Card trick (presented to you as a puzzle)

[Jon Corby]

Here's a card trick my brother and I tweaked and perfected, and performed repeatedly to an adoring crowd of about 5 members of our family for probably well over half an hour, who were continually stumped (my dad might remember this).

The essence of it was kinda presented to me as a puzzle to work out, so I'll do the same here. If you've seen it before, or are a bit of a prick who scampers straight off to Google, don't spoil it for others please.

Basically, here's how the trick works:

My brother leaves the room.
A volunteer takes five cards from a regular 52 card deck, and hands them to me.
I lay four of these cards face up on the floor, and the fifth face down.
My brother returns to the room, and announces what the face down card is.

How?

You'll need to ask questions, which I shall of course answer.

[Matt Morrison]

No idea yet, but these questions seem a good place to start:

• Are/can the cards be shuffled after your brother has left the room?
• Can the five chosen cards be picked out of the deck in any order?
• Can the five chosen cards be shuffled before being handed to you? (actually that's basically the same question I suppose)
• If not, can your brother only announce what the 5th card is, or would it work regardless of which of the five cards are upturned?
• Does your brother's announcing what the fifth card is actually require him to be back in the room looking at the cards? (that's like a kill-all question for making sure there is no 'other' communication between the two of you, including card markings or the way you lay them out etc.)

[Jon Corby]

No idea yet, but these questions seem a good place to start: Good questions.

• Are/can the cards be shuffled after your brother has left the room? YES
  
• Can the five chosen cards be picked out of the deck in any order? YES
  
• Can the five chosen cards be shuffled before being handed to you? (actually that's basically the same question I suppose) YES
  
• If not, can your brother only announce what the 5th card is, or would it work regardless of which of the five cards are upturned? NO I'm choosing the upturned card. For presentation purposes it's best if I don't make this too obvious, although if you do it repeatedly (which you can) people are likely to catch on. Occasionally I do have a choice of which card to leave face down, but not often
• Does your brother's announcing what the fifth card is actually require him to be back in the room looking at the cards? (that's like a kill-all question for making sure there is no 'other' communication between the two of you, including card markings or the way you lay them out etc.) YES

[David Williams]

1. Do you have a glass floor?

2. Does the relative position and orientation of the laid-down cards vary each time you do it?

[Matt Morrison]

This is really fun. I don't want to take over, and nor do I want to ask so many questions that I strip the magic out of it either, so I'll shut up for a while after these ones:

• Do you look at all five cards before deciding which one you will leave face down?
• If so, would the order you lay down the four upturned ones make a difference?
• Is there any maths involved for you or your brother?

[Kirk Bevins]

Is the 5th card face down on the floor as it could be on glass and he could see the reflection.

Does your brother come in and just announce the card or does he spend time before announcing it?

Is your brother allowed to flick through the whole pack to find which card is left?

[Matt Morrison]

I think it's fairly safe to presume the cards aren't floating above a mirrored surface like glass, and nor does Jon's brother get to look through the rest of the pack.

This is Jon Corby we're talking about, not Derren 'Disappointment' Brown.

[Kirk Bevins]

Are the cards arranged in any way? i.e. face up face up face down face up face up? If so this could give your brother a hint as if you have: 6, 7, face down, 10, jack he might know it's between a 7 and 10 numerically. Am out of ideas now.

[Sue Sanders]

If you were to leave the room when your brother comes in, would he still be able to give the answer?

[Jon Corby]

1. Do you have a glass floor? NO

2. Does the relative position and orientation of the laid-down cards vary each time you do it? I lay them down in a line, as you'd expect four cards to be laid down so you could see them, i.e. 1-2-3-4. The fifth card can be anywhere really, somebody can keep hold of it if they like.

[David Williams]

There must be at least two cards the same suit, so you pick one of them and put the other in an agreed spot relative to the others.

You now have three other random cards to use to determine the number, plus the fact that you can also have agreed that the face-down card can be higher/lower/further from 7 than the exposed one.

[Jon Corby]

This is really fun. I don't want to take over, and nor do I want to ask so many questions that I strip the magic out of it either, so I'll shut up for a while after these ones:

• Do you look at all five cards before deciding which one you will leave face down? YES. Again, I perfected this so I could do it pretty much instantly - it appeared as if I was just scattering the four cards down and then concealing the fifth (laying it face down, handing it to an audience member, whatever)
• If so, would the order you lay down the four upturned ones make a difference? Ooh - YES.
• Is there any maths involved for you or your brother? YES, but not difficult. As I said, I'm doing it pretty much instantly as if just scattering down four cards without thinking. Likewise my brother can reply instantly. This does take a little practice though...

