Re: coins in a dark room
 

Ken,

Last time I checked, 18+18=36 not 32.

Mike

Re: coins in a dark room
 

Your answer is right but the explanation is wrong:

The original piles has 32 heads up, so you take 32 from the pile and flip them over. That is your second pile.

If you grabbed all 32 heads then there are no heads left in either pile

if you grabbed 18 heads and 14 tales, then you left 14 heads in the first pile, and when you flip the second pile you have 14 heads in both piles

if you grabbed 19 heads and 13 tales, then you left 13 heads in the first pile, and when you flip the second you have 13 heads and 19 tails

it works for all combinations

Regards,

Mike

Re: coins in a dark room
 

Are you sure thats the riddle? I did a bunch of googling (Try googling a riddle :rolleyes: ) so yeah here is what i came up with.
"You are wearing a blindfold and thick gloves. An infinite number of quarters are laid out before you on a table of infinite area. Someone tells you that 20 of these quarters are tails and the rest are heads. He says that if you can split the quarters into 2 piles where the number of tails quarters is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and the gloves prevent you from feeling which side is heads or tails). How do you partition the quarters so that you can win them all?"
uh i also found a forum based on this riddle. http://www.ocf.berkeley.edu/~wwu/cgi...m=102 8100705

Re: coins in a dark room
 

it was actually just something that my friend said to me
i don't have a clue where he got it from

Re: coins in a dark room
 

Which one of the states is infinately large? :-P
What state is on the back of all the quarters?
Do you know what breed of horse is on the back of the Delaware quarter?



...A quarterhorse.

Re: coins in a dark room
 

Here's the algebraic proof that the solution of grabbing 32 coins and flipping them works.

Let's call our pile of 18 Pile I
Let's call our pile of 32 Pile II

Let a be the number of heads in Pile I
Let b be the number of tails in Pile I
Let x be the number of heads in Pile II
Let y be the number of tails in Pile II

Therefore,

eq1: a + b = 18 (From Pile I)
eq2: x + y = 32 (From Pile II)
eq3: a + x = 32 (From the total number of heads)
eq4: b + y = 18 (From the total number of tails)

Combining eq2 & eq3 we see that,
y - a = 0 --> y = a

Therefore the number of heads in Pile I is the same as the number of tails in Pile II

Now, we flip all 32 coins in Pile II. So all the heads become tails and vice-versa. So now we have y heads and x tails in pile II.

Since a = y, we see that the number of heads in both piles I & II is equal!

Re: coins in a dark room
 

ok, I was in a hurry: 32 / 2 = 16, not 18

32 heads and no tails
31 heads and 1 tail...
down to 18 tails and 14 heads, it comes out the same - no matter which 32 you grab

for the question with infinity coins, you grab 20 and flip them, and make that your second pile

I want to know, how can you tell only 20 were originally tails up, if there is an infinite number of coins in the room?

Re: coins in a dark room
 

You can't, as you can't fit an unlimited amount of something in a finite space.

Not to mention that there is no such thing as infinity coins.