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Lemmiwinks

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You saw a tshirt for 97€.

 

You didn't have the cash so you borrowed 50€ from your mom and 50€ from your dad.

 

50€ + 50€ = 100€

 

You bought the shirt and had 3€ change.

 

You gave your dad 1€ and your mom 1€ and you kept 1€ for yourself.

 

Since you already gave mom and dad 1€ you now owe them 49€ each.

 

49€ + 49€ = 98€ + your 1€ = 99€

 

Where is the missing €???

 

97 / 2 = 48,5

 

50 - 48,5 = 1,5

 

48,5 + 1 = 49,5

 

49,5 + 49,5 = 99

 

They agree that if the first prisoner says black it means there is an odd number of black hats, but if he says white there is an even number of black hats. Knowing this, each other prisoner can count how many black hats are in front of him and because they have heard how many other hats have been told before, he can then know whether his own hat is black or not according to the parity initially set.

 

The only prisoner guessing the color of his own hat is the first one.

 

If the hat colours are even then only the first might loose for himself, otherwise it's still a guess i think... for instance imagine only the first or the first few to have white hats, it could be a drama.
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You are presented with 12 marbles. You are told that 11 of them are identical, but one of them is either lighter or heavier than the others. You have a pair of scales - the old-fashioned sort that will only tell you whether the items placed on one end are heavier than, lighter than or the same mass as the items placed on the other end. You are allowed to use the scales three times, and no more.

 

How do you identify which marble is the odd one out, and whether it is lighter or heavier than the others?

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You are presented with 12 marbles. You are told that 11 of them are identical, but one of them is either lighter or heavier than the others. You have a pair of scales - the old-fashioned sort that will only tell you whether the items placed on one end are heavier than, lighter than or the same mass as the items placed on the other end. You are allowed to use the scales three times, and no more.

 

How do you identify which marble is the odd one out, and whether it is lighter or heavier than the others?

 

umm... you pick it up in your hand and tell if it's heavier or not? lol

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umm... you pick it up in your hand and tell if it's heavier or not? lol

 

No.

 

When you say old-fashioned you mean something like that Posted Image ? :ph34r:

 

I had something more like this in mind:

 

Posted Image

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umm... you pick it up in your hand and tell if it's heavier or not? lol

 

Throw them to yours brothers head and ask him which one pained him more or lesser than the others ;) .
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If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 25%

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If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 25%

 

question? what question?

 

You are presented with 12 marbles. You are told that 11 of them are identical, but one of them is either lighter or heavier than the others. You have a pair of scales - the old-fashioned sort that will only tell you whether the items placed on one end are heavier than, lighter than or the same mass as the items placed on the other end. You are allowed to use the scales three times, and no more.

 

How do you identify which marble is the odd one out, and whether it is lighter or heavier than the others?

 

I haven't forgotten you Rotwang...
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Since A) and D) answers are different choices, i would say i had a 25% to score.

 

But if the question refers to the answer then i would say a 50% since my answer would be 25%, otherwise 25% for the rest answers.

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50%

 

nop

 

E) 100%

 

nop

 

Since A) and D) answers are different choices, i would say i had a 25% to score.

 

But if the question refers to the answer then i would say a 50% since my answer would be 25%, otherwise 25% for the rest answers.

 

you are on the right path , give me one answer :)
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None of the answers is correct. If A and D were correct then you would have a 50% chance of guessing right by choosing one at random, so B would be correct and A and D would be wrong. If B or C were correct you would have a 25% chance of guessing right by choosing one at random, so A and D would be correct and B and C would be wrong. So every one of the choices given is self-refuting. Therefore the correct answer is 0%.

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None of the answers is correct. If A and D were correct then you would have a 50% chance of guessing right by choosing one at random, so B would be correct and A and D would be wrong. If B or C were correct you would have a 25% change of guessing right by choosing one at random, so A and D would be correct and B and C would be wrong. So every one of the choices given is self-refuting. Therefore the correct answer is 0%.

 

I was thinking about this : since there are only 3 distinct answers and each one is equally possible, the overall probability should be independent of how you reach them (i.e. how they are organized in multiple choices). So overall probability of hitting the right answer should be 1/3.. just becuase 25% appaears twice doesnt reduce the probability of it being wrong

 

Probability of 25% being the right answer is 1/3 and probability of you picking A or D is 1/2

Probability of 50% being the right answer is 1/3 and probability of you picking B is 1/4

Probability of 75% being the right answer is 1/3 and probability of you picking C is 1/4

So, overall: (1/3*1/2) + (1/3*1/4) + (1/3*1/4) = 1/3 ~ 33.33%

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I was thinking about this : since there are only 3 distinct answers and each one is equally possible,

Each one is "equally possible" in the sense that each one is wrong.

