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...

  • $\begingroup$ That's impossible! You can't make a pile of 10 cards faced down into 50 cards faced down! What exactly does: separate those two piles into two piles mean? $\endgroup$ Jan 25, 2015 at 22:41
  • $\begingroup$ Depending which side a card shows it is called faced up or faced down. For example each card has a red and a blue side. If you pick up a card and you can its red side, then it is called faced up and faced down otherwise.. $\endgroup$
    – Imago
    Jan 25, 2015 at 22:51
  • $\begingroup$ @Bob I think the man can flip face-down cards into face-up cards and visa versa. $\endgroup$
    – Kevin
    Jan 25, 2015 at 23:00
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    $\begingroup$ Is this what you meant to ask? puzzles.nigelcoldwell.co.uk/thirtysix.htm $\endgroup$
    – Quark
    Jan 25, 2015 at 23:37
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    $\begingroup$ I assume he has to know that the cards are right and stop? If not, he can just spend eternity flipping cards in every permutation, right? Eventually he'll hit a "correct" state, even if he doesn't know it. $\endgroup$
    – Set Big O
    Jan 26, 2015 at 2:49

5 Answers 5


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.

  • $\begingroup$ @MaskedMan why not? You take the top ten cards and flip them: in your example this would give all 100 cards facing up. $\endgroup$ Jan 11, 2019 at 16:11

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.


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}

  • $\begingroup$ isn't this what frodoskywalker already answered? $\endgroup$ Aug 28, 2016 at 15:29

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.


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.

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

  • $\begingroup$ Is this method intended to produce the correct piles eventually, at some unknown time? $\endgroup$ Jan 26, 2015 at 11:47
  • $\begingroup$ I was thinking about how many times you have to repeat the process in order to make sure. There is maybe a connection to 1*2*2*5*5 = 100. $\endgroup$
    – Avigrail
    Jan 26, 2015 at 11:48
  • $\begingroup$ But is this intended to produce the desired result at the end of the process, or at some point in the process? $\endgroup$ Jan 26, 2015 at 13:04
  • $\begingroup$ There is a process with a certain amount of moves. After all the moves were executed, the piles are prepared. $\endgroup$
    – Avigrail
    Jan 26, 2015 at 13:23
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    $\begingroup$ I understand what you're saying, but you don't give any reason why you think this will work. $\endgroup$ Jan 26, 2015 at 13:57

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