c4countdown

You have a deck of cards, 42 of which are face-down and 10 are face-up. You must divide the deck into two piles containing the same number of face-up cards. You are blind.

Clarifications:
* It's not a trick question.
* You must indicate when you're finished, and you only get one chance.

Shuffle them, count out 26 for each pile, and hope for the best.

(i.e. absolutely no idea if there's no physical differences between the two sides)

Tear every card in half, and put each half in a separate pile?

Keep taking the top card off the deck and putting it into a second pile, at some point they'll both have 5 face-up cards (although you won't know to stop at that point). Does that count?

Liam Tiernan wrote:Tear every card in half, and put each half in a separate pile?

Nice idea, but no. It's not a trick question. You can manipulate the cards but you can't destroy them, eat them or breed them.

Innis Carson wrote:Keep taking the top card off the deck and putting it into a second pile, at some point they'll both have 5 face-up cards (although you won't know to stop at that point). Does that count?

That's a fair solution to the problem as posed, I will clarify the wording.

Does it have to be an equal pile?

Marc Meakin wrote:Does it have to be an equal pile?

Nope.

Think I have it:

Deal 10 cards out as if you were turning them face up (i.e. any card facing down will now be facing up, but any card facing up will now be facing down). Then put all the remaining cards in the other pile (face down from your blind point of view). If all the cards you dealt out (in the first 10), were initially down, they would now be up so there would be ten in each. For each card that was initially facing up in the 10 you dealt, you would have one less card facing up after dealing, but you would correspondingly have one less in the other pile, i.e.

- Deal 10 cards "face up", 2 of which were originally up, leaves you with 8 up.
- The other pile must have the other 8 remaining cards that were face up originally.

Brilliant puzzle, I think this works:

Deal 42 cards in pile 1, then turn the rest upside down and place them in pile 2.

Both Bob and Kieran nailed it exactly.

This is one of the best puzzles I've seen for ages. I will pester my source for more.

Charlie Reams wrote: This is one of the best puzzles I've seen for ages. I will pester my source for more.

Agreed, that is an awesome puzzle.

Agreed! Great puzzle. Never seen anything quite like that one before. Now determined to demonstrate it, with a blindfold, sometime.

When reading the puzzle, I quite thought that it was clearly impossible.

Looking forward to more.

Charlie Reams wrote:Both Bob and Kieran nailed it exactly.

This is one of the best puzzles I've seen for ages. I will pester my source for more.

Yeah, not bad actually. Unfortunately I was too lazy/thick to give it any proper thought and read the solutions.

Howard Somerset wrote:Agreed! Great puzzle. Never seen anything quite like that one before. Now determined to demonstrate it, with a blindfold, sometime.

When reading the puzzle, I quite thought that it was clearly impossible.

Looking forward to more.

All of this. Bit mad.

[I haven't looked at any answers or suggested answers yet, cos I want to try it on my own]

I presume you're not allowed to enlist the aid of a sighted person?

Alice Moore wrote:[I haven't looked at any answers or suggested answers yet, cos I want to try it on my own]

I presume you're not allowed to enlist the aid of a sighted person?

Like Clare Sudbery?

(Dude, everyone knows.)

I spent a while thinking about manipulating pairs in various complicated ways, then ran out of time so had to admit defeat. Such a simple solution! This is a lovely puzzle.

I was distracted by stuff I learnt from another puzzle, which is also satisfying, although not quite so simple:

I hand you a deck of cards. I ask you to cut it. I stand well back and allow you to cut the deck wherever you please. I take the half you have removed and put it on the bottom of the deck. We repeat this process twice more.

I then ask you to split the deck roughly into two halves. I ask you to turn one of those halves upside down and riffle-shuffle (this is where the two halves are placed next to each other (slightly overlapping and with the ends curving up), then mingled together by riffling through both half-packs at once so their ends get kind of tangled up). We now have a pack whose cards are a mixture of face-up and face-down. I don't expect you to do a perfect riffle shuffle, so there will be clumps of face-up (and face-down) cards together.

I now place the deck behind my back. I cannot see them. I announce that I can show you two cards - one will be red and the other black: I do it successfully. I do the same again. Then I anounce that I will show you four cards, one of each suit. I can do this too. And again. And again, and again. Then I pull off three more red-black pairs. Then I announce that I will pull off (still behind my back) 13 cards, and although they may not be in order, they will contain exactly one each of A, 2, 3... up to Jack, Queen, King. Finally I hand you the rest of the pack - there are 13 cards left - and again, they consist of all (and only) the values from Ace through to King.

