5

Long ago someone demonstrated the following card trick with a standard 52-card deck:

(1) A volunteer selects 5 cards from a shuffled deck, which the performer does not see.

(2) The assistant puts exactly 4 of those in a pile and passes it to the performer.

(3) The performer can then identify the 5th card.

There are a few ways to execute this trick, one of which goes as follows.

(1) The assistant finds a suit common to 2 of the 5 cards, with values $x$ and $y$ modulo $13$.

(2) He then removes one of them, say $x$, such that $x-y \in [1..6]$ modulo $13$.

(3) He then puts the other one at the top of the pile.

(4) And he puts the 3 other cards after that in a permutation corresponding to $x-y$ modulo $13$.

Since (1) works by pigeonhole, (2) works because $\frac{13-1}{2} = 6$, and (4) works because there are 6 possible permutations of 3 cards, the trick succeeds. My question is whether the same trick can work with more than 52 cards. It is not obvious to me how to do so, nor compute the maximum number of (pairwise distinct) cards possible. Also, it seems to me that even if the trick can be made to work for $n$ cards, it may not work for $n-1$ cards! Any ideas?

user21820
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    The problem isn't the number of cards, it's how many are in each suit and how many suits there are. So the correct generalization isn't "an $n$ card deck", it's "a deck of $n$ suits of $m$ cards each". – Jack M May 18 '14 at 14:01
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    I have seen a writeup on how to do it with a $120$ card deck. The order the cards are stacked can pick one of $24.$ The selection of which of the five will be kept gets the other factor of 5. I don't remember how to do the selection part. It was in a Martin Gardner column – Ross Millikan May 18 '14 at 14:12
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    @JackM: No, the trick can clearly be done with $52$ pairwise distinguishable cards. The suits are irrelevant although they help the performer and his assistant execute the trick faster. – user21820 May 18 '14 at 14:21
  • @RossMillikan: How can that be possible? I thought information would be lost because the performer does not know which of the $5$ were removed. – user21820 May 18 '14 at 14:24
  • @user21820 The suits are certainly not irrelevant. If the deck has no suits, I don't even know what the question means. That is, it's not clear how to generalize step (1) to the case of a suitless deck. – Jack M May 18 '14 at 14:39
  • @JackM: What I mean is that the trick is described in the first half, with 3 steps that the audience can see. The second half describes how to execute the trick for a standard deck of $52$ cards. The suit is used for convenience but the trick could have easily been carried out on any deck of $52$ distinct cards. – user21820 May 18 '14 at 14:43
  • @user21820 Oh, I see. So you want an injective function on the collection of $5$ card subsets of the deck, mapping $X$ to a $4$ card subset of $X$. – Jack M May 18 '14 at 15:01
  • @JackM: Yes, and the first paper that Barry linked to was the perfect answer. – user21820 May 18 '14 at 15:04

1 Answers1

5

For the $5$-card trick, a $124$-card deck is the maximum. Papers by Michael Kleber, Colm Mulcahy, and Shai Simonson and Tara Holm are good places to look for explanations, as well as some history of the trick.

Barry Cipra
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  • The Colm Mulcahy link is broke. – Hans Aug 16 '19 at 07:14
  • @Hans, thanks for the alert. https://web.northeastern.edu/seigen/11Magic/FitchCheneyFiveCardCount/FitchCheneyFiveCardTrick.html will connect to a copy of the Mulcahy article, along with other goodies. – Barry Cipra Aug 16 '19 at 13:47