[ExI] Fashion
spike
spike at rainier66.com
Fri May 10 13:59:10 UTC 2013
On Behalf Of Anders Sandberg
Subject: Re: [ExI] Fashion
On 2013-05-09 15:41, Stefano Vaj wrote:
.
>>.Hey, I spend more than a little time every day deciding what to wear, how
to put things together in original ways, what else I may need and how much I
do not care about the embarrassment of being unconventionally overdressed
for the occasion. :-)
>.Yes, but you do live in Milano and you do have great taste. And a mutual
friend mentioned an amazing shirt collection...
EXCELLENT Anders me lad! Finally! Someone writes about fashion in a
language I can understand! This is ground-breaking work sir. If fashion
magazines would have articles with this kind of material, I might actually
read them, and perhaps put together some spreadsheets. If the models work
right, I could lose my geek status, by becoming a snappy dresser! Of course
it would be done by formula rather than by the traditional means, so that
might actually strengthen my status. It isn't clear how it would work.
spike
Anders posted the following fascinating analysis:
>.A mathematical model of dressing: suppose you have probability p of
selecting something that looks good, and probability q of noticing when you
have a bad combination. So after the first try you have a nice combination
with probability p, leave with something ugly with probability (1-p)(1-q),
and do a new try with probability (1-p)q. Then the total probability of
ending up with something nice will be P = p + (1-p)q( p + (1-p)q ( p + ...
)))). The series S=1+x(1+x(1+...)))) must fulfill S=1+x(S), or S=1/(1-x) (it
is a geometric series after all), so P = p/(1-q(1-p)).
For example, if p=0.5 and q=0.25, P=0.5/(1-.125)=0.57. If you have a sharper
eye, q=0.5, and now P=0.66.
But, how long does it take? The number of steps is distributed as a
geometric random variable with parameter (1-p)q. So you should expect to do
1/(1-(1-p)q) trials before you finish. In the above p=q=0.5 case you would
hence on average try 1.33 times.
What is the most efficient level of critical scrutiny? We could model the
utility as the probability of dressing nicely divided by the number of steps
it takes:
U=p[ (1+(1-p)q) / (1-q(1-p)) ]
Obviously the utility goes up as p increases. It is a bit less harder to
see, but the bracketed expression is also monotonically increasing: higher q
means better utility. Hence one should be as critical as one can.
Another utility model says that nice appearance has value N and sloppy
appearance has zero value, and lost time has value -1 per step. In that case
the total utility will be U=N(p/(1-q(1-p))) - 1/(1-q(1-p)) = (Np -
1)/(1-q(1-p))
So the utility goes up for higher q if N>1/p. If the value of looking neat
across the day is better than losing a few minutes in the morning (how many
minutes lost depends on your sense of style p) you should hence be a
critical dresser. If if isn't, just throw on anything that fits and is clean
enough.
However, the wisdom of XKCD strikes again: by thinking through these
considerations now, you can optimize your dressing for the foreseeable
future. And you can save a lot of time by increasing p, especially if it is
really low (it is both likely easy to increase, and it reduces the time
spent iterating).
But rationally, in situations where variation doesn't matter, it might be
best to use memoization: just spend enough time to find a really perfect
combination (or a set of combinations) and store them for later. That way
dressing can be solved in O(1) time for everyday use.
(Currently going for a spring look with a tan suit and slate-blue shirt that
I think complements my cryonics necklace nicely.) Anders
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.extropy.org/pipermail/extropy-chat/attachments/20130510/1ce1994f/attachment.html>
More information about the extropy-chat
mailing list