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### Big Data Engineer, New Hampshire at Fidelity Investments was asked...

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Dec 15, 2011
 Let's play a game - I give you a 12 sided die and will pay you whatever the die lands on. If you are unhappy with the roll, you can choose to roll another two 6-sided dice and I wil pay you the sum of the two dice. How much are you willing to pay to play this game.11 Answers1/2 expected value of rolling one die + 1/2 expected value of rolling two dice = \$8.25The expected value of the second roll is 8.5. I would choose to roll the second pair if the first die lands less than or equal to 8. The expected value of this strategy is 8.5 *2/3 + 9/12 +10/12+11/12 +12/12. This is 9.1667.That is wrong. The expected value of the second roll is 7. So I would choose to roll the second if I get less than 7 in the first die. The expected value of this strategy is 8.25.Show More Responses12 sided die: Expected Val 13/2=6.5 2 6-sided die: (2*1+3*2+4*3+5*4+6*5+7*6+8*7+9*4+10*3+11*2+12*1)*1/6*1/6=268/36=77/9 First time rolling side 12, will stop to roll the 2 6-side one if the number is >=9; If the number is <=8, then roll the second dies. (9+10+11+12)/12+(8/12)*77/9=9.2None of these answers are correct It says that if you roll the second, then whatever comes up will be your payment Strategy: Roll the 2 D6 in the case where you see a number on the D12 that is less than 7 because E(2*D6) = 7. This happens half the time so it will be .5 * E(D12) + .5 * E(2*D6) = 6.75The answer is .5*9.5+.5*7=4.75+3.5=8.25. the way to get it is you roll again if you get less than 7 on the first roll.@Christine your approach is right, the only thing is that if you have 2 6-sided die then the prob of outcome of 8 is 5 ((2,6), (6,2), (3,5), (5,3) and (4,4)). Thus you will get (2*1+3*2+4*3+5*4+6*5+7*6+8*5+9*4+10*3+11*2+12*1)*1/6*1/6 = (14*(1+2+3+4+5) +7*6) /36=14*18/36=7. Overall we get 7*7/12 + (8+9+10+11+12)/12=(49+50)/12=99/12=8.25Suppose your strategy is to to be "happy" if the first roll is 7 or higher and otherwise go on to the second round. The expected value is: 1/2 * [(7 + 8 + 9 + 10 + 11 + 12) / 6] + 1/2 * [(1 + 2 + 3 + 4 + 5 + 6) / 6 + 3.5] = 8.25.isn't answer is obvious? if roll less or equal than 6, roll again; otherwise accept it so expected value is 1/2*6.5 (if roll again)+1/2*9.5 (9.5 is the expected value of 7, 8, 9, 10, 11, 12) =16/2=8@sys: the expected value of rolling again is E(6-sided) + E(6-sided) = 3.5+3.5 = 7. .5 * 7 + .5 * 9.5 = 8.25E(rolling 6 dice twice ) = 2*E(6 dice one) = 2*(1+6)/2 = 7 So I reroll only if 12 dice gave me 6 or less. So with this strategy Let X payoff then E(X) = p(stick with 12)*E(X|sitck with 12 die) + p(reroll)*E(X|reroll) = 1/2 * (7+12)/2 + 1/2 * 7 = 4.75 + 3.5 = 8.25

May 14, 2011