Over the weekend I will be giving a couple of talks at the Cheltenham Science Festival. Do come along and say hello. So, to the 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?

As ever, please do **NOT** post your answers, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

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.

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Easy one!

I could do this one blindfolded

I’m not sure I understand the question, but the way I’ve interpreted it, I have a solution.

It seems ambiguous as to when the man is allowed to start interacting with the cards – either before 10 of them randomly turned over or after. If it’s the former then I have a solution, if it’s the latter then it’s a harder puzzle and I haven’t worked it out yet (but I like it).

As the man is given the cards *before* some are turned over, I suspect it’s the former.

I interpret it as being the harder puzzle. It is solveable!

I’ve heard this one before… It is definitely the latter. I’m curious what your answer to the former would be, though. He can’t mark the cards or anything like that…

Is it a man who has come to fix the blinds?

Or can he really not see?

Or is it a man who has come to fix the blinds who is also severely visually impaired?

These questions are so ambiguous

The problem is ambiguously stated. However, since there’s really only one assumption that will allow a blind man to solve it, I’m pretty sure I got it in about 90 seconds.

Got it. Sadly took more than one stop on the train. About 5 mins.

I have 2 solutions, but I know one of them is THE answer. The other solution is not very good and does not work with every pack of cards.

Took me about 15 seconds to figure out, and a minute to check.

Had heard something like this before but couldn’t recall the solution instantly. Took a couple of minutes to work it out again.

Are there meant to be two piles of five cards, two piles of twenty six cards or (for completeness) two piles of some other number?

As I wear a mask this question should be right up my street, but I’m not sure that I understand it.

two piles and ensure that that each pile has the same number of cards facing up

Thank you M.

So two piles, each one card high would be a solution – as long as each pile had the same number of cards facing up?

Dividing the cards into 2 piles means that all the cards have to be used in just those 2 piles. If you made 2 piles of 1 card each then there would be another 50 cards in a 3rd pile, which wouldn’t satisfy the requirement.

A pile of 1 card plus a pile of t’other 51 cards would be fine in theory, so long as you can also meet the ‘face up’ requirements.

Simples! Took about 30 seconds. There is more than one answer, of course, depending on the scope of solution available, but there is ONE way the blind mand can do this unaided🙂

And without “seeing” the deck prior.

I know this puzzle (in another form), so no time at all to solve it.

I’m afraid I can’t even see a puzzle here – assuming the man can count.

Its not you that can’t see, its the man

I’d love to hear your solution then, because it’s blatantly wrong.

exactly what i thought stu!

I have two perfectly valid answers, but I suspect that neither is the required one.

two seconds to solve, if I have it right!

i spent ten minutes considering weather the blinded person could engage the assistence of their guide dog or this. i think they could do this with training, there are many very clevver dogs around, think of Pudsley and Ashley for e.g.

but i think mr wiseman want’s a mathematical trick solution so i create one. is simple. just deal into two piles while alternatively turning cards over. not hard!

a) As asked, don’t post your solution in here, it will be posted on Monday.

b) It’s not correct (quite).

aleck, thank you for your comments. but remember still, this is the internet. mr wiseman has asked for help and i have offered help. i cannot garantee to offer my help to any pacific deadlines of his. if he wants my help it is freely available but by my schedule.

also to withhold my publication is also censorship and that too is antthetical to the internet.

He hasn’t asked for help, he’s posed a puzzle for those who are interested to have a go at solving. The only thing he’s asked is that people not post solutions. Since you’ve proceeded to not only do that but claim that a request not to spoil the puzzle for others is “censorship”, I can only assume that help is the furthest thing from your mind.

This was on Dara O’Briain’s Maths show a couple of weeks ago. It relies on being ambiguously worded; the piles don’t have to be of equal size, just have to have the same number of face-up cards.

What’s ambiguous about the wording? The puzzle seems well-defined to me.

If some people presume there are requirements that aren’t mentioned in the puzzle, that’s their mistake, not a problem with the puzzle wording.

You don’t say whether or not the randomly chosen cards are returned to the pack!

I think it’s a logical assumption that they are!

Ill-defined problems with Richard are the norm. Here’s a (perhaps)clearer statement of the problem which has the solution he’s looking for.

You have 52 cards in one pile, 10 of which, scattered at random through the deck, are face up. With a blindfold on, divide the pack into two piles each containing the same number of face up cards.

I don’t think I’ve seen this before, and it took me about 3 mins to work it out.

wow, that helps a lot. makes much more sense!

That’s the “real” puzzle that I assumed before, and removes the ambiguity, thanks! Now to just think of an answer.

You’ve pretty much restated it verbatim. This does not add enough value to give you the right to slag off Richard’s puzzle like that.

Verbatim? Not remotely.

Ivan as stated by you the puzzle is insoluble

You will be telling us next you can divide the cards into two piles each with the same number of spades on whilst wearing a blind fold

ivan’s statement of the problem is exactly correct. 10 cards are randomly located faced up in the deck. Without looking separate the deck into two piles each of which has the same number of faced up cards. It’s definitely solvable. When you see the answer you will slap yourself for not getting it.

Ivan/Niva

I stand corrected.

I worked out a solution in practice and slapped myself – I still haven’t worked it out in theory, however, which is more annoying to me.

Think I got it, took me a couple of minutes spread through the morning…

I don’t see the ambiguity in the question some people have sugested. Think I’ve got it, fairly simple! Assuming my answer is correct. Lol

I think it’s very ambiguous. The blind man is given the pack of cards… if he chooses ten cards randomly and then turns them face up, he now has 42 cards faced down and 10 cards faced up, which by my reckoning would be fairly easy to split into two piles containing an equal amount of upturned cards!

