# [ExI] Fashion

Anders Sandberg anders at aleph.se
Fri May 10 10:18:15 UTC 2013

```On 2013-05-09 15:41, Stefano Vaj wrote:
> On 5 May 2013 09:50, Giulio Prisco <giulio at gmail.com
> <mailto:giulio at gmail.com>> wrote:
>
>     My own coolness standard is simple: I don't think about coolness at
>     all. I tend to buy cheap but solid clothes that may last for decades,
>     with simple colors without labels and decorations. When I wake up, I
>     wear the first clean things that I find. The message that I try to
>     give is "there is no message here."
>
>
> 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...

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.)
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