Probability on a Multiple-Choice exam Options

[Paradox]

I am utterly convinced that this concept works. Don't even bother responding unless you thoroughly read through each scenario. If you disagree please PM and quote which part exactly you disagree with and your justification, so that we can discuss that part intelligently. If you make no attempt to do so I will ignore your rebuttal.
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Given:

You are taking a multiple-choice test. There are 3 incorrect answers in each problem and 1 correct answer.

Before you read each question you randomly guess.

Lets say your random guess turns out to be A

therefore:

25% chance the correct answer is among -> A
75% chance the correct answer is among -> B C D

In other words, there is a 25% chance you chose correctly. There is also a 75% chance you chose incorrectly and the answer is either B,C, or D. It is more likely you chose incorrectly.

Now we can take this information and apply it to different scenarios:

(I left out scenarios that would have identical percentages in the end.)

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Scenario 1

You know for sure B and C are wrong after reading the question. So now A or D could be the correct answer, but you are unsure which.

25% chance the answer is among -> A
75% chance the answer is among -> B C D

Conclusion:

25% chance the answer is among -> A
75% chance the answer is among -> D

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Scenario 2

You know for sure B is wrong after reading the question. So now A, C, or D could be the correct answer, but you are unsure which.

25% chance the answer is among -> A
75% chance the answer is among -> B C D

therefore:

25% chance the answer is among -> A
75% chance the answer is among -> C D

Conclusion:

25% chance the answer is among -> A
37.5% chance the answer is among -> C
37.5% chance the answer is among -> D

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Scenario 3

You know for sure A is wrong after reading the question. So now B, C, or D could be the correct answer, but you are unsure which.

25% chance the answer is among -> A
75% chance the answer is among -> B C D

therefore:

0% chance the answer is among -> A
100% chance the answer is among -> B C D

therefore:

100% chance the answer is among -> B C D

Conclusion:

33.33% chance the answer is among -> B
33.33% chance the answer is among -> C
33.33% chance the answer is among -> D

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Scenario 4

After reading the question you are unable to eliminate any incorrect answers. So now A, B, C, or D could be the correct answer, but you are unsure which.

25% chance the answer is among -> A
75% chance the answer is among -> B C D

therefore:

100% chance the answer is among -> A B C D

Conclusion:

25% chance the answer is among -> A
25% chance the answer is among -> B
25% chance the answer is among -> C
25% chance the answer is among -> D

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Scenario 5

After reading the question you eliminate A, B, and C.

25% chance the answer is among -> A
75% chance the answer is among -> B C D

therefore:

0% chance the answer is among -> A
100% chance the answer is among -> B C D

therefore:

100% chance the answer is among -> B C D

Conclusion:

100% chance the answer is among -> D

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Conclusion

From this we gather that Scenario 1 is similar to the Monty Hall problem. The answer is much more likely to be D.

Scenario 2 is better than if we eliminated incorrect answers first and ended up with 33.33% among A C D.

We would have the same conclusion for Scenario 3,4 and 5 regardless of using this concept or not.

Therefore, this concept actually helps us. If we did not use it we would not have Scenarios 1 and 2. Scenarios 1 and 2 give us a slightly better idea of what the answer might be.

[coolmaninsano]

This is all bullshit.

[xlro]

The probabilities you give in your scenarios are not correct, because you can not rule out your Scenarios 3 and 5 from hapenning. The mistake you make is that you don't consider with which probablities you reach Scenario 1-5 (+ I think this grouping is not the best, if you want to calculate them). Looking at how and with what chance you reach which configuration is essential here, please try to do an analysis for n=4, and you should notice the problem.

see also: http://harddrop.com/forums/index.php?showt...32743&st=90

A lot of discussion, since I last checked this thread - I found this to be a nice question actually.
(to all calling "offtopic": just consider this a probability theory homework ;) )

The imo only interesting scenario to look at is the one that paradox was arguing about a couple times: The answers you rule out for sure are all in the non-chosen group and are not the correct answer itself. (I)
I found some posts actually to go a bit besides that point. E.g. increasing the number of options ofc decreases your chances, since you actually update a hidden assumption, with each increase of the total number of options. This assumption (being able to rule out all but two options) might be still reasonable for n=3,4 and a normal test taker (that is what we want to look at), but surely not for n=10 or more. So all this shows is that a test becomes extremly hard if the number of options per question is huge and you just have to rely on random guessing.

Nevertheless I'll try to maybe convince paradox that this method does not increase your chances, even if you only look at the cases when you have reached (I). The problem with this method lies in the different ways you can reach (I).

