On Friday I set this puzzle…..

A blind man is given a deck of 52 playing cards.  All of the cards are face down and the man is told that this is the case.  Ten of the cards are then randomly chosen and turned face up.  How can the man divide the cards into two piles and ensure that that each pile has the same number of cards facing up?

If you have not tried to solve it, have a go now.  For everyone else the answer is after the break.

The man divides the cards into piles containing 42 cards and 10 cards, and then turns the whole pile of 10 cards over.  Did you solve it?  Any other solutions?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle (UK here and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.


  1. Oh! It’s obvious now, and I feel silly that I didn’t get it. I probably would have got it if I had thought about it over the weekend for a few hours, instead of doing useful normal things.

  2. My solution was to take hold of the pack, count 10 off into one pile by turning each one over, then put the rest of the pack down to make another pile. Amounts to the same thing.

    What had initially tripped me up was that I was looking to create two EVEN piles of cards. Impossible.

  3. I didn’t understand the wording of the question at all. It wasn’t clear to me that the 10 cards were turned face up and randomly distributed throughout the deck.

    This solution is a good one, and I doubt I would have got there.

    1. I’m with you Rob. It was not at all clear to me from the original wording that the face up cards were put back into the deck before the blind man dealt out the deck into two piles. The puzzle should have read something like: “A blind man is given a deck of 52 playing cards. All of the cards are face down and the man is told that this is the case. Ten of the cards are then randomly chosen, turned face up, and replaced at random into the deck. How can the man divide the deck into two piles to ensure that that each pile has the same number of cards facing up?”

  4. Yes, got it, but only after commenter ivan had defined the puzzle much clearer. The original obfuscations left open the possibility for all kinds of ludicrous solutions.

    1. That was one solution I came up with too. I also thought that he could ask the person who handed him the cards to do it for him too.

    2. This was my solution, too.
      My understanding of the word ‘divide’ is to split X into Y parts, each containing the same amount. hence why 10/3 is always 3.33333 and not 3, 3, 4.

  5. Got it after about 20 seconds, and it took a few minutes to check if it worked. I like the answer.

    I also had an alternative answer that is not very good: most packs of cards face and back feel different. Just feel the cards and pile them up the right way.

  6. The solution can be proven mathematically. Take 10 cards from the deck. If there are x faced-up cards in these 10 cards, then there must be 10-x faced-up cards in the 42 other cards. Turn the 10 cards over and you will find 10-x faced-up cards. This will actually work for any number of faced-up cards. If there are n faced-up cards, take n cards from the deck and flip them over to find n-x faced-up cards in the n cards and in the other 52-n cards.

    1. If 10 cards have been turned up (and returned to the deck), then their position in the deck does not matter using the solution.

      The pile of 10 cards will contain X cards that are face up. The pile of 42 cards will contain 10-X that are face up.

      If you then flip every card in the pile of 10, the original cards that were face up (X) will now be face down and the other cards (10-X) will now be face up. Now both piles, 42 and 10, will have 10-X cards face up.

    2. That doesn’t make any more sense at all.

      > “The pile of 10 cards will contain X cards that are face up. The pile of 42 cards will contain 10-X that are face up.”

      There is absolutely no way you can be sure of that. You might have 10 face-up in the first pile and zero in the latter.

    3. Dave:

      >> “The pile of 10 cards will contain X cards that are face up. The pile of 42 cards will contain 10-X that are face up.”

      >There is absolutely no way you can be sure of that. You might have 10 face-up in the first pile and zero in the latter.

      What is 10-0?

    4. Actually Spamsac, staying true with the explanation given and Dave’s question, 10 face-up in the first plie would mean X=10 so the second pile would contain 10-X or 10-10 which equals 0. Once you flip second pile over, you now have 10 face-up like the first pile.

    5. David: Take 10 cards from the deck. If there are no face-up cards in your hand, there must be 10 face up cards left in the deck. When you turn all your cards over, you also have 10 face-up cards. Ta da!

      But say you happened to get 3 face-up cards in your hand. That leaves 7 face-up cards in the deck. Turn all your cards over and you also have 7 face-up cards. Ta da!

      Whatever number of face-up cards you have in your hand, subtracted from 10, equals the number of face-up cards left in the deck. And it also equals the number you have in your hand when you turn all your cards over.

  7. My solution was to make two piles each containing how many cards you like. Then hold them so there sides are facing up. Then both piles have no cards facing up! Some of the answers of these puzzles or so stupid I thought I had a decent shot at being right!

