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Topic: Logic/Math puzzles

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All Forums : [General] : General Trivia > Logic/Math puzzles
8 JAN 2003 at 9:53pm

szcax

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Ok all you JA math lovers... Let's see what you're made of! I'm a big fan of math trivia. Here's my favorite. Feel free to post your own favorites.  I'd love to hear them.

1)Circle A has a circumference of 3 inches. Circle B is tanget to circle A and has a circumference of 1 inch. If circle B starts rolling around circle A, how many revolutions will it make before returning to its starting position?

2)You're on a game show. The game consists of 3 doors, behind one of which is a million dollars, the other two contain nothing. You tell the game show host that you choose door #2. He tells you that he will reveal one of the doors that doesn't hold the prize and opens door number 1. Then he says "Now that you know door one is a dud, you may switch to door number three. Would you like to?" Should you? Does it matter?
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8 JAN 2003 at 9:57pm

Agustín Cordes

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1) 3 revolutions

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8 JAN 2003 at 10:03pm

Agustín Cordes

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2) In the first place you have a 33% chance of chosing the right door. After the host opens a wrong door, you have 50% chance of chosing the right door. It doesn't really matters if you change your choice or not.

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8 JAN 2003 at 10:32pm

MichalN

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Originally Posted By Rael (8 JAN 2003 10:02pm)
2) In the first place you have a 33% chance of chosing the right door. After the host opens a wrong door, you have 50% chance of chosing the right door. It doesn't really matters if you change your choice or not.

Here's an alternate explanation: in the first round, you have a 1/3 (not exactly 33%) chance of choosing the right door. In the second round, there is 66% chance that the door you didn't choose is the right one, hence you should change your choice.

Let me explain. Suppose we label the doors A, B, C. Any of them can be the prize door with equal probability. Let's say the first choice is always A and second choice is whichever of B, C is left.

There is 1/3 probability that A was the right one and you will erroneously switch away from it. However, there is 2/3 probability that the right door is B or C and you will get the right one.

So yes, you should switch and yes, it does matter.
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8 JAN 2003 at 10:45pm

Agustín Cordes

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Originally Posted By MichalN (8 JAN 2003 10:32pm)
(not exactly 33%)

You're right, it's 33.3333333333333333333...%

Anyway, I liked your explanation better.

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9 JAN 2003 at 12:00am

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Here's an alternate explanation: in the first round, you have a 1/3 (not exactly 33%) chance of choosing the right door. In the second round, there is 66% chance that the door you didn't choose is the right one, hence you should change your choice.

Seems to me that's it a only 50 % chance - once the other door is revealed to be empty -  



What?


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9 JAN 2003 at 12:41am

Aya

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Originally Posted By MichalN (8 JAN 2003 10:32pm)
So yes, you should switch and yes, it does matter.

no it doesn't matter! at first you got 1 out of 3 probability to get it right... when one of the wrong doors is opened then there's a 1/2 probability that you chose the right door... whether you change your selection is not at all important, the probability will still be 1/2

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9 JAN 2003 at 12:51am

MichalN

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Originally Posted By Aya_Brea (9 JAN 2003 12:41am)
no it doesn't matter! at first you got 1 out of 3 probability to get it right... when one of the wrong doors is opened then there's a 1/2 probability that you chose the right door...

No. If you don't change your selection, the probability stays at 1/3.

whether you change your selection is not at all important, the probability will still be 1/2

I beg to differ. We'll see whar the arbiter (szcax) has to say but I still think what I said was right.

What you are forgetting is that you are receiving extra information about the two doors you didn't choose initially.

I say that the probabilities of getting the prize are like this:

- Choose one and stick with that selection - 1/3
- Choose one and randomly choose another - 1/2
- Choose one and then the other - 2/3

I think we all agree on the first two.

There is 2/3 chance you got the initial choice wrong. However, if you were indeed wrong the first time (and there is 2/3 probability you were!), you will be right the second time, turning the disadvantage into your advantage.
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9 JAN 2003 at 12:55am

Agustín Cordes

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Originally Posted By MichalN (9 JAN 2003 12:51am)
What you are forgetting is that you are receiving extra information about the two doors you didn't choose initially.

