Take this quiz if you're smart

X-Chick

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Seriously, not to see who is smarter than who, but to see if this guy's logic is faulty in his questions and if his explanations to his answers make any sense to you. There's one question in particular that I still can't understand how this guy thinks his answer is right. :huh:

Quiz
 
Which answer do you disagree with?
 
it's not so much an answer as all of them and how he explains his logic.
 
3 out of 5 isn't to bad I guess.
 
I got the last three. I was close to picking the first, but I couldn't quite justify the answer in my head, although I figured it would be the weird one.
 
I don't have the patience for this **** right now. Maybe later.
 
apparently none of the answers are simple.:csad:
 
I only got 1, but I'm also on vicodin.

This guy still doesn't make sense though...
 
meh, there's formal language and understood language. I only got one and it was by accident i think

i'm willing to let to other ones slide but can't get the coloured card one...

if the odds of getting the first card as red have already been accounted for, then surely the remaining probability of getting the next red should be 0.5.

i have yet to read his reasoning but these sorts of problems come up in basic statistic problems, there are even formulae to deal with them.

I think his wording of the question is somewhat misleading into making you believe one probability factor has already been dealt with.
 
I'm too tired to deal with probability equations right now. I say this guy sucks.
 
2/5. I could've gotten three correct, but oh well. Question nr. 5 is bull****.
 
If I take half an hour (or more) to do the quiz, I'll have no time left to post in the Hype... :csad:

(not to mention that's too much thinking for me first thing in the morning :o )
 
4/5 Just call me Wily E. Coyote...super genius. LOL!

I'm actually very good at these types of quizzes
 
meh, there's formal language and understood language. I only got one and it was by accident i think

i'm willing to let to other ones slide but can't get the coloured card one...

if the odds of getting the first card as red have already been accounted for, then surely the remaining probability of getting the next red should be 0.5.

i have yet to read his reasoning but these sorts of problems come up in basic statistic problems, there are even formulae to deal with them.

I think his wording of the question is somewhat misleading into making you believe one probability factor has already been dealt with.


I expected the last two questions to throw a lot of people off, because it usually does. I've seen it before, and know how to explain it a little better. So let me give it a try. (Both questions are really the same problem, but I'm going to focus on problem 5 because it's easier to visualize.)

Question 5 is essentially this: You are one of three men randomly chosen to die. The name of one of the men chosen to live is given to you. What is the probability you were chosen to die?

This is the classic Monte Hall problem (for those Let's Make a Deal fans out there). It's a little easier to understand the explanation when you look at it this way: You're a contestant on a game show. On stage are three doors. Behind two doors are stupid prizes, but behind one door is a million dollars. You pick a door, we'll say door number 3. Monte (the show's host) has them open door number 1 and behind it is a frying pan. So the million dollars is either behind your door, or door number 2. Monte asks if you want to switch your choice. Should you?

The answer is yes, because, even though only two doors are still closed, there's only a 1/3 chance your door (number 3) is correct, and a 2/3 chance the million is behind door number 2.


And the reason this is true is because you made your choice out of three doors. Opening one of them doesn't change the fact that you picked one of three doors, not one of two. Now, had Monte opened a door, then had you pick, you would indeed have a 50-50 chance of getting it right (because you're only picking from two available doors). And in the original question of men chosen to die, you've implicitly chosen yourself (as the winning door), so there's a 1/3 chance you'll die, and a 2/3 chance it's the other guy.


If it still doesn't make sense, expand the problem and it's even easier to see the logic behind it. Let's there isn't 3 doors on stage, let's say there are 30 doors (1 with the million dollars, and 29 with bad prizes). Now Monte has you pick a door. You take number 15. Then Monte has every door except 15 and 28 opened, showing bad prizes. So the million bucks is either behind the door you picked (15) or the one Monte didn't open (28). Do you still think you just happened to pick the right door out of thirty? Or is it more likely that Monte didn't open number 28 because it really has the million dollars behind it? That's right, you should switch your choice because you only had a 1/30 chance of guessing right, and a 29/30 chance it's behind the other door.


It's just much harder to see the reasoning when there are only three options because our brains are 'confused' by what looks like a 50-50 choice.
 
And the reason this is true is because you made your choice out of three doors. Opening one of them doesn't change the fact that you picked one of three doors, not one of two. Now, had Monte opened a door, then had you pick, you would indeed have a 50-50 chance of getting it right (because you're only picking from two available doors). And in the original question of men chosen to die, you've implicitly chosen yourself (as the winning door), so there's a 1/3 chance you'll die, and a 2/3 chance it's the other guy.
this is what happens in the question though, Tom or whoever recieves this information that he is not being killed before we are asked what is the probability it is him getting killed.

I don't know why the elimination of one character makes one character twice as likely to die than another after knowledge has been shared since the probability of one person dying has now become 0, therefore the rest of the odds should surely be shared out equally.

there is a difference between saying this

what is the probabity of (B) occuring given that (A) has occured before (looking at total probability)

than saying

Once (A) has occured, what is the probability of then (B) happening (individually looking at the second based on assumptions of the first).

The first one, as you state takes account of the first events, while the second one solely looks at the probability of the second and it no longer becomes a statistical problem, rather one of pure wording put together to trip people up rather than stimulate thought.
 

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