Hello, I am not very good at this type of stuff. I thought up something and I was wondering if someone can look over it and tell me if my justifications are correct or not. I am not very familiar with how things are written out when referring to probability.

If there is something wrong with my justifications try to put it in a way that makes sense to me.

If these conclusions are correct, please put it in a standard way that it would be written.

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.)

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

A lot of that looks wrong to me. For instance in scenario 1 since you eliminate B and C and are only left with A and D there is a 50% chance for either. You should read up on conditional probability.

Another way to see why your way was wrong is that there is absolutely no reason why you need to split it up like
25% -> A
75% -> B,C,D
beforehand and could have just as easily split it up like
25% -> A
25% -> B
25% -> C
25% -> D
which if you then apply your method should give you the correct results each time.

A lot of that looks wrong to me. For instance in scenario 1 since you eliminate B and C and are only left with A and D there is a 50% chance for either. You should read up on conditional probability.

I'm afraid I'll have to correct you there Dason (there's a first, I never thought would happen ).

Maybe I can explain it easily with this (Game Theory):
You're on a gameshow, you have four doors, only one has a prize in it. You choose one (now called group A), knowing there is a 75 percent chance the prize was in one of the other 3 doors (now called group B). Oke, so now we have 2 groups one group A 25% winning chance and group B with a combined 75% winning chance. Now the game show host opens two empty doors (in B) and gives you the chance to switch. [At this moment every thing changes, and you should now think in terms of 'strategies' and not in terms of doors and respective probabilities] - Great! the sucker just gave you a chance to switch to the 75% win chance group.

Key is a game theoretical approach of thinking in strategies. Its very useful for instance to define all outcomes of the two strategies (lets look at all outcomes after you have chosen 'A'). The 'fitness' of the switch strategy is 3 x higher:

As you can see this is very reminiscent of the Monty Hall problem. Logically a nasty one. I've been mistaken before with these.

You should be able to prove this one using bayes theorem as shown on the wiki (anyone interested in adapting the proof for this one?).

The wiki reminded me of a simulation I once made like this for an evolutionary strategy problem during my BSc years.

I adapted it to this problem just for the fun of it (in R code):

Code:

# all possible answers
test=c('A','B', 'C', 'D')
# the simulation program, strat = your strategy either switch or stay
simtest=function(N=10000,strat='switch'){
#vector to say results
results=rep(NA, N)
for(i in 1:N){
# assign a random correct choice
correct=sample(test,1)
# assign a random choice
chose=sample(test,1)
# id wrong answers
false=test[which(test!=correct&test!=chose)]
# id switch possibilities
residuals=test[which(test!=false[1]&test!=false[2])]
# which strategy
if(strat=='switch'){chose=residuals[which(residuals!=chose)]}
results[i]=ifelse(chose==correct,1,0)
}
results}
# running sims
a=simtest()
b=simtest(strat='stay')
#plotting results
barplot(c(sum(a),sum(b))/10000,names.arg=c('switch','stay'),ylab='win chance',ylim=c(0,1))

I have a lot of people that disagree with me about this. Actually everyone I have mentioned this to disagrees with me.

disagreements:

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.

I can see what the code does easily, the modeling error is here:

# id wrong answers
false=test[which(test!=correct&test!=chose)]

this is NOT what we do when we take a test, and that is exactly the difference between taking the test and a Monty Hall Problem where some third entity with all the knowledge removes doors/options, but NEVER our initial choice (your Scenario 3 and 5).

since this CAN happen, the correct line would be:

false=test[which(test!=correct)]

He also modeled the situation in C. I don't know if you know that language or not.

I have a lot of people that disagree with me about this. Actually everyone I have mentioned this to disagrees with me.

Monty Hall problems are famous for their counter intuitive nature. There are records of Nobel prize winners writing against correct answers in print (making complete asses out of themselves). Oh If only the Reverend was still here to preach!

Plus, Who are you quoting? Is it a typo. Or are these cross posts from other fora?

this is NOT what we do when we take a test, and that is exactly the difference between taking the test and a Monty Hall Problem where some third entity with all the knowledge removes doors/options, but NEVER our initial choice (your Scenario 3 and 5).

