You are in a dark room.
There are 52 cards on the table all mixed up. 13 of the cards are facing up while the rest are facing down. You are asked to seperate the cards into two piles with the number of cards facing up in each pile being the same. You can flip the cards over as you wish but of course you can't see which way the cards are facing, it's dark in the room. Is it possible to do?

I don't see a method that works whatever 13 in the pack are the wrong way.
Unless it is a trick - do you place the pack on its side? (0 face up & 0 down).

Originally posted by iamatiger I don't see a method that works whatever 13 in the pack are the wrong way.
Unless it is a trick - do you place the pack on its side? (0 face up & 0 down).

Originally posted by Mephisto2 What if you take 13 cards in one pile, and then turn them all over? The remaining 39 are left as they are. That should do the job.

edit. if x= the number of cards with face 'up' among the 13 chosen, then
pile of 13: x (Up) + (13-x) (Down)
pile of 39: (13-x) Up + (26+x) (Down)

turn the 13 over, leaves: x (Down) + (13-x) (Up)
QED

Doh! I was assuming that the number of cards in the two piles had to be equal. Nice one.