Unformatted text preview: , for all the event s we will charge them $x. Hence, EV[charge] = Pr{ all }*charge price = 1 *x In order to gain the profit, EV[charge] > EV[pay] 1*x > 1/6*100 x > 100/6 = 16.67 Thus, the charge should be more than $ 16.67 and the minimum value is $ 16.67 . B. You allow the person to roll one die before paying you, and if they like the number they pay you and roll the second die. Now what minimum value of xx results in a profitable operation? From the previous question, the set of outcomes for sum equal 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} One can notice that no matter what the first dice is, the second dice has to be one of 6 possible outcomes, therefore Pr{sum = 7} is still 1/6 . EV[charge] > EV[pay] x > 1/6*100 x > 16.67 Thus, the charge should be more than $ 16.67 and the minimum value is $ 16.67 ....
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 Spring '09
 Probability theory, Dice, minimum value

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