[Closed] Probability/Multiple Choice

My brain started to smoke a bit after 20 seconds so Immagonna wait and join in later. 🙂

0%. But I hate probability. So could be wrong, but I don't think so.

1 in 3 chance

Assuming 25, 50 and 60 % are equally like to be correct then

(1/3 * 1/4) + (1/3)*(1/4) + 1/3*(1/2) = 1/12 + 1/12 + 1/6 = 4/12 = 33.333%

But surely you do not need the maths and it is obvious?

EDIT - few dodgy ( )

50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%....

Either no one else has understood the question properly, or I'm thinking about this too deeply.

GJP - I got 33.3333........% (or 1 in 3) without any maths. There are three possible answers and you are picking one of them. Surely?

So, if you answer any four answer MC question at random, there's a one in four chance of getting it right i.e. 25%. But two of the answers are 25%. So is it now 50%?

I think the confusion (or twist/brain ****) is, is that the actual answer, isn't one of the answers.

It's impossible to choose an answer 'at random' surely.

RealMan = 0% isn't an option, so...

50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%, no... 25%, no... 50%....

No darcy - if the actual answer is 25%, then you have a 50% chance. if it isn't, then you don't.

*edit* - that was my first approach anyway, but isn't correct

[i]*pulls up a chair*[/i]

ATSR - yes you are correct and that is the intuitive approach - I needed to derive it from first principles - you are probably a lot younger than me and my brain is ****ed

Although to my credit I actually stated the key assumption that all unique answers were equally likely to be correct whereas it was implicit within your intuitive thinking

RealMan = 0% isn't an option, so...

If you choose an answer at random, what is the probability you will be correct.

So you choose an answer, it won't be correct => probability = 0%.

Or something.

This statement is false, innit?

b)
50 % it is either right or wrong the number of choices is irrelevant I suspect the exact wording of the Q is critical to the answer.

There is no question

Surely there is no spoon...

Anyway, it's a paradox... of similar making to the one with the book that lists all books which don't reference themselves.

50 % it is either right or wrong the number of choices is irrelevant I suspect the exact wording of the Q is critical to the answer.

Yeah, you also have a 50% chance of winning the lottery, cause you can either win or lose, right?

Yes well spotted.

You cannot deny that each week i either win it or i dont there is no other chance. Obviously the odds/probabilities heavily favour one outcome but there are still only two chances hence 50%
I think the key is the word chance here and it is the semantics of the Q that is the issue here. It cannot be probability as others suggest as that would give 33% which is not an option.
Anyway I have more chance of being right than you as I went for an answer that was at least there 😛

It cannot be probability as others suggest as that would give 33% which is not an option.

So there is zero chance you'll be right?

So there is zero chance you'll be right?

There's two questions, with a similar set of answers, which is confusing. Compare,

If you choose an answer to this question at random, what is the chance you will be correct?
a) A
b) B
c) C
d) A

Answer, 33%.

Probably.

It's self referential, so it's a paradox and hence cannot be answered.

I'd be impressed if someone's come up with a new one though - there must be a recognised and named version of this problem surely? Game theory anyone? CharlieMungus to the forum, please.

So there is zero chance you'll be right?

that is some cruel mocking of my maths ability there grahamS 😥

Yes I should have realised its the opposite of what I think as this does seem to be the rule with any maths question on here

I did a bit of Googling.

I've seen suggestions that it's paradoxical and there is no answer, with "[url= http://en.wikipedia.org/wiki/Russell%27s_paradox#Informal_presentation ]Russell's Paradox[/url]" posited. Frankly that's beyond me, set theory to that degree makes me want to kill kittens.

I also found this analogy, which is ostensibly where I was going with it, only a bit better written:

There is a box containing two red balls, one blue ball, and one yellow ball. I secretly write down one of these colours on a piece of paper.

If you pick a random ball from the box, what is the probability that the ball is the colour written on the piece of paper?

Answer, 33%.

I'm convinced this is correct now, though the maths was breaking my head for a while.

Using the 'balls' analogy, cos it's easier; you pick a ball at random, what's the probability that it matches what's written on the paper? The key here is that [i]the contents of your ball bag doesn't matter(*).[/i] Whether you've got two red balls, a hundred red balls, or in fact only red balls, what are the chances of it matching a 1-in-3 random choice? It's 33%.

(*) - never thought I'd find myself typing that

Ooh, I think I've got it. Is the answer to this question no?

No.

I'm now questioning my own theory actually - if there are 100 red balls, one blue ball and one yellow ball, the chance of picking the yellow ball is 1 in 102.

Therefore if the answer is yellow (or blue), it's 1 in 102, whereas if the answer is red, you have a 100 in 102 chance?

Aaaaarrrgggghhhh!!!!!

Do I need to stand on a conveyor belt for this one?

Just for the record, anyone that thinks a plane on a conveyor belt would take off is a dumbass.

thankyou Cougar 🙂

it only took one post!

I want to win a goat. Which choice do I make for that?

A looong time since I was at school - can you add probabilities together like fractions, because if so, it revives the 1/3 theory

Red - 2/4
Yellow - 1/4
Blue - 1/4

Add together = 4/12 or 1/3

Works for 5 balls too, with 3 being red

Red - 3/5
Yellow - 1/5
Blue - 1/5

Add together = 5/15 or 1/3

Or is that just a load of nonsense?

I want to win a goat. Which choice do I make for that?

just don't switch doors when Monty give's the chance too.

Or is that just a load of nonsense?

It's nonsense. When adding fractions you don't add the denomenator, only the numerator.

Or is that just a load of nonsense?

It bears out my theory, so it's golden as far as I'm concerned. (Though as GF says, that's not how you add fractions, not that it matters)

Is it the fact that 2 of the numbers are the same rather than what the actual numbers are? There's nothing to say that any of them are 'correct'

So if the question is : if you choose one of 4 options at random and 1 of them is correct your chances are 1/4. If two of them are correct/the same (as the question sort of implies) then it's 2/4. If 2 are the same but incorrect then it's still 1/4.... is it?

Ha, yeah - of course. Stoopid me, I knew that anyway. Doh!

If you pick a random ball from the box, what is the probability that the ball is the colour written on the piece of paper?

Cougar that's not the same question.

In this case the correct ball depends on the ball you pick.. so it's a paradox.

s it the fact that 2 of the numbers are the same rather than what the actual numbers are?

it's both.

the two numbers the same change the probability of picking the correct answer from 25% to 50%. the correct answer becomse 50%. yet 50% is now not the correct answer as it only appears once thus making the correct answer 25% and so on...

[i]In this case the correct ball depends on the ball you pick[/i]

this.

is it the fact that 2 of the numbers are the same rather than what the actual numbers are?

it's both.

the two numbers the same change the probability of picking the correct answer from 25% to 50%. the correct answer becomse 50%. yet 50% is now not the correct answer as it only appears once thus making the correct answer 25% and so on...

Only if 25% is the correct answer to start with. Couldn't the question just as well say:

a) blue
b) yellow
c) red
d) blue

So the odds of the one you pick being the 'right' one depends on what the right one is, which we don't know? Or have I totally missed the point here?

EDIT: Someone's already said this haven't they?

No, you are probably right Mr Salmon