Question 1202738
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A classmate offers a game for you to play. You roll a single six-sided die, 
and if the roll is 3 or higher, you win $3. If you roll a 1 or a 2, you owe him $2. 
Little do you know, the die is loaded. The probability that you'll roll a 1 is 0.53, 
and the rest of the values have equal probabilities. What is the expected value of this game?
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From the condition, the probability density distribution function is 

    P(1) = 0.53;  P(2) = P(3) = P(4) = P(5) = P(6) = {{{(1-0.53)/5}}} = 0.094.


Therefore, the expected value of the game is

    -2*P(1) - 2*P(2) + 3*(P(3)+P(4)+P(5)+P(6)) = -2*0.53 - 2*0.094 + 3*4*0.094 = -0.12 dollars.


It means that you will lose 12 cents, in average, in each game, if you play this game many times.
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Solved, with complete explanations.