Actuarial Outpost Optimal strategy for complicated true/false questions
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#1
01-27-2019, 02:58 PM
 thefrumactuary CAS Join Date: Feb 2016 Location: New York City Studying for IFM March 2019 College: Stony Brook University Alumni Favorite beer: Bayesian Posts: 15
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

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
01-27-2019, 03:04 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 6,147

I didn't read any of that but.... what about the choices I and III only or II and III only?
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#3
01-27-2019, 03:17 PM
 thefrumactuary CAS Join Date: Feb 2016 Location: New York City Studying for IFM March 2019 College: Stony Brook University Alumni Favorite beer: Bayesian Posts: 15

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
01-27-2019, 03:22 PM
 thefrumactuary CAS Join Date: Feb 2016 Location: New York City Studying for IFM March 2019 College: Stony Brook University Alumni Favorite beer: Bayesian Posts: 15

Quote:
 Originally Posted by Actuarialsuck I didn't read any of that but.... what about the choices I and III only or II and III only?
Interesting: the math would come out to
A : 0.3103
B : 0.1379
C : 0.1379
D : 0.2069
E : 0.2069
Total : 1

 Tags #actuary, #ifm, #logic, #soaprobs