Let's use a little restricted example setup to make the problem more graspable: 3 options per question, you initially pick one option, you then always can rule out exactly one option each time, and the rule-out you perform is fail-safe, meaning you never pick the correct answer, cause you're 100% sure your ruled-out option is not correct. I.e. the rule-out is uniformly over all n-1 (=2) wrong answers. Your strategy then is: If we reached (I), we always switch our pick.

There are 3 scenarios now:

• A) Your first random pick is the correct answer. (p1 = 1/3)
  This means after you rule-out one of the two other options, you reach (I), 100% of the time.
  => using the strategy (switch pick), gets you a wrong answer with p2 = 1 (always)
  total for this case: pa = p1 * p2 = 1/3
• B) Your first random pick is not correct. (p1 = 2/3)
  50% of the time you rule-out the wrong answer, which is not your pick and we reach (I). (p2 = 1/2)
  => using the strategy (switch pick) always gets you the right answer
  total for this case: pb = p1 * p2 = 1/3
• C) Your first random pick is not correct. (p1 = 2/3)
  50% of the time you rule-out the wrong answer which is your pick and we do not reach (I). (p2 = 1/2)
  Now you have the situation that you have a 50/50 chance of picking between two left over answers. (and no strategy for that)
  total for this case: pc = p1 * p2 = 1/3
  and pc = pc1 + pc2, with
  pc1 = 1/6 (C1 = chose correct answer of the two remaining)
  pc2 = 1/6 (C2 = chose wrong answer of the two remaining)

So all in all you always have a 50/50 chance for the right answer (using the switching strategy or not), under the assumption we made (always ruling out one wrong answer), which makes sense, if we formulate (I) in the first place.

This is ofc not a complete proof for this relatively general problem, since it uses many assumptions (which I all tried to state before in the example setup description) - but I think it shows the main logic behind this, and where the error in thinking lies, i.e. multiple paths to (I).

If you have doubts, just try to apply the same analysis for n=4 (your initial setup). You will see that you also end up with 50/50 for two rule-outs or 1/3 chance for 1 rule-out respectively - essentially confirming that the initial random pick changes nothing for your chances, after you add your gained information from your rule-outs, apart from the normal improvement you get because of them. This is because having the chance that you rule-out your initial random pick, splits the option for "initial pick is not the correct answer" into two sub-branches (see: B and C), in contrast to the Monty Hall Problem.

btw, if you want quick practical "proof" and you know any simple programming / scripting language, write yourself a simple simulation of the problem (handful lines of code), and let it run for a couple of iterations (1 mil.+). You will directly see that the chances to get the right answer are with or without switching strategy always 1/(n-r), where n is the number of options per question and r the number of options you can rule out 100%.

[DAS44]

You really do love trolling don't you?

[Paradox]

Don't even bother responding unless you thoroughly read through each scenario. If you disagree please PM and quote which part exactly you disagree with and your justification, so that we can discuss that part intelligently. If you make no attempt to do so I will ignore your rebuttal.

[coolmaninsano]

We already discussed how this is not true.

[xlro]

I read through your scenarios, told you where you should start to look to correctly model the situation and see no need to PM anything, that's also not the point of a forum imo.

If you really want to investigate that problem yourself and are open to the possibility that you have a mistake hidden somewhere I'd suggest that you start by writing a little simulation for yourself and then try to explain the results / cold numbers (that don't lie ) to yourself, first. It's only a hint - I'm def. not here to have a heated argument and get outraged over you not wanting to accept my analysis or anything, just here to help.

[Maii04]

this is a tetris forum
not a probability one

[coolmaninsano]

Maii there's a reason for the GENERAL subsection.

[KeroKai]

Either I'm missing something major or... I should have stopped reading after scenario 1.

Edit: Right... I re-read the Monthy Hall Problem so can sort of understand better now where you are coming from.

Would you be suggesting that in Scenario 1 situations: It'd be more advantageous to randomly select one, rule out the options and then switch then it is to examine the problems, and then decide between the final two options. Giving the sensation it is 50-50.

Anyhow...
Applying to real life situations when people have vague ideas of what could be the answer. Surely this sort of knowledge would end up as going against the idea that first impressions/intuition is correct. There's also the general problem that people tend to dislike swapping... I suppose part of this could be explained that a person swapping and getting the incorrect answer/prize will always be more disappointed than the individual who doesn't swap but loses.

Hm... I wonder if this idea is true. It'd be quite an interesting dissertation topic to study in psychology.
Then again, it seems to be quite a simple idea where it's obvious that people in general dislike change for the fear that things can go wrong after they have made their original decision.