HW2_Sol09

HW2_Sol09 - for all the event s we will charge them \$x...

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TELCOM 2120: Network Performance Fall 200 9 Homework 2: Solution Problem 1 (10 points) You have two dice. You charge a person \$xx to roll the dice, if the dice total seven you pay them \$100. A. What minimum value of xx results in a profitable operation for you? Basically, we can analyze the outcome for this gambling as two events as follows: 1. sum of two dices = 7 - Finding the probability for this event: The set of outcome for sum equal 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} from totally 36 possible outcomes. Thus, Pr{sum =7}= 6/36 = 1/6 2. sum of two dices 7 - Finding the probability for this event: Pr{sum 7}= 1- Pr{sum = 7}= 5/6 Next, we have to find the expected value (EV) for each event as follows: According to the event 1(sum = 7), we need to pay them \$100. Hence, EV[pay] = Pr{sum =7}*pay price = 1/6*100 However
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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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