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#1




Optimal strategy for complicated true/false questions
The goal of this thread is to determine that there is an optimal strategy for guessing on multiple true/false questions
Say a question says I. Yellow has 6 letters II. Red has 3 letters III. Blue has 5 letters A) I only (not II and not III) B) II only C) III only D) I and II only E) All are true Assuming (proof by contradiction follows) that each answer choice is 20 % correct Now A,D,E > I is true... P(I) = 60 % Now B,D,E > II is true... P(II) = 60 % Now C,E > III is true... P(III) = 40 % Assuming I,II, and III statements are independent Then P(I)(1  P(II))(1  P(III)) = choice A and P(I)P(II)P(III) = choice E based on these probabilities: A) 60 % * (1  60 %) * (1  40 %) = A : 0.144 B : 0.144 C : 0.064 D : 0.216 E : 0.144 Total: 0.712 The exam writers are stuck with 5 choices, so some combinations of {A,B,C} such as {} (not A not B not C) are left out. These excluded probabilities from the sample space result in a Total less than 1. Scaling the probabilities to add to 1: P(A*) = P(A)/total = A : 0.2022 B : 0.2022 C : 0.0899 D : 0.3034 E : 0.2022 Total : 1 Therefore by proof of contradiction, there exists an optimal strategy. Just by studying the answer choices, you can come up with a strategy that gets the question right more than 20 % of the time. I would be interested to see if experience proves this right, if we had a large test bank with answers. 
#2




I didn't read any of that but.... what about the choices I and III only or II and III only?

#3




There would still exist an optimal strategy if:
A) I B) II C) III D) I and II E) I and III Assume A,B,C,D,E equal chance of 20 % each P(I) = 60 %, A,D,E P(II) = 40 % B,D P(III) = 40 % C,E And the math that follows would result in at least one choice being higher 
#4




Quote:
A : 0.3103 B : 0.1379 C : 0.1379 D : 0.2069 E : 0.2069 Total : 1 
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#actuary, #ifm, #logic, #soaprobs 
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