I suppose this is what they call a blog. Except that blogs
are supposed to be updated more often than this is.
From Mark Abley's book
Spoken Here
about endangered languages:
The Inuktitut language goes further, much further.
Inuit distinguish
utsimavaa (he or she knows from experience),
sanatuuq (he or she knows how to do something),
qaujimavaa (he or she knows about something),
nalujunnaipaa (he or she is not ignorant of something),
nalunaiqpaa (he or she is no longer unaware of something),
and two or three other verbs that mean, roughly speaking, "know".
Or, as Mark Abley inexplicably failed to put it,
the Inuit have eight different
words for know.
On a recent trip abroad, I had to choose whether to pack everything
into one bag or take two. One factor: how much longer will I wait in
baggage reclaim if I take two bags? Making the assumption (implausible
for a couple of reasons, but never mind) that bags arrive independently
at random times during some fixed interval, it turns out that if you
have n bags then the expected (i.e., average) arrival time of
bag k is k/(n+1) of the way through that
interval. In particular, on average your last bag arrives n/(n+1)
of the way through. So if you take two bags instead of one then you wait,
on average, for 2/3 of the worst-case time instead of 1/2;
you wait 4/3 times as long.
This isn't difficult to prove, but the simplest proof I can find
still involves doing some integrals. Isn't there a one-line argument
that makes it obvious?
Update, 2007-12-09: A friend emailed me with
a nice intuitive proof for the case n=2. I haven't been
able to make it work rigorously for arbitrary n, but reflecting
on why not leads to this, which isn't a one-liner but at least
involves few ideas and no integrals:
Suppose your bags (numbered in advance, rather than in order
of eventual arrival) arrive at times
t1...tn.
They're independent and uniformly distributed.
Now, suppose you learn that t1
lies in a certain interval; then
the distribution of t1 conditional
on this new discovery is uniform in that interval,
and the distribution of all the other tj
is what it was before.
Learning that t1=t
for some particular t *and that t1
is the smallest of all the tj*
is just a matter of learning that t1
is in [0,t] and that all the others are in [t,1];
therefore, all the other tj
are now uniformly distributed in [t1,1],
and this is true whatever the value of t was so it's
always true.
At this point we've reduced the situation for n
to the situation for n-1 and a simple application of
induction finishes the job.