The puzzle is this: How did I stack the deck to ensure that no matter what you did during the set-up of the trick, I would always be able to pull what I wanted from behind my back?

Alice Moore wrote:I spent a while thinking about manipulating pairs in various complicated ways, then ran out of time so had to admit defeat. Such a simple solution! This is a lovely puzzle.

I was distracted by stuff I learnt from another puzzle, which is also satisfying, although not quite so simple:

I hand you a deck of cards. I ask you to cut it. I stand well back and allow you to cut the deck wherever you please. I take the half you have removed and put it on the bottom of the deck. We repeat this process twice more.

I then ask you to split the deck roughly into two halves. I ask you to turn one of those halves upside down and riffle-shuffle (this is where the two halves are placed next to each other (slightly overlapping and with the ends curving up), then mingled together by riffling through both half-packs at once so their ends get kind of tangled up). We now have a pack whose cards are a mixture of face-up and face-down. I don't expect you to do a perfect riffle shuffle, so there will be clumps of face-up (and face-down) cards together.

I now place the deck behind my back. I cannot see them. I announce that I can show you two cards - one will be red and the other black: I do it successfully. I do the same again. Then I anounce that I will show you four cards, one of each suit. I can do this too. And again. And again, and again. Then I pull off three more red-black pairs. Then I announce that I will pull off (still behind my back) 13 cards, and although they may not be in order, they will contain exactly one each of A, 2, 3... up to Jack, Queen, King. Finally I hand you the rest of the pack - there are 13 cards left - and again, they consist of all (and only) the values from Ace through to King.

The puzzle is this: How did I stack the deck to ensure that no matter what you did during the set-up of the trick, I would always be able to pull what I wanted from behind my back?

I presume you're not allowed to enlist the aid of somebody to stand behind your back and tell you what the cards are?

Group them so that each of the four thirteens contains one of each of A, 2, 3 etc. in the same order. Also arrange such that they are S, H, C, D, S, H, C, D etc. Cutting changes nothing. Shuffling in this way is the equivalent of starting at one random card and working up or down from that randomly. So just take off the top. It wouldn't work if you started by taking an odd number of pairs. It also wouldn't work if you tried to take the thirteen any earlier (though you could take the thirteen off the bottom at any time).

Did you leave out a step after the riffle-shuffle ? Deal out the cards in two piles (face up & face down), then put the 2 piles together again, effectively undoing the shuffle. Then Davids solution above would work.

No. What I'm saying is that taking the top card off the shuffled pack is the same as if you didn't complete the shuffle, but left the two halves as separate piles and took the top card off one or other at random. For example, the two piles could be as follows (top on the left)

Face down. H C D S H C D S etc
Face up. S D C H S D C H etc.

However you pick them you'll get a red and a black the first time and the second time. This leaves you with a similar position to the start point. Pick four more and you'll have one of each suit. And taking thirteen from the bottom if you've ordered them right will have one each of A, 2, 3 etc. But as there are only 26 left when you take out 13 the same is true whether you start at the top or the bottom.

David Williams wrote:No. What I'm saying is that taking the top card off the shuffled pack is the same as if you didn't complete the shuffle, but left the two halves as separate piles and took the top card off one or other at random. For example, the two piles could be as follows (top on the left)

Face down. H C D S H C D S etc
Face up. S D C H S D C H etc.

However you pick them you'll get a red and a black the first time and the second time. This leaves you with a similar position to the start point. Pick four more and you'll have one of each suit. And taking thirteen from the bottom if you've ordered them right will have one each of A, 2, 3 etc. But as there are only 26 left when you take out 13 the same is true whether you start at the top or the bottom.

Still can't figure out how you beat the randomness of the shuffle that way.

Try it!

David's right. It's called the Gilbreath Principle and it's pretty amazing. More here.

I was shown this trick by James Grime, who works at Cambridge Uni on something called the Enigma Project. I managed to work out the answer but hadn't heard of the Gilbreath Principle.

He's a great speaker - watch out for him at conferences etc, particularly if he's doing card tricks - which he also does on YouTube.