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Question 1200328: A person must pay $4 to play a certain game at the casino. Each player has a probability of 0.12 of winning $15, for a net gain of $11 (the net gain is the amount won 15 minus the cost of playing 4).
Each player has a probability of 0.88 of losing the game, for a net loss of $4 (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?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
X = net winnings from the player's perspective
We start off with this probability distribution
Let's introduce a third column labeled X*P(X)
We'll multiply each X and P(X) value along a given row.
| X | P(X) | X*P(X) | | 11 | 0.12 | 1.32 | | -4 | 0.88 | -3.52 |
Add up the values in that third column
1.32 + (-3.52) = -2.20
The player expects to lose $2.20 on average per game.
Over the course of a large number of games, the player is expected to lose a lot of money.
For instance, if s/he plays 1000 games, then that player is expected to lose around 2200 dollars (since 2.2*1000 = 2200). This is an average.
This makes sense because the odds are rigged in favor of the casino. Otherwise the casino wouldn't make any money.
Answer: Expected value = -2.20
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