[Jon Corby]

Are the cards arranged in any way? i.e. face up face up face down face up face up? If so this could give your brother a hint as if you have: 6, 7, face down, 10, jack he might know it's between a 7 and 10 numerically. Am out of ideas now. NO. Not as such - the four cards lay side by side. The fifth can be anywhere, face down, in the volunteer's pocket, whatever.

[Jon Corby]

If you were to leave the room when your brother comes in, would he still be able to give the answer? YES. I assume what you're getting at is do I give him some kind of verbal/visual cue. I don't. Obviously as you repeat the trick, people want you not to say anything, not to look at him, not to hold the pack, etc etc. You can eliminate all these as they request, once I've put the four cards down my part is over.

[Jon Corby]

There must be at least two cards the same suit, so you pick one of them and put the other in an agreed spot relative to the others. VERY GOOD.

You now have three other random cards to use to determine the number, plus the fact that you can also have agreed that the face-down card can be higher/lower/further from 7 than the exposed one. Very close, can you be a bit clearer though?

Edit: I'm heading home from work now so questions will remained unanswered for a few hours, sorry. DW is very close above though, we just need to be a bit more precise about what we're doing with the other three cards.

[David Williams]

You now have three other random cards to use to determine the number, plus the fact that you can also have agreed that the face-down card can be higher/lower/further from 7 than the exposed one. Very close, can you be a bit clearer though?

If I could work out the answer, then I'd be clearer! If you've got, say H4, H5, C, D, S you might put the H4 in position 1 so we know we're looking for a heart higher than 4. But there are only six different ways of arranging the other three cards. Close, but not close enough.

[Matt Morrison]

Haha I just made this amazing system up for working out the value (just value, not suit) of the card, spent 15 minutes writing it out then when I went to submit the post Jon had already said David was close. Arse. Back to the drawing board.

[Matt Morrison]

There must be at least two cards the same suit, so you pick one of them and put the other in an agreed spot relative to the others. VERY GOOD.

You now have three other random cards to use to determine the number, plus the fact that you can also have agreed that the face-down card can be higher/lower/further from 7 than the exposed one. Very close, can you be a bit clearer though?

Er, surely this gets fucked up when someone picks out five cards of the same suit, no?

[Gavin Chipper]

Basically, here's how the trick works:

My brother leaves the room.

Easy. I could do that.

A volunteer takes five cards from a regular 52 card deck, and hands them to me.

Not exactly rocket science.

I lay four of these cards face up on the floor, and the fifth face down.

Again, a piece of piss.

My brother returns to the room, and announces what the face down card is.

This is the only hard part so with only four parts to this in total, your brother only needs to be 33% better than someone with no training at all to do this.

[Kirk Bevins]

I was gonna post what David said about 4 suits 5 cards. I'd worked that out long ago but couldn't find how to phrase it.

[Paul Howe]

There must be at least two cards the same suit, so you pick one of them and put the other in an agreed spot relative to the others.

You now have three other random cards to use to determine the number, plus the fact that you can also have agreed that the face-down card can be higher/lower/further from 7 than the exposed one.

Is this not pretty much the answer?

Pick the first card so that: a) the first card has another card in the same suit, and b) the other card is no more than 6 places "higher" than the first one (you might have to go round in a loop)

so e.g if you had a Q and 2 (same suit), you'd turn the Q up as the first card and have the 2 as the card to be guessed. There can never be more than a gap of 6 between cards with the looping (7+6=13) so you have 3 cards remaining to communicate a number between 1 and 6, which can easily be done since there are 3! = 6 ways of arranging the cards.

Bit of a garbled explanation, but it looks right to me!

[Phil Reynolds]

There must be at least two cards the same suit, so you pick one of them and put the other in an agreed spot relative to the others. VERY GOOD.

You now have three other random cards to use to determine the number, plus the fact that you can also have agreed that the face-down card can be higher/lower/further from 7 than the exposed one. Very close, can you be a bit clearer though?

Er, surely this gets fucked up when someone picks out five cards of the same suit, no?

Why? All Jon's brother needs to know is that the concealed card is of the same suit as (say) the first exposed card. The suits of the other three cards are irrelevant. What we're after is some formula that allows you to use the values of any three cards to determine the value of a fourth. (Edit: like the one described by Paul Howe...)

[Gavin Chipper]

Without working out the particulars of how this is done, there are 120 combinations of how you can order the five cards and only 52 cards. So it's certainly well possible (you wouldn't even need to choose which card to leave face down) - the difficult bit would be working out a way of doing it quickly.

Let's base it on David's method then - face down card goes two to the right (or whatever) of card of same suit. Although they're in a line, you can pretend it loops round. Then if the face down card is on position one (of five) it means it's ace to 3, two means 4-6, three is 7-9, four is 10 or jack, five is queen or king. So we've already got it down to a maximum of three possible cards.