Uh-oh, I think I see a problem in your logic becuase you do choose one of the remaining doors initially. Read carefuly.

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9 JAN 2003 at 12:58am

szcax

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Originally Posted By MichalN (9 JAN 2003 12:51am)

I beg to differ. We'll see whar the arbiter (szcax) has to say but I still think what I said was right.


And interfere with this great squabble? I think not! I'll let people mull about until tommorrow. Don't forget, the answer to the first one hasn't been figured out yet. It's not 3  


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9 JAN 2003 at 1:04am

MichalN

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Originally Posted By szcax (9 JAN 2003 12:58am)
And interfere with this great squabble?

Scientific discourse you mean?


Don't forget, the answer to the first one hasn't been figured out yet. It's not 3  

Maybe the question was formulated ambiguously. Revolutions around which axis?
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9 JAN 2003 at 1:26am

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the probabilities change according to the sample... no matter how many doors there are (x) you'll have 1/x chance to get it right... if some of the wrong doors are revealed (y) and you get to guess again, no matter what you guess, your probability will now be 1/x-y... if you don't guess again it doesn't matter which and how many doors are opened, you always have 1/x chance to get it right

You have gotten the attention of the mysterious lady. She turns to face you. Her face is devoid of any flesh. You are frozen with horror as she begins ripping your body into a bloody mess.


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9 JAN 2003 at 1:29am

Aya

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on this specific example the question "Now that you know door one is a dud, you may switch to door number three. Would you like to?" is a trick question... when this question is uttered it can immediately be translated as "Now that you know door one is a dud, which door do you want, 2 or 3?" answering no to the first is like answering 2 to the second, answering yes is like answering 3... so your probability changes to 1/2

You have gotten the attention of the mysterious lady. She turns to face you. Her face is devoid of any flesh. You are frozen with horror as she begins ripping your body into a bloody mess.


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9 JAN 2003 at 1:54am

MichalN

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Originally Posted By Aya_Brea (9 JAN 2003 1:25am)
the probabilities change according to the sample... no matter how many doors there are (x) you'll have 1/x chance to get it right... if some of the wrong doors are revealed (y) and you get to guess again, no matter what you guess, your probability will now be 1/x-y... if you don't guess again it doesn't matter which and how many doors are opened, you always have 1/x chance to get it right

Exactly! The point is, I'm saying you shouldn't guess the second time and should choose a door other than your initial selection. Then there is ((x-1)/(x))*(1/(x-y-1)) probability of success, instead of 1/x. I think


Which for x=3 and y=1 works out to 2/3. For x=4 and y=1 it would be 0.375 which is still better than 0.25.
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9 JAN 2003 at 2:06am

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but whichever door you choose the second time it will be 1 door out of 2... if you say you'll stick to your selection it's 1/(3-1)... if you say you'll change your choise again it's 1/(3-1)... both are an entirely new selection with a new sample

according to your example if there were 100 doors and the host opened 98 then if you changed your selection you'd have a 0.99 probability to be right when selecting 1 out of 2!!!
of course not... 1 out of 2 is always 0.5

You have gotten the attention of the mysterious lady. She turns to face you. Her face is devoid of any flesh. You are frozen with horror as she begins ripping your body into a bloody mess.


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9 JAN 2003 at 2:27am

szcax

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Originally Posted By MichalN (9 JAN 2003 1:04am)

Revolutions around which axis?


The question is asking how many times the smaller circle will make a complete rotation (If it's a quarter, one rotation would mean our friend George Washington did one somersault) before it gets to its initial point on the large circle.
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9 JAN 2003 at 2:46am

MichalN

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Originally Posted By Aya_Brea (9 JAN 2003 2:05am)
but whichever door you choose the second time it will be 1 door out of 2... if you say you'll stick to your selection it's 1/(3-1)... if you say you'll change your choise again it's 1/(3-1)... both are an entirely new selection with a new sample

No! I'm saying that the second time you randomly choose a door other than the one you selected first. In case of three door, there is only one such door so the second selection isn't very random at all.

according to your example if there were 100 doors and the host opened 98 then if you changed your selection you'd have a 0.99 probability to be right when selecting 1 out of 2!!!
of course not... 1 out of 2 is always 0.5

Of course yes! If you randomly select one of the remaining doors then yes, there is 1/2 chance of success.

