# dark room problem

ilya936
Posers and Puzzles 08 Sep '05 11:53
1. 08 Sep '05 11:53
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?
2. 08 Sep '05 11:57
Bah... That's not even challenging. Not only is it possible but is rather easy to do it.
3. 08 Sep '05 13:38
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).
4. 08 Sep '05 13:50
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).
i forgot to mention. NO TRICKS
5. 08 Sep '05 14:27
You burn them. And then pile the ashes into two seperate piles. And then go play some chess.

Oh and doing things by yourself, in the dark, makes you go blind. ðŸ˜‰
6. 08 Sep '05 18:201 edit
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
7. 08 Sep '05 21:44
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.