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Captain Centipede


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Ooh, tricky!
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I guarantee you won't get the answer right.*


Three cards are in a hat. One is red on both sides, one is white on both sides, and one is red on one side and white on the other. I draw a card from the hat, and drop it on the table. The upward-facing side is red. What are the odds that the downward-side is also red?


I'll be back with the answer later.


 


*Not a guarantee.



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The Lone Wolf


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0.5 ?

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Hitmanx2x


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50%?

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Keith


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I ****ing hate these questions.... you lot just taking **** coz i do statistics.....

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Lil Miss


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is it a third? 33.333%?

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Lostintheplot


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I think it's two thirds.


I was half asleep when I did it.


 p(A|B)=p(A&B)/p(B)


Therefore p(Red underneath|red on top)= p(red on both sides)/p(red on top) = 1/3 / 2/3 = 2/3


So, two thirds. Do I win a prize?



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Nebula


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Well..............altogether = 6 (3 red sides, 3 white sides) 1 red is already taken up. so it wud be about.....2/5............? ....or 2/6 if u still count the red already facing up...
lol.....im retaking GCSE maths...the chances r high that im wrong

-- Edited by Nebula at 17:36, 2005-03-26

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Keith


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It's 50% chance because it has an option of only between two cards

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Lostintheplot


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That seems obvious. Seeing as Ross guaranteed* we'd get it wrong, I'd say it's not a half. But still, a half seems right.


 


 


*not a guarantee



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Captain Centipede


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Okay, I think enough people have answered now. Plus I can't be bothered to wait any longer, so if you haven't had a try already, tough.


The winner(s) and correct answer are...


 


 


 


 


 


 


 


 


 


 


 


 


Monster with 2/3!


It's one of those things that's hard to do, but when it's explained to you it seems obvious. The complicated equation wasn't really necessary, as it's quite simple if you set your mind to it (I thought it was 0.5 at first as well).


The white card is irrelevant. Basically, we've got 3 red sides we could have face up: 2 sides on the red/red card and 1 side on the red/white card (so we know the probability is out of 3). Whichever side of the red/red card is facing up, the other side is going to be red too. However, if you got the third red side facing up (as in the one on the red/white), the other side is going to be white.


So both situations with the red/red card lead to red being on the other side too. There is no way the red/white card can have a red on the other side as well. Hence, in 2 out of 3 situations, the other side will be red too.


I hope I explained that well.


By the way, don't worry if you got it wrong. It's a puzzle that's fooled even the most intelligent minds for decades. (Doesn't mean you're a genius though Monster - but well done, most people don't get it, including me.)



-- Edited by Captain Centipede at 20:25, 2005-03-26

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Nebula


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Well i half got it right....i sed 2/5.......or 2/6........and 2/6 is the same as 1/3 is it not!!

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Captain Centipede


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It is indeed, but the answer is 2/3.

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Lostintheplot


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quote:
Originally posted by: Captain Centipede
"The white card is irrelevant. Basically, we've got 3 red sides we could have face up: 2 sides on the red/red card and 1 side on the red/white card (so we know the probability is out of 3). "
Except it's now out of two seeing as one red side has been put face up...

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Captain Centipede


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Yes, but it could be any of the 3 sides, you see.

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Nebula


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oh AHA, i mis-read it. lol

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Lostintheplot


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quote:
Originally posted by: Captain Centipede
"Yes, but it could be any of the 3 sides, you see."

Yes...leaving how many Ross?


 


Twatting stupid problems...and these things too



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Captain Centipede


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What do you mean leaving how many?

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Nebula


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Your confusing the hell outta me!!!



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Lostintheplot


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quote:
Originally posted by: Captain Centipede
"What do you mean leaving how many?"
If there are three red sides, and we have one face up, how many red sides are unacounted for?

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Captain Centipede


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2...why?

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