 9 (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.

Expected Value = $

If a person plays this game a very large number of times over the years, do we expect him/her to come out financially ahead or behind?

ahead

behind

Answer 1

0.11is probability of winning and acquiring $15 and 0.89 is the probability of losing and losing 49.

Expected value = 0.11*$15 - $9*0.89 = -$6.36 thus since the expected value is a negative value if one plays this game many times over a number of years they will be behind.