All the cards can be assigned a rank based on number and suit - this would be easy enough. Of the three face-up cards left, if the lowest ranked one is on the far left, then the face-down card is ace, 4, 7, 10 or queen (the only one it can be), if it's in the middle it's 2, 5, 8, jack or king and if it's on the right it's 3, 6 or 9.

Edit - or what Paul said.

[Phil Reynolds]

you have 3 cards remaining to communicate a number between 1 and 6, which can easily be done since there are 3! = 6 ways of arranging the cards.

Can you clarify a bit how this part works? There are indeed six ways of arranging three cards; but they could be any three cards from the pack. How do you ensure that you can communicate a number from 1 to 6 using any of the possible combinations of three cards in a deck?

[Matt Morrison]

There must be at least two cards the same suit, so you pick one of them and put the other in an agreed spot relative to the others. VERY GOOD.

Er, surely this gets fucked up when someone picks out five cards of the same suit, no?

Why? All Jon's brother needs to know is that the concealed card is of the same suit as (say) the first exposed card. The suits of the other three cards are irrelevant. What we're after is some formula that allows you to use the values of any three cards to determine the value of a fourth. (Edit: like the one described by Paul Howe...)

Yeah, I guess. I was just thinking the answer would be a bit more clever than just being the same suit as the first exposed card, as communicating the suit in that way would be utterly obvious to anyone who'd seen the trick more than say twice. Of course the card value bit is more clever so I guess that makes up for it.

[Matt Morrison]

you have 3 cards remaining to communicate a number between 1 and 6, which can easily be done since there are 3! = 6 ways of arranging the cards.

Can you clarify a bit how this part works? There are indeed six ways of arranging three cards; but they could be any three cards from the pack. How do you ensure that you can communicate a number from 1 to 6 using any of the possible combinations of three cards in a deck?

Yeah right, including four of the same value (hope I'm not being stupid this time like I was with the all-the-same-suit thing).

[David Williams]

Pick the first card so that: a) the first card has another card in the same suit, and b) the other card is no more than 6 places "higher" than the first one (you might have to go round in a loop)

Yes! Going round in a loop was what I missed. Then rank all the rest of the cards like in Bridge - 2C, 3C, . . . AC, 2D, 3D up to AS, and there are six combinations, exactly what you need.

[Gavin Chipper]

you have 3 cards remaining to communicate a number between 1 and 6, which can easily be done since there are 3! = 6 ways of arranging the cards.

Can you clarify a bit how this part works? There are indeed six ways of arranging three cards; but they could be any three cards from the pack. How do you ensure that you can communicate a number from 1 to 6 using any of the possible combinations of three cards in a deck?

Yeah right, including four of the same value (hope I'm not being stupid this time like I was with the all-the-same-suit thing).

You can just say that certain suits rank more than others.

[Jon Corby]

Back.

And yep, pretty much as Paul (and then others) have described. Of the 5 cards, (at least) 2 are of the same suit. You can add a maximum of 6 to one to reach the other (if it's a 2 and an 8, add 6 to the 2; if it's a 2 and a 9, add 6 to the 9 to get back round to the 2). So of the four cards you lay, one is a "control" card, which you add either 1,2,3,4,5 or 6 to. This number is communicated by the other 3 cards. Since they can be any 3, the only thing you can do is say one card is lowest (1), one is middle (2) and one is highest (3), and communicate the number with the pattern (ie. 123 = +1, 132 = +2, 213 = +3, 231 = +4, 312 = +5, 321 = +6). Use the rank of the card firstly to order, then an agreed sequence of suits (clubs, diamonds, hearts, spades [low->high] is the obvious) to differentiate cards of the same rank.

Example: cards are 3S, 6S, 7C, 8H, 10H. We have a choice here, we can either hide the 6S or the 10H. We'll choose the 10H, and assume that we've agreed position one will be the control card. So we lay 8H 3S 7C 6S. They take the control card as 8H, note that the remaining 3 cards are in the sequence 1-3-2 (lowest-highest-middle), so add 2, giving the missing card as 10H. Easy!

With a bit of practice, you can do it so it just appears you're taking the cards, glancing at them, and chucking 4 down without much thought (I think once out of probably 50+ times we did it I messed up so he was one out on the card). I think it's a little too obvious for an intelligent audience if your control card is always in the same location, so we developed a system whereby we'd converse once the trick was over, and whichever number (1-4) I said last before he left the room again was the position for the control card next time. You can actually be as blatant as signalling a number one to four with fingers, as people don't tend to notice that, as they think you can't be discussing anything useful as the trick hasn't started.

There's so much to whittle away as well depending how you start - if you start with 5 random cards from the deck, they'll want to pick their own (as someone above suggested, people often choose 1,2,3,4,5 of the same rank - far from making it difficult this actually gives you the greatest choice!), they'll want you to look away, they'll want you not to speak, they'll want to hide the card themselves etc etc. Hours of fun.