BUT! There is 99% chance that your initial selection was wrong, ie. one of the 99 doors you didn't choose contains the prize. And it that case, there is 99% chance (overall) that the other remaining door is the right one.

You seem to be overlooking the fact that there are two selections going on and for the second one you've received extra information.
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9 JAN 2003 at 2:48am

MichalN

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Originally Posted By szcax (9 JAN 2003 2:27am)
The question is asking how many times the smaller circle will make a complete rotation (If it's a quarter, one rotation would mean our friend George Washington did one somersault) before it gets to its initial point on the large circle.

That is not a very clear answer... So I will be more explicit. Are you asking about rotation around the axis of the smaller circle, the larger circle or both?

Of course, this question may be the answer

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9 JAN 2003 at 3:01am

szcax

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Ok I think I see what you mean.... the big circle is totally still. The question is asking how many times the small circle rotates around it's own axis as it rolls around the circumference of the big circle...
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9 JAN 2003 at 3:09am

Agustín Cordes

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Originally Posted By szcax (9 JAN 2003 3:00am)
Ok I think I see what you mean.... the big circle is totally still. The question is asking how many times the small circle rotates around it's own axis as it rolls around the circumference of the big circle...

Unless my understanding of English is mistaken due to my lack of sleep, it's still three revolutions - the circumference of the big circle is three times the circumference of the small circle.

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9 JAN 2003 at 3:15am

MichalN

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Originally Posted By Rael (9 JAN 2003 3:09am)
Unless my understanding of English is mistaken due to my lack of sleep, it's still three revolutions - the circumference of the big circle is three times the circumference of the small circle.

That's what I'd say too. Three times around its own axis and once around the axis of the larger circle (obviously).
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9 JAN 2003 at 4:07am

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I'm confused (SNAFU) -  

What?


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9 JAN 2003 at 5:57am

MichalN

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Originally Posted By InlandAZ (9 JAN 2003 4:07am)
I'm confused (SNAFU) -  

Confused by what exactly? I dare say there's a lot to be confused by in this thread
Or are you just generally confused?

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9 JAN 2003 at 1:17pm

Aya

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Originally Posted By MichalN (9 JAN 2003 2:46am)
Of course yes! If you randomly select one of the remaining doors then yes, there is 1/2 chance of success.

BUT! There is 99% chance that your initial selection was wrong, ie. one of the 99 doors you didn't choose contains the prize. And it that case, there is 99% chance (overall) that the other remaining door is the right one.

You seem to be overlooking the fact that there are two selections going on and for the second one you've received extra information.

practically you might be right i think... i see your point... it's more obvious with 100 doors than it is with 3!
but mathematically i think you're wrong cause the probability during the second choice is 1/2 cause it's a separate choice... but when you're based to the fact that your first choice was 99% wrong (or 66% for the 3 doors) then yes you theoretically have a bigger chance if you change your selection

You have gotten the attention of the mysterious lady. She turns to face you. Her face is devoid of any flesh. You are frozen with horror as she begins ripping your body into a bloody mess.


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9 JAN 2003 at 6:58pm

MichalN

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Originally Posted By Aya_Brea (9 JAN 2003 1:17pm)
practically you might be right i think... i see your point... it's more obvious with 100 doors than it is with 3!

Aha! Yes, the 100 door example seems clearer. Thanks for coming up with it


but mathematically i think you're wrong cause the probability during the second choice is 1/2 cause it's a separate choice...

Only if you disregard what you know about the doors. If you randomly choose from the remaining two doors then there's no question, the chance of success is 50%. What I'm saying is that the second choice should not be random at all.

but when you're based to the fact that your first choice was 99% wrong (or 66% for the 3 doors) then yes you theoretically have a bigger chance if you change your selection

Exactly.
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