71 lines
6.4 KiB
TeX
71 lines
6.4 KiB
TeX
% !TEX TS-program = xelatex
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\documentclass{../leelavati}
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\usepackage{framed}
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\begin{document}
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\selectlanguage{english}
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\setstretch{1.1}
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\title{\LARGE The Fitch Trick and Variants}
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\author{\AuthorDual{ନୀଲ୍ଧାରା ମିଶ୍ର}{Neeldhara Misra}}
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\date{}
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\maketitle
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\setstretch{1.2}
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\begin{abstract}
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The magician selects a volunteer to shuffle the deck and choose any five cards. The magician examines the cards, hides one of the five cards, sets the remaining four cards in a faceup row, and has the volunteer retrieve an accomplice from outside the room. The accomplice briefly examines the cards and identifies the missing card.
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\end{abstract}
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\setstretch{1.1}
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This is a trick that is to be performed by Aalaya (a magician) and Badarika (an accomplice) with a standard deck of $52$ cards. Badarika steps out of the room and leaves her phone, smartwatches, etc. behind, so nobody can communicate with her. Then the audience picks out five cards out of the deck and gives it to Aalaya. Aalaya then arranges four out of these five cards in a row\footnote{The cards can also be placed in a stack, if the audience worries that Aalaya is signaling by angularities in the layout of the cards.} and Badarika comes back and takes a good look at the four cards on the table, and then identifies the missing card.
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How is this possible?
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\begin{framed}
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...image here.
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\vspace{140pt}
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\end{framed}
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\newpage
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Notice that Badarika has to guess one out of 48 possible cards, since the deck has 52 cards of which four are face up. On the other hand, Aalaya can arrange the four cards in $4! = 24$ different orders, which is fewer than 48. This seems like we are falling short by a factor of two: but notice that Aalaya actually has a point of leverage that we have not accounted for: she can \emph{choose} the card that she wants to set aside.
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So what is the right way of thinking about a signaling strategy? On the one hand, there are: $${52 \choose 5} = 2,598,960$$ possibilities for what can happen at the trick. Call this set $\mathcal{P}$. Now the number of ordered sequences of four cards, say $\mathcal{Q}$, is $$52\cdot 51\cdot 50\cdot 49 = 6,497,400.$$
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As an example, let $\alpha = \langle 4S, 5H, 2C, 7C, 6D \rangle \in \mathcal{P}$ be the subset of cards handed to Aalaya by the audience. Now Aalaya has $120$ possibilities for what she can do with these cards. These possibilities are all recorded in $\mathcal{Q}$. What we want is a way to map every element $\alpha$ in $\mathcal{P}$ (what the audience does) to a distinct element $\beta$ in $\mathcal{Q}$ (what Aalaya would do). Now Badarika comes and sees $\beta$ and maps it back to $\alpha$ and figures out the missing card.
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Can you figure out such a mapping that Aalaya and Badarika can use to perform and impress? You can use math to argue that such a mapping exists \emph{in principle}, but more usefully, you can also try to construct an explicit mapping that both Aalaya and Badarika can use in practice to impress!
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\newpage
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Incidentally, if Aalaya is facing a tough crowd, they might not let her pick the hidden card, and might insisting on making the choice for her. Then indeed Aalaya has only $24$ signals: but turns out there is a way out even in this scenario.
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One way out, for example is the following: Aalaya can attempt to use her 24 signals in a way that narrows down the possibilities for the hidden card to just 2. Once Badarika sees the 4 cards and figures out the two possibilities, she runs through the whole deck and finds the two cards and casually places one on the top of the deck and the other at the bottom. After this setup, she can distance herself from the deck and ask the audience what the card was.
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If the audience shouts out the card at the top, then Badarika can snap her fingers and pull up the top card, and if the audience indicates the other card, then Badarika can simply flip the whole deck over with a dramatic flourish instead.
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You may have also noticed that there is a difference between the number of signals ($6,497,400$) and the number of possibilities ($2,598,960$) that we calculated earier. A natural question then is if we can make a more impressive trick with the same constraint on Aalaya and Badarika: that is the effect remains that four cards are shown in a fixed order and a fifth card is guessed, but can the audience make their pick from a larger deck?
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If the audience is choosing from a deck of size $n$ in general, then there are: ${n \choose 5}$ possibilities for what can happen at the trick and the number of ordered sequences of four cards is $n\cdot (n-1)\cdot (n-2)\cdot (n-3)$. We need, at the very least, that the space of the number of signals is at least as large as the space of possibilities.
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In other words, we would need:
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$$\binom{n}{5} \leqslant n(n-1)(n-2)(n-3),$$
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which, upon simplification, works out to $n \leqslant 124$. This, of course, only tells us that $n \leqslant 124$ is a necessary condition for the card trick to work in the form that we have discussed --- it may or may not be sufficient. Our next challenge will be to show card trick does, in fact, work with a deck containing as many as 124 cards. So once you have figured out how to do it with a standard deck, either in principle or in practice, see if you can extend your arguments to larger and larger decks: perhaps all the way to 124?
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A variation of this trick, as performed by Elwyn Berlekamp, is to use a deck of size 64, with the additional feat of guessing the result of a coin flipped by an audience member. If you can figure out a way of doing the 124-variant, you will probably also have some ideas for the version with the coin.
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You may also wonder if the reveal should necessarily involve an ordering of the four cards. A variation of the trick suggested by Colm Mulcahy involves revealing the cards in an order but also flipping some over, so that some cards are face up and some are face down. It turns out that with some clever observation, you can signal the hidden card using only three cards rather than four! This makes for great banter, as discarding a card is an excellent excuse for some drama.
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In its most general version, this trick has the audience pulling out $p$ cards out of a deck of $n$ and the magician revealing $q$ of the $p$ cards. For what values of $n$, $p$ and $q$ is this feasible? And for a feasible setup, how many distinct strategies are there?
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{\small
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\vskip 0.05in
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\affiliation[1.6em]{Associate Professor, CSE; IIT Gandhinagar}
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\email{neeldhara.misra@gmail.com}
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}
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\end{document}
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