So there you go. That didn't take you guys long at all, you're all too smart. Now all you need to do is find a similarly smart accomplice.

[Derek Hazell]

So there you go. That didn't take you guys long at all, you're all too smart.

It would probably have been solved before you'd even written it, if Kieran wasn't taking a break from the forum to concentrate on Apterous!

[Kirk Bevins]

I don't get it. Anyone care to explain it in an easier way than Jon?

[Jon Corby]

I don't get it. Anyone care to explain it in an easier way than Jon?

Sorry, I am notoriously shit at explaining things. I'll give it another go if you like - which bit don't you understand?

• Of any 5 cards you pick from a 52-card deck, you will always be able to add 1,2,3,4,5 or 6 to one of the cards (call this one CARD A) to get to one of the others (call this one CARD B) in the same suit (wrapping round the ranks if necessary - eg counting +5 from J gives Q,K,A,2,3)
• Of the remaining three cards, you can identify one as being lowest (call it CARD C), one as middle (D), one as highest (E). For any tied cards, use the standard suit order (low->high = clubs, diamonds, hearts, spades) to separate them.
• These three cards can be arranged in 6 ways, which communicates how many (1,2,3,4,5 or 6) to add to CARD A to reach CARD B. CDE = +1, CED = +2, DCE = +3, DEC = +4, ECD = +5, EDC = +6. I accept that people who aren't used to permutations (and possibly programming) might think this is a tricky list to remember, but it's second nature to me to order and rank perms like this.
• You need to have an agreed location in the four cards that CARD A will be placed. For ease of example we'll have it in position one, but it's best not to keep it static if doing the trick multiple times.
• If we need to add 1 to card A to reach B, we would lay A C D E [B is upturned, or in someone's pocket, or whatever].
• If we need to add 2, we would lay A C E D. And so on.

That's probably worse, isn't it?

[Sue Sanders]

I've passed my eyes over all those words.

I think I'll stick to my more reliable party trick of being able to lick my own nipples

[Matt Morrison]

I've passed my eyes over all those words.

I think I'll stick to my more reliable party trick of being able to lick my own nipples

pics or it didn't happen

[Sue Sanders]

I've passed my eyes over all those words.

I think I'll stick to my more reliable party trick of being able to lick my own nipples

pics or it didn't happen

Please don't, Matt, because I do have pics and in moments of bravado, I do things I later regret!!!

[Kirk Bevins]

I don't get it. Anyone care to explain it in an easier way than Jon?

Sorry, I am notoriously shit at explaining things. I'll give it another go if you like - which bit don't you understand?

• Of any 5 cards you pick from a 52-card deck, you will always be able to add 1,2,3,4,5 or 6 to one of the cards (call this one CARD A) to get to one of the others (call this one CARD B) in the same suit (wrapping round the ranks if necessary - eg counting +5 from J gives Q,K,A,2,3)
• Of the remaining three cards, you can identify one as being lowest (call it CARD C), one as middle (D), one as highest (E). For any tied cards, use the standard suit order (low->high = clubs, diamonds, hearts, spades) to separate them.
• These three cards can be arranged in 6 ways, which communicates how many (1,2,3,4,5 or 6) to add to CARD A to reach CARD B. CDE = +1, CED = +2, DCE = +3, DEC = +4, ECD = +5, EDC = +6. I accept that people who aren't used to permutations (and possibly programming) might think this is a tricky list to remember, but it's second nature to me to order and rank perms like this.
• You need to have an agreed location in the four cards that CARD A will be placed. For ease of example we'll have it in position one, but it's best not to keep it static if doing the trick multiple times.
• If we need to add 1 to card A to reach B, we would lay A C D E [B is upturned, or in someone's pocket, or whatever].
• If we need to add 2, we would lay A C E D. And so on.

That's probably worse, isn't it?

Thanks Jon for taking the time to explain it. I understood point 1 and then point 2. When you said "tied cards" do you mean of the same number, e.g. 5 of diamonds and 5 of spades? I think I actually need a pack of cards to help understand this because I just can't picture what you're supposed to do.

[Jon Corby]

Thanks Jon for taking the time to explain it. I understood point 1 and then point 2. When you said "tied cards" do you mean of the same number, e.g. 5 of diamonds and 5 of spades? I think I actually need a pack of cards to help understand this because I just can't picture what you're supposed to do.

Yes mate. I mean, you can decide however you like to order them, but the most sensible seems to be by rank (2,3,4...K,Q,A) and then by suit.

You can try Googling, you will find the trick around (I don't know it has a name, so use your imagination for what to search on - it is out there!) and I think there are sites which will explain it better than me. And have nice examples. And possibly a GUI where you can try it out for yourself.