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Monty Hall Problem

Let’s observe the hypothetical situation:

1) In multiple choice section the candidate chooses a random answer from a, b, and c (e.g. a).

2) 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%? 

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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%. 

Magina, the Antimage II Never ever give up!!

Samuraicode wrote:

Let’s observe the hypothetical situation:

1) In multiple choice section the candidate chooses a random answer from a, b, and c (e.g. a).

2) 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%? 

Hence only if he is instructed by somebody who knows the right answer beforehand, as explained in my previous comment 

Magina, the Antimage II Never ever give up!!

“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.

Samuraicode wrote:

“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 

Magina, the Antimage II Never ever give up!!

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).

wlfgngpck wrote:

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 

Magina, the Antimage II Never ever give up!!

Deal_Clincher wrote:

Samuraicode wrote:

“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 

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

Magina, the Antimage II Never ever give up!!

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.

Nenorr wrote:

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 : )

Samuraicode wrote:

Let’s observe the hypothetical situation:

1) In multiple choice section the candidate chooses a random answer from a, b, and c (e.g. a).

2) 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%? 

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 laugh

wlfgngpck wrote:

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).

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.

Edbert wrote:

wlfgngpck wrote:

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).

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.

But, the solution makes sense with three doors, too.

tickersu wrote:

Edbert wrote:

wlfgngpck wrote:

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).

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.

But, the solution makes sense with three doors, too.

I once saw a Prof explaining this problem to a class of undergrads, and after he was done, a student raised his hand and said “I disagree.”  And before the student was allowed to explain why, the Prof just said “This is math, not politics!  If you disagree, then you’re just wrong.”  Kinda mean, but was pretty funny. :)

Samuraicode is correct, under the specific conditions he describes.

Let’s say that before even reading the answer choices to a question, you pick A.  Then you look at the answer choices.  You don’t know whether A or C is correct, but you are 100% certain that B is wrong.  So you eliminate B.

When you initially guessed A, you had a 1/3 chance of A being correct.  That means there is a 2/3 chance the answer is not A.  Therefore, if you switch your answer to “Not A”, you have a 2/3 chance of being correct.  “Not A”  means either B or C.  But since you are 100% sure B is wrong, the only possible way “Not A” can be correct is for C to be correct.  Therefore, your chance of being correct increases to 2/3 if you switch from A to C.

However, it is important to note that this only works in the exact scenario described above.

If you read the answers, eliminate B as wrong, and then pick A from the remaining two answers, you will not increase your chance of being correct if you switch to C.  Because you are initially making your choice of A from only 2 options, either one has a 1/2 chance of being correct.

More important, if after first picking A you’re not 100% sure that B is wrong, you cannot be sure that you will improve your chances of being right by switching to C.

Finally, even if you first pick A and then eliminate B, it is often the case that you will have an informed opinion on whether A or C is more likely correct.  If you follow your informed opinion that A is more likely correct than C, you would probably be better off not switching to C in those cases.

So this approach does work in theory, but in practice you most likely would not be able to apply it consistently over a large enough sample size of questions on a single CFA exam to gain an appreciable advantage.  You would need to be 100% certain that one of the 2 choices you did not initially pick is wrong, and you would also have to ignore any sense you might have that your initial choice is more likely to be correct than the choice you could switch to.  And there would have to have enough of these questions for your results to reflect the statistical average probability.

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lol this is nonsense, the entire premise of the monty hall problem is the fact that the presenter knows exactly what the right answer is and will always give you new information after your initial choice is made. 

What you’re not considering is that if you suddenly are able to eliminate an answer later in the test after initially guessing you’re just as likely to eliminate your guess as you are one of the other two choices, versus monty hall which will always eliminate a wrong answer that was not your original guess. 

Example in exam if you always switch

One third of the time your original guess was right which means you will be wrong since you later switch, one third of the time your original guess is wrong and you eliminate that guess – this will lead to a conditional probability of .5 and the last 3rd of the time you will have guessed wrong, eliminated the other wrong answer, and therefore have a conditional probability of 1. 

(1/3)*(0) + (1/3)*(.5) + (1/3)(1) = .5

Example if you never switch (unless of course you later eliminate your original guess)

two thirds of the time you will guess wrong and have a .5 probability of later eliminating your original guess which leads to a conditional probability of .5 and one third of the time you guessed right which leads to a conditional probability of 1

(2/3)*(.5)*(.5) + (1/3)*(1) = .5