A person must pay $10 to play a certain game at the casino. Each player has a pr

Question

A person must pay $10 to play a certain game at the casino. Each player has a probability of 0.11 of winning $13, for a net gain of $3 (the net gain is the amount won 13 minus the cost of playing 10). Each player has a probability of 0.89 of losing the game, for a net loss of $10 (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?

Explanation / Answer

let x=net gain

x=3, with probability=0.11

x=-10, with probability=0.89

so probability distribution of x=net gaian is given as

f(x)=0.11, x=3

=0.89, x=-10

Expected Value for the player=E(x)=sum(x*p(x))=0.11*3+0.89*(-10)=-$8.57

Since the expected value is negative so , If a person plays this game a very large number of times over the years, we expect him/her to come out financially behind.

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