 

the overall probability should be independent of how you reach them (i.e. how they are organized in multiple choices). So overall probability of hitting the right answer should be 1/3..

No, this doesn't follow. There is some ambiguity in the question, in the sense that there are infinitely many ways to choose one of the four given answers at random (one can assign probabilities to each of A, B, C and D any way one likes, provided that they sum to 1), but the natural interpretation is that the question means for you to pick one of A, B, C or D with equal probability for each - that's what people usually mean when they talk about picking one of a finite number of choices "at random" without any further qualification.

 

just becuase 25% appaears twice doesnt reduce the probability of it being wrong

I don't know what you mean by this. It's either is wrong or it isn't, probability doesn't come into it.

 

Probability of 25% being the right answer is 1/3 and probability of you picking A or D is 1/2

Probability of 50% being the right answer is 1/3 and probability of you picking B is 1/4

Probability of 75% being the right answer is 1/3 and probability of you picking C is 1/4

In saying that the probability of 25% being right is 1/3, you are assuming that each of the three answers is equally likely to be correct, and that exactly one of them is correct (as opposed to, e.g. none of them). That isn't the case.

 

So, overall: (1/3*1/2) + (1/3*1/4) + (1/3*1/4) = 1/3 ~ 33.33%

But that doesn't make any sense. If the answer were 1/3 then there would be a 1/3 probability of picking 1/3 by choosing one of A, B, C or D at random. But none of A, B, C or D is 1/3.

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Rotwang you are stopping at randomly selecting but not pursuing "choosing randomly and correctly" ... in the sense the probability of selecting 25% as the correct answer yields a probability of 16.67 % and not 50 (1/2*1/3) .. and if you were to calculate the probability of selecting the right answer it could either be 25, 50 or 75 , by adding another 25 gives you the illusion of another choice but actually there are just three to choose from .

 

Framing the question another way :

 

If you were to choose one of the below answers to "which pet do i have at home?" at random, what would be the chance that you got it right?

 

A) Monkey B ) Monkey C) Monkey D) Dog E) Cat

 

Now just because monkey is appearing 3 times doesn't make it any more or less likely to be the correct answer than Dog or cat , right ? Each has an equal probability of being the correct answer but BUT because it appears thrice it makes it more likely to be the right "selection" (3/5*1/3) = 20% as opposed to the other two (1/5*1/3) = 6.67 %

 

Much more interesting is if we were to ask the same question in terms of flipping of a coin , if we had 999 heads and a 1 tail and i asked the you the chance of randomly selecting the right answer, what would you say ?

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Upon reflection, I think you've gone about it as if the question were slightly different. Consider this question:

 

I have picked one of red, green or blue at random. If you were to choose one of the below answers to the question of which colour I picked at random, what would be the chance that you got it right? A) red B ) green C) blue D) red

Note that there are two questions at issue here. Question 1 is "Which colour did I pick?". Question 2 has the form "What is the probability that you will correctly guess the answer to Question 1 if you go about choosing your answer a certain way?". The answer to the second question has no effect on the answer to the first, so we can go about answering both questions by first figuring out what the possible answers and respective probabilities are for the first question, and then applying that knowledge to answer the second question.

 

But the problem you posed was

 

If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 25%

Note that here there is only one question, which refers to itself. There is only Question 1, where Question 1 has the form "What is the probability that you will correctly guess the answer to Question 1 if you go about choosing your answer a certain way?". You're trying to apply the approach we can use on the first two questions, which worked there, to the last question.

 

edit: I notice that you posted a reply while I was writing this one, and your reply illustrates that you were making the mistake that I thought you were. The problem about pets you posed is exactly analogous to the "red, green, blue" example - it involves two different questions. But the original question only refers to itself. Your method of approaching it is to assume that there's a second question that asks you to calculate the probability of correctly guessing the answer to the first. But since they are really the same question, the two answers you have given (25% or 50% or 75%, each with probability 1/3 for the first question, 33.33% for the second question) contradict one another.

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