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<div class="moz-cite-prefix">On 2013-05-09 15:41, Stefano Vaj wrote:<br>
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cite="mid:CAPoR7a5JiF2ONa5wk2A3F6YrS2UMVnL4ExnaTnazNcvqssNNvQ@mail.gmail.com"
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<div class="gmail_quote">On 5 May 2013 09:50, Giulio Prisco <span
dir="ltr"><<a moz-do-not-send="true"
href="mailto:giulio@gmail.com" target="_blank">giulio@gmail.com</a>></span>
wrote:<br>
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My own coolness standard is simple: I don't think about
coolness at<br>
all. I tend to buy cheap but solid clothes that may last
for decades,<br>
with simple colors without labels and decorations. When I
wake up, I<br>
wear the first clean things that I find. The message that
I try to<br>
give is "there is no message here."</blockquote>
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<div>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. :-)<br>
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<br>
Yes, but you do live in Milano and you do have great taste. And a
mutual friend mentioned an amazing shirt collection...<br>
<br>
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)).<br>
<br>
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. <br>
<br>
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. <br>
<br>
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: <br>
U=p[ (1+(1-p)q) / (1-q(1-p)) ] <br>
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.<br>
<br>
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))<br>
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.<br>
<br>
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). <br>
<br>
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.<br>
<br>
<br>
(Currently going for a spring look with a tan suit and slate-blue
shirt that I think complements my cryonics necklace nicely.)<br>
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