Add new card trick article!
This commit is contained in:
parent
343a3a1f46
commit
c220dd2fa4

@ 0,0 +1,68 @@





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!

Loading…
Reference in New Issue