Does powerscore not understand statisitcs (or do I)? Forum

[rvadog]

In the LG Bible and the LR Bible powerscore has a section at the end on guessing. In that section they state that you should always pick one letter (e.g. C) and use that everytime because it raises your chances of getting at least 1 out of 5 correct.

That's ridiculous right?

No matter whether you pick all "C" or you randomly choose a letter each time, each question is independent of all the other questions and so each question you have a 1/5 chance of guessing the correct answer.

Right?

[Nova]

Ive always wondered that too.

I agree with you.

[manofjustice]

It's not ridiculous. It probably wrong though.

I believe they did an analysis and show a slightly greater proportion of right answers are Cs. But...

a) if the right answers are randomly selected, it is still almost certain one answer will have a predominate representation among the right answers. But the answers are still randomly selected, so picking that answer when in doubt would not improve your chances of getting a right answer over randomly picking an answer.

b) even if it were true and there was some non-random cause for their being a slightly greater proportion of C right answers (which seems unlikely), the degree of the phenomenon is so slight it would likely have no effect on your scaled score.

You are probably right.

[crazyrobin]

IMHO, statistically, if you choose a random answer for each question, ( let's suppose there are 6 questions) you'll have 1/5^6 chance to get all answer correct.

And you also could end up getting 0 correct.

But if (a big IF) ACs are distributed as exactly as PS describes (A,B,C,D,E has relatively same frequency), you choose the same answer choice, then you could at least get one answer correct ( assume there are 6 answer choices, and the distribution for each answer choice are the same, this means A,B,C,D,E would individually at least appear once)

Conclusion, PS's approach is risk-free if you run out of time.

[manofjustice]

IMHO, statistically, if you choose a random answer for each question, ( let's suppose there are 6 questions) you'll have 1/5^6 chance to get all answer correct.

And you also could end up getting 0 correct.

But if (a big IF) ACs are distributed as exactly as PS describes (A,B,C,D,E has relatively same frequency), you choose the same answer choice, then you could at least get one answer correct ( assume there are 6 answer choices, and the distribution for each answer choice are the same, this means A,B,C,D,E would at least appear once)

Conclusion, PS's approach is risk-free if you run out of time.

I would doubt C's are "distributed" evenly. I think they are "equally probable." Distributed evenly means that every 5 answers, there is a C. In that case, yes: the advantage over choosing C is that you'll get one right, whereas when choosing randomly, you could get 0 right. But where the C's are placed is randomly chosen. If you choose all C's for 5 answers, there is a chance that none of those 5 answers are Cs.

[crazyrobin]

I would doubt C's are "distributed" evenly. I think they are "equally probable." Distributed evenly means that every 5 answers, there is a C. In that case, yes: the advantage over choosing C is that you'll get one right, whereas when choosing randomly, you could get 0 right. But where the C's are placed is randomly chosen. If you choose all C's for 5 answers, there is a chance that none of those 5 answers are Cs.

I agree with you. If all answer choices are equally probable instead of distributed evenly, then you can get 0/5 no matter whatever you choose.