Cards in the darkness

Question

A man is sitting in a dark room. In front of him is a pile of 100 cards. He knows that there are 90 cards faced up and 10 cards faced down. Is it possible for the man to form two piles from these cards such that each pile contains the same number of faced down cards? No loopholes...

Answer 1

Assuming Gerhard's edit reflects what the OP meant to ask, then yes.

Take 10 cards from the deck. You now have a stack of 10 cards with n face down cards and a stack of 90 cards with 10-n face down cards. Now turn the stack of 10 over, to give you 10-n face down cards.

Answer 2

And out-of-the-box solution.

Make 2 roughly equal piles and place them standing one against the other. No card faces down, they all face the one or the other side.

Answer 3

Arbitrarily (randomly) divide the pile (or deck) of cards into two piles of $90$ and $10$. \begin{array}{lcl}\text{Status}&\text{Pile1}&\text{Pile2}\\\text{Total cards}&90&~~10\\\text{Total face-down}&x&~~10-x\hskip1in\end{array}
Now flip all cards in Pile2 to get equal number of face-down cards in both piles. \begin{array}{lcl}\text{Status}&\text{Pile1}&\text{Pile2}\\\text{Total cards}&90&~~10\\\text{Total face-down}&x&~~10-(10-x) = x\end{array}

Answer 4

The key is you have to make any two piles, doesn't necessarily 50/50

Lets call $C = 100$ the number of cards
and $N = 90$ number of face up

For simplicity imagine $N = 1$

You take 1 card at random from the $C$ pile now you have a $pile A = 99$ cards and $pile B = 1$ card

Now you flip the card in $B$ two scenarios occurs
$\cdot$If card $B$ is the face up mean $A$ doesnt have any face up cards after the flip $A$ and $B$ have 0 face up cards
$\cdot$if card $B$ is face down mean $A$ have 1 face up cards after the flip $A$ and $B$ have 1 face up card

The Generalised Case

There are $N$ face up cards in Pile 1. We take $N$ cards from Pile 1 to form Pile 2. Lets say, by chance $F$ of those are face up. This necessarily leaves $N-F$ face up cards in Pile 1.

Pile 2 contains $N$ cards, $F$ of which are face up, meaning $N-F$ are face down. When we flip over Pile 2 these $N-F$ face down card become face up. Both piles contain $N-F$ face up cards.

Answer 5

First he builds two piles by dealing one card at a time. (0,0), (1,0), (1,1), (2,1)...(50,50).

After this, he puts one pile onto the other (without shuffling) and repeats the process, but he is then dealing a different number of cards.

The idea is to follow 1*2*2*5*5 = 100. In other words:

First round, 1 card per pile.
Second round, 2 cards per pile.
...
Last round, 5 cards per pile.

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I tried it and it worked. Maybe I was lucky since I have no idea why it worked (if it does).