Let’s observe the hypothetical situation:
In multiple choice section the candidate chooses a random answer from a, b, and c (e.g. a).
If candidate knows exactly which answer is wrong from the two choices left (e.g. b)
Then is it correct to assume that chances of getting the answer right by switching from a to c will increase from 33% to 67%?
No. Because how could he know B is also incorrect? If he knows A and B is incorrect he can be 100% confident in C. But if he choses to go with A and later is told B is incorrect his chances increase from 1/3 to 1/2 by switching to C…
It’s a precondition that he/she should know for sure that one of the remaining answers is incorrect.
Just work out the cases. In any event, yes in the situation you described (which is the standard Monty Hall problem) the decision to switch wins 2/3 of the time.
Note that this is a common interview question for quantitative finance roles, so probably a good idea to work through the logic of why this is the answer.
in monty hall you only get to know that one of the other choices are goats. Not true when you first choose A or C. You have to choose one and get info that another is a goat.
In Monty Hall you do exactly what I’ve described. First you choose randomly (e.g. c), then if you know the wrong choice (e.g. a) from the other two choices, you eliminate that choice (a) and switch to the other one (b) and the probability of getting the right answer increases.
I can see how that makes sense because the chances of you getting the random choice correct is 33% which means when you do the switch it will only be incorrect if you chose the correct answer the first time so it leaves you a 67% chance of getting it right after the switch.
Now im wondering if you do a complete guess and just switch randomly would your chances increase
67% * 50% = 33.5%…slight increase
In MH there are two i/p sources. One who does not know, second one who knows the answer and he deliberately gives a hint that one of the options that is left is incorrect.
That implies
Case 1 :If you change your option but initially you were incorrect. Prob of Being Correct - 1
Case 2 :If you change your option but initially you were incorrect. Prob of Being Correct - 1
Case 3 :If you change your option but initially you were correct. Prob of Being Correct - 0
Total Prob of being correct = 2/3
But, when you eliminate one of the options and select between one of the rest, your prob of picking up the right answer is 50%.
Hence only if he is instructed by somebody who knows the right answer beforehand, as explained in my previous comment
“In MH there are two i/p sources. One who does not know, second one who knows the answer” this condition is perfectly substituted by the fact that the candidate SHOULD KNOW EXACTLY WHICH OF THE REMAINING CHOICES IS WRONG.
Then his initial probability of selecting the wrong answer is not 33%
He knows Option C is incorrect. He randomly picks C. He will choose one of the rest -> 1/2
He knows Option C is incorrect. He randomly picks A. -> When A is incorrect -> 1, when A is correct -> 0
He knows Option C is incorrect. He randomly picks B. -> When B is incorrect -> 1, when B is correct -> 0
=> Total Prob (5/2)/5 = .5
I’m always amazed that this question generates so many arguments. Honestly it’s a pretty simple problem (which is why it’s a common interview question).
YEP simple stuff
In first case if there is an outsider who could prompted him that out of which A & B one option is incorrect prob of being correct would have gone up to .67
Ouch, seems like you broke the exam. Now everyone can get the necessary 67%.
Nope. Monty Hall is a sequential game where you get new info after initial choice. In a multiple choice question where you know one of three answers are incorrect you go from 1/3 to 1/2. Only other option is if you introduce new info between steps. For instance you know the candidate in front of you is always wrong, and see his choice after you made yours. (Obviously violation and as such a unrealistic scenario).
You make the answer selection such a game though. Without looking at the question, select an answer at random. 33% of the time it will be correct one, 67% incorrect one. Now you read the question and eliminate one answer you are sure is incorrect. Then, switch your answer to the remaining answer. Assuming you have correctly identified the wrong answer, you will have increased the odds in your favor. Basic Monty.
The assumption there is pretty major, though. Candidates who lack knowledge will discard the correct answer a large % of the time.
about 50% of the time : )
I’ve always thought it’s a bit silly the CFA exams only provide three possible answer choices. Imagine taking a test where every question has at least five choices, and sometimes over 15, with many choices separated by one or two peripheral, yet important, facts or concepts.
I think it would still benefit the candidate even if they don’t perfectly know which is correct or incorrect of the remaining. If they choose at random, then read the question and eliminate something not at random (and they have a decent track record), then, they will probably benefit to an extent.
The question is, how will the candidate use a random selection process during the exam? The human mind isn’t capable of this. Do the financial calculators have the ability to generate random numbers between some bounds? That would be useful, set it to an integer on [1,3] and pick based on that.
Edit: So the BA II plus has a RNG over (0,1) on a uniform distribution, so basically, just round to 2-3 decimal places and define cutoffs. I wonder how this would perform in a simulation of the above scenario 
it’s because the example scenario only has 3 doors. it’s much easier to understand the concept if the example goes from 3 to 100 doors.
there are 100 doors, 99 goats and 1 car, and the contestant picks a door number. the host then flips open 98 doors that aren’t the chosen door and another door, revealing that behind the 98 doors are goats. now suddenly the solution makes sense.