Logically and thus mathematically it is (highly) analog to the Monty Hall problem, I will even go so far as to say it IS a Monty Hall problem (with certainty). The fancy story around it can change, don't let that fool you, here the ground rules are the same. Its a non argument that a third entry reveals a false choice (as you can see I'm only considering scenario 1, the rest are not as interesting) and in Scen 1 you also NEVER think your initial choice is false. So as you explained it: you first make a random choice, then see that two answers are definitely wrong (ergo the third entity is completely irrelevant here) and your random one is not one of them, now you need to decide whether to switch. Its classic Monty Hall probability problem. See also the Bertrand problem.

since this CAN happen, the correct line would be:

false=test[which(test!=correct)]

He also modeled the situation in C. I don't know if you know that language or not.

No I did not model sit C. I modeled scenario one, there are no mistakes in the code. By changing it to "false=test[which(test!=correct)]" it became a scenario closer to 3 (or C as you call it). The language is R, as clearly specified in my post.

Yeah I posted this on another forum. I posted it in a homework help thread. The forum doesn't actually specialize in this that is why I went here. Most people called me an idiot/stupid/troll. Not one person agreed with me. If you would be kind enough to help my side of the argument that would be nice.

Its a tetris forum, go figure! From the first response you can see the general opinion. That is a separate thread than the homework help thread (I made its own thread because it was getting pretty crazy)

*Oh and I meant that he wrote out the situation in C (the programming language) in a way he thought was correct

I probably should have read that initial post a little better. I skimmed the first part and then looked at scenario 1 without the added background that you guess at the answer before looking at the question which does turn it basically into a Monty Hall problem.

Whelp, I guess I looked like an ass. On the plus side TheEcologist wasn't as foolish as I and did an really good job explaining it.

Don't worry about it, 50 other people did it to me before you.

I actually reached a conclusion to this dilemma!

For instance in scenario 1 since you eliminate B and C and are only left with A and D there is a 50% chance for either

I bet you will be very interested to hear that it does indeed end up being 50%. Not because there are only two answers left but if you factor in the chance that you will eliminate your initial guess, you end up with 50%. Even after considering the Monty Hall problem! Amazing.

I believe that this is NOT a Monty Hall problem. The big difference is that in the Monty problems, the revelation of the goat (or what have you) provides information about the other choices. This is because you know that the moderator (or logical equivalent) will not show you where the car is, as to do so is contrary to the spirit of the game. Thus, when the moderator gives you information, you know that it is based on knowledge of where the payoff is.

In the scenario here, the items are independent. Knowing that, say, C is incorrect gives you no information about the other items. (The case of the badly constructed item is another matter altogether.)

In short, when Monty shows a goat, he gives information about the other door 2/3 of the time (in the classic 3-door problem). When you eliminate one test item, you are only getting information about that item.

Correct, only scenario 1 is the Monty hall problem. I didn't create all the scenarios so I didn't see why it ends up being a 50/50 at the end.

It turns out when you eliminate two answers:

If your initial guess is right (25% chance) you will always be wrong if you switch

If your initial guess is wrong (75% chance)
1/3 of the time you will eliminate two answers that aren't your initial pick. You will be right 100% of the time if you switch.

2/3 of the time you will eliminate your initial pick and another answer (In this case, since you eliminated your initial guess you must use intuitive guessing and therefore have a 50% chance of guessing right)

(0.5 + 0.5 + 1.0 ) /3 = .6666

.6666 * .75 = .5

Therefore you still have a 50% chance of guessing right in the end, because unlike the monty hall problem you have to consider that you might eliminate your initial guess.

I have a dare for anyone who thinks that this is a Monty Hall problem.

Next time you write a multiple choice test, randomly select an answer. Work hard to eliminate two of the remaining answers, then switch to the fourth solution. Do you really think you'll get 75% of the items correct?

Given you always eliminate two answers you know are wrong, and it is never your random initial guess, then yes you will get around 75% correct.

The problem in a realistic case is there are three possible scenarios that you will be in if you eliminate 2 answers and your initial guess was wrong. Two of which you will eliminate your initial guess, and in one you will not eliminate your initial guess and switch to the right answer.

Whats crazy is 25% of the time you will guess right, 75% of the time you will guess wrong initially and switching will help you. But 2/3 of the time you will eliminate your initial guess.

2/3 * .75 = .50 so your chances are the same surprisingly.

The implausibility is that you can always eliminate exactly two of the non-choices. This is only possible if you always know at least two incorrect responses and that in the cases where you know exactly two, you never make one of them your random guess.