Brainteaser/Puzzle Corner (Part 5)

Game. Set. Match. Your right, nice example btw, it does take a little while to wrap ones head around it. But for the record BIzzaro's last example was correct it just didnt apply to the problem. So was mine now that I look at it.

Its always nice having logic/math duels whether you win or lose because either way you come out with something you didnt have before.

 

Why don't you guys check out Z-BosoN's riddle thread, it is very interesting

 

The Discs.

There are three discs in a drawer.

One has 1 black side and 1 white side,
One has 2 black sides,
One has 2 white sides.

You stick your hand in a drawer and randomly choose a disc. The front side of the disc is black. What is the probability of the back side of the disc being black as well?

 

1/3... there are three discs.

 

lol wtf... read the rest of the problem

 

well... you know that one side is black.

it asks for the probability that the other side of the disc is black.

this means that the disc must have 2 black sides.

1 of the three discs has 2 black sides.

therefore, 1/3 probability that the back is black too.

 

Well that is clearly flawed...

if you find out that one side is already black then you don't count the white disc.

 

rofl.. wow i feel like an idiot... am i allowed to guess again?

 

Guess as many as you want

 

yay, the 2 year old gets to guess again!

the probability that the other side is black is 1/2, because 2 of the discs have at least 1 black side. This one black side is the one you saw at first, but only one of the 2 has another black side. Therefore, 1 (favorable outcome) out of (/) 2 (all possible outcomes), or 1/2.

 

I'm trying to figure out what the catch is. 1/2 is the obvious answer, but it just seems too obvious( no offense EMR ).

I'm going to say it is zero because once you turn to look at the back side it will become the front side. I know, it is a stupid answer but there has got to be a catch to this.

 

That's not the catch, there is no grammatical catch.

Just logic. And the 1/2 logic is wrong.

 

If one half is wrong I think I need to revise my interpetation of the question.

P1 = P[ draw a disc with a black front showing ] = 1/2
P2 = P[ a disc with black front has a black back ] = 1/2

P[ drawing a disc with a black front and having a black back ] = P1 * P2 = 1/2 * 1/2 = 1/4

 

Pulling a both-sided black disk is 1/3, how can the probability be less than that?

 

you would have a 50/50 chance of getting the other side of the disk to be black because of the two with black sides, one has a white underside, and one has a black underside. By looking at the black side you picked up, you have no way of telling what color the other side is, but you know that half of the ones with a black side have black on both sides, so you know your chances of getting the one with both black sides is 50%

 

And... nope

When I give the answer, if no one gets it, I will get eternally slapped.

This one defies common sense even more than the curtain one.

 

100% when you turn the disk over to see the other side, the black side of the disk that was facing you is now the back side, meaning that the back side is guaranteed to be black.

 

I'll try and draw a tree diagram

                                                ___
                                              /        back is black P[1/2]
     ___                             ___/
   /        front is black P[1/2]     \
 /                                            \ ___  back is white P[1/2]
/
\                                                ___
 \                                            /        back is black P[1/2]
   \ ___  front is white P[1/2]___/
                                            \
                                              \ ___  back is white P[1/2]

As you can see the odds of drawing a disc with the front side black and the back side black is 1/4.  This is because the disc with a black and white side acts like 2 seperate discs on the first draw since drawing it with one face turned one way is considered different than having that face turned the other way.

I just realized how to do it. This is wrong, well its right but it doesn't answer the question.

 

yeah you will have a 50% chance of getting the disk with both side being black.
you have 3 disks but one is white all over so that one is out because you pulled out a disk that have at least one black side.
you have one thats half white and half black so it could be that one
and you have one thats all black so it could be that one as well.
so it would be 50%

@ eonmaster: 100% is not right, there are two disks that have at least one black sides all you know is that the disk you are holding have its front side black, you dont know if the back side is black or white.

 

lurker I already said that and it was wrong, I think my new answer at the top may be right, though.