From c220dd2fa47933679b40c59eb14a3b338ee1eee7 Mon Sep 17 00:00:00 2001
From: "Niels G. W. Serup" <ngws@metanohi.name>
Date: Fri, 22 Mar 2024 19:40:09 +0100
Subject: [PATCH] Add new card trick article!

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+---
+abstract: A mathematical proof.
+---
+
+# In Defense of Poor Card Tricks
+
+I have four opinions about the art of manipulating cards for magic
+tricks:
+
+1. I like doing card tricks.
+2. I don't like learning card tricks.
+3. I like making new card tricks.
+4. I'm not very patient with making new card tricks.
+
+My core belief in these opinions result in one of two things happening
+when I'm performing card tricks in front of friends and family: either
+
+- the card trick works but is not very sophisticated, so they figure it
+  out; or
+- the card trick is very clever or complicated but completely fails.
+
+I think that both these results would typically be considered failures,
+but I have found a way to think of them as successes, which I'll detail
+here.
+
+The first insight is that *the point* of doing a card trick is
+*to entertain*.  What's not immediately clear from this insight is *whom*
+the entertainment is aimed at.  The ones receiving the card trick?  The
+magician who's performing it?  Both?
+
+I think it's only fair that both the giver and recipient of a card trick
+gets joy out of it.  Describing this more formally, we can say that the
+total entertainment value $E_T = E_M + E_R$, where $E_M$ is the
+magician's entertainment, and $E_R$ is the recipient's entertainment.
+
+For the recipient, we say that
+
+- any card trick that fails or is easy to figure out has a value of $0$,
+  and
+- any card trick that succeeds has a value of $1$.
+
+The value must be within this bound.
+
+For the magician, the math is more complicated.  We assume the worst
+case, where the magician has no idea what they're doing, and so whether
+the card trick works or not is up to pure luck.  This makes it harder to
+succeed (in the classical sense), but also hightens the enjoyment when
+the trick actually *does* work.
+
+Assuming an evenly spread out distribution, the chance of success is
+$1/52$.  To be fair to the math of the recipient, we use the same upper
+enjoyment bound of $1$ for the magician, just in the context of a random
+variable instead.
+
+So if the magician succeeds, $E_M = 52$, but in all other cases $E_M =
+0$, so $E[E_M] = 1$ (the expected value).  That is, the magician is
+getting more enjoyment out of a successful trick than the recipient,
+because only the magician knows how hard it is to achieve this success.
+
+We assume that the recipient *expects* a 50% chance of success, in which
+case we also have $E[E_R] = 1$.  Even though this expectation will be
+wrong if the recipient only ever receives card tricks from the poor
+magician, we assume that this poor performance is amortized by other,
+better card trick magicians.
+
+In the end, we can see that, by a big margin, $max(E_M) > max(E_R)$, and
+so it is mathematically valid to perform unpracticed, not very good card
+tricks!