is this mathematically sound?

[YouMama]

if a 4 option multiple choice question is given to a person. They have a 25% chance of getting it right (if they dont know the answer and cant eliminate any of the options)... is taking into account that they are gonna get the question wrong 75% of the time, then picking another option, mathematically benifical in anyway?

[tto827]

I'm 99.99999% (if we round that's 100%) sure that it does not, and that they have a 25% chance no matter what.

This may be circular reasoning here, a math whiz can clarify, but my proof would be something like this.

Assuming they eliminate their first instinctive guess because it is wrong 75% of the time (and were correct that their first guess was wrong), they would have a 1/3 chance of getting the answer correct. If we take that 1/3 likelihood, and multiply it by the 3/4 chance that his first assumption had of being correct, we end up right back where we started at 25%.


And on a similar topic, this tripped me out in 8th grade when I learned, but its a fun little math trick on the mind.

http://montyhallproblem.com/

[YouMama]

ya i know the monty hall problem... but that problem assumes monty hall knows which door the prize is behind and eliminates one of the ones he knows is wrong... not true in the op problem

I know the monty hall is a mathematical illusion

-- but i like your explanation, i couldnt reason that 4 sh-t. ... thanks

[tto827]

Ya I know the Monty was completely different, its just such a wacky problem I like bringing it up whenever dealing with hypothetical math issues.

If you needed an answer to this question for something besides your own knowledge, I hope someone can proof it better than me, or verify mine, because I very well may be ignoring something very important.

[TheCentaur]

I'm 99.99999% (if we round that's 100%) sure that it does not, and that they have a 25% chance no matter what.

This may be circular reasoning here, a math whiz can clarify, but my proof would be something like this.

Assuming they eliminate their first instinctive guess because it is wrong 75% of the time (and were correct that their first guess was wrong), they would have a 1/3 chance of getting the answer correct. If we take that 1/3 likelihood, and multiply it by the 3/4 chance that his first assumption had of being correct, we end up right back where we started at 25%.


And on a similar topic, this tripped me out in 8th grade when I learned, but its a fun little math trick on the mind.
http://montyhallproblem.com/

I agree and think this is right.

Think of it this way too:

In 100 questions with 4 choices $1 was bet each time, paying even money with a correct choice.

In the first scenario of keeping all four choices your results would be
25*$1 - 75*1= -$50

In the second scenario in which we eliminate the first choice because it is wrong 75% of the time we get
a(75*.33333*$1) - b(75*.66666*$1) - c(25*$1) = -$50


a-One third of the time in the 75 trials that contain a correct answer after elimination of the first choice a correct choice will be made from the 3 remaining for a win of $1

b-Two thirds of the time in the 75 trials that contain a correct answer after elimination of the first choice an incorrect choice will be made from the 3 remaining for a loss of $1

c-100% of the time in the 25 trials that DO NOT contain a correct choice because it was eliminated there will be a loss of $1


Each strategy results in a loss of $50