Puzzle Time
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Cats and George have devised a clever card trick. While Cats is out of the room, audience members pick five random cards from a bridge deck and hand them to George. He looks them over, pulls one out, and calls Cats into the room. Cats is handed the four remaining cards (all face down) and proceeds to guess correctly the identity of the pulled card.
How do they do it, with nothing for Cats to base her guess on other than the four cards she is handed?
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Questions:
- It's a 52 card deck, right?
- Cats cannot see the position from which the card is pulled, right?
- Can Cats see the values of the cards after they are handed to her?
- Can George arrange the cards in a certain order (encoding the card value with that order) before he hands them over?
I'm confused by the "all face down" remark and am not sure whether they it applies only to the moment of "hand over" or whether she can never look at them.
In the latter case, I'm out of ideas. In the former case, I'd say that George can maybe encode the card value by the order in which the remaining cards are sorted. On the other hand, there are only 4! = 24 different orders of four cards, so it's not clear how to encode a number between 1 and 52. My trick would easily work with six cards being pulled and five cards being handed over (giving 5! = 120 different orders, which is enough to specify a card between 1 and 52).
However, a possible way to do this with 5 cards only is to make use of the fact that George can choose the card. For instance, he could always choose the lowest or highed or "middle" card. That could presumably be used to narrow down the information that must be transmitted via the card order by one bit or two, such that at most 24 possibilities remain.
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Sorry that remark was confusing. George can order the cards as he likes and Cats can look at them.
The face down remark simply means that he can’t convey additional information based on varying whether individual cards are face up or down.
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Yes
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So yes to all 4 of your questions. The ‘face up’ point just means the 4! combinations doesn’t become 2^4 * 4! by varying whether each card is face up or down.
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This is probably a cheat ... but if I can exploit the “orientation” of how a card is placed (e.g., the angle the card is placed against the edge of the table), one card can encode arbitrarily many possibilities, limited only by George’s ability to place a card at the desired orientation precisely and Cats’ ability to tell apart one orientation from another (think of it like a Shannon limit of sort).
Thinking about encoding many possibilities with four cards more conventionally, I don’t think we need to encode 52 possibilities, but rather at most 48 for any given hand since the four cards laid out can be eliminated, so we just need one more “dimension” to double that 4! permutations. I have not work it out yet.
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Yeah that’s a cheat. You have the order of the cards. That’s all.
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I'm not quite there yet, but I think I'm on the right track. That said, maybe I'm making this too complicated.
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Let's give the card names from 1 to 52.
The selected cards are c-1 to c-5, whereby c-1 < c-2 < c-3 < c-4 < c-5.
He removes c3.
Now George proceeds as follows:
Let's call the 24 different orders in which four cards can be arranged s-1 to s-24.
Case 1: c-2 and c-4 < 26.
In this case, c-3 must be between 1 and 26, but not c-2 or c-4. This means, at most 24 possibilities remain. If c-3 was the n'th card (not counting c-2 and c-4), choose order s-n.
Case 2: c-2 and c-4 >= 26.
This case is symmetrical to case 1.
Case 3: c-2 < 26 and c-4 >= 26.
Here I'm stuck.
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It would all be easy if somehow one additional bit of information could be transmitted, which would bump up the 24 sorting orders to 48, which would be enough (since only 48 cards remain).
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I will say this much - it’s definitely the case that the selection within the five by George has to be strategic as well as the ordering of the other four.
DON’T let them eat cake.
- RFK Jr -
@jon-nyc said in Puzzle Time:
I will say this much - it’s definitely the case that the selection within the five by George has to be strategic as well as the ordering of the other four.
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On that point, I’m thinking the selection of that card has to be suite based. Since there are only four suites and George gets five cards, it is guaranteed that at least two cards will come from the same suite.
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Got it now, I think.
Here's how to get that extra bit of information out of the card choice.
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Now consider the colors (spades, clubs etc.) of the cards.
The notation 4.1 means that there is one color that shows up four times and 1 color that shows up once.
For the five cards there are these possibilities.
a) 4.1
b) 3.2
c) 3.1.1
d) 2.2.1
e) 2.1.1.1For four cards, there are these possibilities:
i) 4.0
ii) 3.1
iii) 2.2
iv) 2.1.1
v) 1.1.1.1From each of a) to e) you can choose a card to reach either one of i), iii) or v), or one of ii) and iv).
Let's call i), iii) and v) the "zero" case and ii) and iv) the "one" case.
There's the extra bit. Together with the 24 sorting orders I can then encode a number between 1 and 48.
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How would you do the encoding of the 48? What kind of prima facie coding scheme would you use when you don’t know in advance which 48 cards you’ll be distinguishing among?
Ax - you’re on the right track. Or a right track at least.
DON’T let them eat cake.
- RFK Jr -
@jon-nyc said in Puzzle Time:
What kind of prima facie coding scheme would you use when you don’t know in advance which 48 cards you’ll be distinguishing among?
That part's easy.
From the 52 cards, subtract the 4 cards that were handed to Cats. Sort them. Name them 1 to 48.
The number encoding is: Name the 24 sorting orders s1 to s24. If you transmit 0 via the method I described above, sorting s-n stands for card n. If you transmit 1, it stands for 24+n.
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Right but the need to choose a particular card to implement your ‘bit’ could conflict with your ability to represent the number, couldn't it?
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Lol.
There’s a much easier way which people could easily do in their head in real time to make the trick work.
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Considering 5 random distinct numbers from 1 to 52, the magician could choose either the highest or the lowest number to keep secret, while ordering the remaining four to identify a number between 1 and 24, since there are 24 orderings of four distinct things A,B,C, and D where A is the lowest number, B the second lowest, etc. The magician would choose either the highest or lowest of the 5 numbers, whichever one was within 24 of the remaining highest number, considering 52 to roll over to 1 as numbers increase. This should always be possible. The assistant would identify the hidden number as the indicated offset from the highest of their four cards, rolling over from 52 to 1 if necessary.
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