nothing to do with trains

[whole diary]

[Mookiemoo]

ok the lucky three is done after the quiz there are 40 questions in the quiz now say you answer question 4 correct question 2 wrong and question 40 wrong, if the quiz master pulled out 4 correct 2 wrong and 40 wrong you would win but if for example they pulled out 40 correct you would lose

Right - so there are 40 questions in the quiz then at the end there is a random draw and it has to match the answer you give for that Q!


I PAss

I'm not a statitistian (sp?)

Because I do not know what you answered in the quiz

Pass - I suspect about the same chances of he national lottery

[6 OF 2 redundant adjunct of unimatrix 01]

well bear in mind that once the quiz is over and the lucky three starts there are only 2 answers for each question correct or incorrect

[eightf48544]

ok the lucky three is done after the quiz there are 40 questions in the quiz now say you answer question 4 correct question 2 wrong and question 40 wrong, if the quiz master pulled out 4 correct 2 wrong and 40 wrong you would win but if for example they pulled out 40 correct you would lose

Right - so there are 40 questions in the quiz then at the end there is a random draw and it has to match the answer you give for that Q!


I PAss

I'm not a statitistian (sp?)

Because I do not know what you answered in the quiz

Pass - I suspect about the same chances of he national lottery

I think I get what relex109 is saying about the lucky three so here goes.

Assumming you are reasonably good at quizes then you will have more questions right than wrong so drawing  "Question  N wrong" is more likely to eliminate you as opposed to drawing "Question N right". So do the odds will vary in proportion the number of questions you have right out of 40?

If you got all 40 original questions right then it's as relex 109 says:

i know that if there were only 3 questions then the chance would be 1/8 as there are 8 combinations across the 3 questions

yyy
nnn
yny
nyn
yyn
nny
nyy
ynn

but is the fact that there are 40 original questions out of which the 3 are selected irelevent as my friend belives

i really have a headache over this 

The same odds would apply if you got all 40 original answers wrong.

But how you work out the odds for any other combination of right and wrong quiz answers? There are now 6 out of the 8 possible selections which afffect the odds. If you have more right than wrong in your original answers then the 3 selections with 2 rights should increase your chance of winning, but the 3 selections with 2 wrongs are more likely to cause you to lose.

So do they even out?

Are we back to simple penny tossing?

Selection 1 Question X  win or lose i.e. 50:50 (if you lose on this round then you are out and further selections are irrelevant)
Selection 2 Question Y  win or lose i.e. 50:50 (out if you lose)
Selection 2 Question Z  win or lose i.e. 50:50 (out if you lose)

So is it back to 0.5*0.5*0.5= 0125 or 1/8?

Where it all started.

My brain hurts?


















 

[6 OF 2 redundant adjunct of unimatrix 01]

i really cant take my mind off this, who would have thought i would get so obsessed about mathmatics

[Mookiemoo]

ok the lucky three is done after the quiz there are 40 questions in the quiz now say you answer question 4 correct question 2 wrong and question 40 wrong, if the quiz master pulled out 4 correct 2 wrong and 40 wrong you would win but if for example they pulled out 40 correct you would lose

Right - so there are 40 questions in the quiz then at the end there is a random draw and it has to match the answer you give for that Q!


I PAss

I'm not a statitistian (sp?)

Because I do not know what you answered in the quiz

Pass - I suspect about the same chances of he national lottery

I think I get what relex109 is saying about the lucky three so here goes.

Assumming you are reasonably good at quizes then you will have more questions right than wrong so drawing  "Question  N wrong" is more likely to eliminate you as opposed to drawing "Question N right". So do the odds will vary in proportion the number of questions you have right out of 40?

If you got all 40 original questions right then it's as relex 109 says:

i know that if there were only 3 questions then the chance would be 1/8 as there are 8 combinations across the 3 questions

yyy
nnn
yny
nyn
yyn
nny
nyy
ynn

but is the fact that there are 40 original questions out of which the 3 are selected irelevent as my friend belives

i really have a headache over this 

The same odds would apply if you got all 40 original answers wrong.

But how you work out the odds for any other combination of right and wrong quiz answers? There are now 6 out of the 8 possible selections which afffect the odds. If you have more right than wrong in your original answers then the 3 selections with 2 rights should increase your chance of winning, but the 3 selections with 2 wrongs are more likely to cause you to lose.

So do they even out?

Are we back to simple penny tossing?

Selection 1 Question X  win or lose i.e. 50:50 (if you lose on this round then you are out and further selections are irrelevant)
Selection 2 Question Y  win or lose i.e. 50:50 (out if you lose)
Selection 2 Question Z  win or lose i.e. 50:50 (out if you lose)

So is it back to 0.5*0.5*0.5= 0125 or 1/8?

Where it all started.

My brain hurts?


















 

I'm going to refer this to a maths and computing forum I frequent for their learned opinon!

Will post results

[eightf48544]

Thanks Mookiemoo It's got me beat will look forward to the results.

[thetrout]

Did we ever solve this...?!

[6 OF 2 redundant adjunct of unimatrix 01]

the more i think about it, the more i belive it is actually 1/8 so you have a 12.5% chance of winning it...

im pretty sure now that the fact that the 3 questions picked out of the 40 original doesnt matter so you are just left with a 50/50 chance on each of the 3 questions as you have to get all three matched it means 0.5x0.5x0.5=0.125----- 12.5% or 1 in 8

my head still hurts tho

[eightf48544]

Having tried several ways through the problem, see previous posts, I agree with relex109 that 1/8 seems to be the answer.

However, if someone could supply a definative answer it might stop my head hurting.