Ivans wording of the puzzle above is much clearer.

Seen this one before, nice puzzle though. While perhaps the wording could be clearer, to make the puzzle easier, there are no ambiguities.

Got it in about five seconds, but I have seen something like it before!

took me a while and I had to try it out with a pack of cards to make sure it always works

I took a deck of cards and the first try revealed not only the answer but furthermore provides in each pile exactly 5 cards facing up. I have no idea why that is, but it works every time. I also found Richard’s solution now.Took me about half an hour until I picked up that deck of cards.

Hm, the answer that I’ve thought of wouldn’t guarantee 5 in each pile, but would at least make sure that they’re even. Maybe I’m wrong!

Think I have the answer now I’ve had time to think about it, but would like a pack of cards to test it out🙂

Oh yes, that’s right, I just happened to have a pack in my desk at work. Handy. Confirmed it🙂

Lies, I thought I had it but was able to find out variations where it doesn’t work. So still stumped. Good puzzle this.

I don’t feel smart today. The only solution I can think of involves destruction, and judging by the comments of people ‘trying’ and confirming their solutions, I doubt I’m right.

Surely any form of dividing (dealing, turning) the cards would not guarantee an even split… Taking 26 and 26 wouldn’t work as you could have them all in the same pile. Dealing alternately wouldn’t work as the cards could be all placed at odd or even intervals, again giving all in the same pile. It seems I’m stumped!

Another puzzle that relies more on ambiguous presentation than it does on a clever solution.

Color me disappointed.

But, Ivan, the puzzle doesn’t say the blind man is handed a deck with ten cards already turned face up and mixed into the deck.. It says “Ten of the cards are then randomly chosen and turned face up.” The blind man has already been given the deck, so isn’t he the one who turns ten of the cards face up?

I think that was Ivan’s point regarding the ambiguity of the puzzle… it doesn’t say that the 10 randomly selected cards are returned into the pack either. If the blind man is the one that turns ten cards faced up, then the puzzle becomes extremely easy (he knows he has 10 face up cards in one hand and 42 face down cards in the other). Surely therefore the puzzle is *meant* to be be as Ivan describes it?

In fact, if “ten of the cards are then randomly chosen and turned faced up”, the blind man could just have easily fanned them out for someone else to select which would then leave him with 42 face down cards… again, pretty easy to split into two piles.

Think I’ve got it but I may be cheating, but says nowhere in the rules that I can’t use this method, will check Monday.

I think some people may be getting confused by the wording of a man being given a pack of cards – it makes no essential difference to the problem but is probably easier to visualise if it’s specified that the cards are not in a solid pile but already spread out on a table face down, before ten of them are turned face up. Then let the blnf man get dividing…

Or even the blind man…

“Ten of the cards are then randomly chosen and turned face up.” To remove any ambiguity, this should have been written “Ten of the cards are then randomly chosen and turned face up and reinserted randomly back into the deck.” If the ten face-up cards are taken out and placed in a separate pile the problem would be too easy. The real problem is just a little bit harder, but not much. If you have the correct solution it will work for any even number of face-up cards up to half the deck!

The solution I’ve got also works for odd numbers of cards, and more than half the deck. I’m intrigued if you’ve got a different solution which doesn’t.

@Smylers – By golly, you’re right! The method does work as you’ve indicated. Don’t know how I missed that. We must have the same solution then. Thanks for the correction.

I have an answer. It’s a stretch, but it works given the wording of the question. We’ll see.

Unusually, I am completely baffled by this puzzle, and I have no idea what the solution might be – even after reading all the comments so far😦

I think he’s given us this one before?!

Oddly enough, I knew “the trick”, but had to figure out the details of it again … that actually took about 5 minutes🙂

I think I’ve got an answer. I’ve no idea how long it’s taken me, because I spent half an hour looking for a pack of cards to check it works. Couldn’t find one.

Finally got it after finding a pack of cards. Nice puzzle simple logic needed in the end.

Read it early and went most of the day thinking it’s impossible to solve head on, without using a silly solution (destroyed cards can’t face up!).

But I gave it a more serious try and figured out how it’s different from for example locating all spades in the deck. Took me 15-20 minutes I guess.

Pretty cool puzzle. Difficult to engage if you get caught with the wrong impression of the problem.

Well it’s taken a while, but I’ve thought of a solution that satisfies the rules, but would take a long time to set up (and wouldn’t warrant checking with a real deck, so the answer isn’t the same as other people here).

If the task is to somehow identify which 10 were turned, I don’t see how it’s possible unless the blind man can handle (and manipulate) the deck before the 10 cards are turned.

Therefore I’m assume the answer is a clever trick involving turning cards… But again, I can’t think of a way to do that which would always work. Sometimes, sure, but not in every case.

Alright, I’m rambling now. Roll on Monday and the solution!

Got it in about five minutes – I had to visualise it.

I was stuck on this one for a while. Unless it is a completely new deck, you can usually feel which way the cards are bent and thus figure out which way they are facing. Took me 5-10 minutes to figure out the real solution and palmface myself…

lot of smart people around here😉 finally got it, but that took more then a few minutes🙂 If you know the answer though, it is quite an elegant one

I’m in the same boat as Thick Mick. Move over, Thick. I got it but I can’t explain why it works. And I’m not sure it works 100% of the time.