    1. Which way is up?
      Dormouse where are you?
      You are usually dogmatic in pronouncing on this sort of thing.

    2. That was also my solution. With the cards standing up on their borders there are no face up or down cards.

    3. This was my second solution. More elegant as it actually divided the cards in two. The idea was to make two, 26 card castles.

  8. My blind guy had marked the cards beforehand (he cheats at cards) and did it that way. His blind brother was more honest and had common sense to ask his sighted girlfriend for help.

  9. I did something different: I take all cards into one pile (randomly distributed within are 10 cards facing up) and start distributing them one by one into two piles. One card on the first pile, second card on the second pile, third card on the first pile, fourth card on the second pile, and so on… here’s the twist – all cards for the first pile I turn and all cards for the second pile I just place them the way they come from the initial pile. After distributing all cards I have two piles with each exactly five cards facing up.

    I don’t know why that works…

    1. I tried that at first, but it only worked once, and it didn’t make sense to me why it would anyway. Came up with Richard’s solution after another 2 or 3 minutes

    2. i also supplied the same solution as goliath, and i was told off by commentators for supplying a soln before they wanted to read one.

      the solution works, i have tested it twice with two different decks of cards (one with blue backs and one with red backs), it seems to work for all colours so far.

    3. You were told off, Dharmaruci, because Richard clearly asked his readers *not* to share their solutions (right or wrong) in the comments of the original post, in which the problem was first posed.

      Incidentally, I’m not sure I understand how this solution could even work in the first place. The first pile, in which all the cards are turned over, must have at least 16 face-up cards (not 5), because it contains 26 cards total, and no more than 10 could be face down (since that’s the number of cards that were face-up in the deck before being turned over and placed into pile 1). And in that extreme case, the second pile will contain no face-up cards at all (since they all went into the first pile). Am I missing something in your description? How can this possibly work at all, let alone be guaranteed to work every time?

    4. @Goliath — Your method fails. By dealing the deck as you have described is no different than just splitting the deck in half with 26 cards in each pile since the face-up cards are randomly distributed in the deck. If pile #1 has x faced-up cards, then pile #2 will have 10-x faced up cards. Flip pile #1 over and you’ll have 26-x faced-up cards. How is 10-x ever equal to 26-x? Cannot happen. In fact, your methods ALWAYS fails.

    5. Quite right you all are… I had a number of lucky tries (6 in a row to be precise) and wrongfully assumed it might be a solution to the problem. Have now tried it again and go only 2 out of 10 tries with equally distributed facing up cards. Thanks for correcting me.

    6. I’m still confused about exactly what you were doing, though, because, as Nivea articulately explained, the method you described fails every time. You will end up with 10-x cards face-up in one pile, and 26-x cards face-up in the other, where x is the number of face-up cards that originally went into pile 1. There is no value of x that can make 10-x equal to 26-x.

  10. “Ten of the cards are then randomly chosen and turned face up.” – I assumed that because this didn’t say that the blind man turned the cards over himself, that someone else did it. I also assumed that they were randomly turned face up at their position in the pack, meaning the blind man had a deck with 10 random cards facing up.

    As the blind man was given the pack in advance, my solution was for him to tear off the top right corner of all the cards, then when they were turned over they can only have the top left or bottom right corner with the tear….

    You know what they say about “assume”…..

    1. +1. “Dara O’Briain’s School Of Hard Sums” on Dave, worth a watch if you’re confused about it. Exactly how I solved it too!

  11. ooh lovverly logic, I missed it on friday but enjoyed a quick brainboggle with my buttie, got it in about four boggles

  12. I have to admit I only thought I got it right. I seem to have a tendency to read into the puzzle what isn’t there. I was looking for five upturned cards in each pile, I do like this puzzle.

    1. But it doesn’t say the blind man cannot ask someone else to tell him which side is up…

  13. I divided the pack into two even halves each with the same number of face up cards. Thinking trick question, I went for cutting all of the cards down the middle. Thinking along those lines (trick question) I gave no further thought as a puzzle. Love the answer though…. and it saves on cards!

  14. I assumed the blind man would look through the cards and make sure 5 were face up in each pile… then he would go and fix the blind.

  15. My solution was simpler – all the blind man needs to do is to tear the cards in half – thus producing two piles of cards with an even number of face up / face down cards. Works every time – ruins the cards though 😀

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