Covering dots on a Table [closed] - math

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In the Communications of the ACM, August 2008 "Puzzled" column, Peter Winkler asked the following question:
On the table before us are 10 dots,
and in our pocket are 10 $1 coins.
Prove the coins can be placed on the
table (no two overlapping) in such a
way that all dots are covered. Figure
2 shows a valid placement of the coins
for this particular set of dots; they
are transparent so we can see them.
The three coins at the bottom are not
needed.
In the following issue, he presented his proof:
We had to show that any 10 dots on a
table can be covered by
non-overlapping $1 coins, in a problem
devised by Naoki Inaba and sent to me
by his friend, Hirokazu Iwasawa, both
puzzle mavens in Japan.
The key is to note that packing disks
arranged in a honeycomb pattern cover
more than 90% of the plane. But how do
we know they do? A disk of radius one
fits inside a regular hexagon made up
of six equilateral triangles of
altitude one. Since each such triangle
has area sqrt(3)/3, the hexagon
itself has area 2*sqrt(3); since the
hexagons tile the plane in a honeycomb
pattern, the disks, each with area π,
cover π /(2*sqrt(3)) ~ .9069 of the
plane's surface.
It follows that if the disks are
placed randomly on the plane, the
probability that any particular point
is covered is .9069. Therefore, if we
randomly place lots of $1 coins
(borrowed) on the table in a hexagonal
pattern, on average, 9.069 of our 10
points will be covered, meaning at
least some of the time all 10 will be
covered. (We need at most only 10
coins so give back the rest.)
What does it mean that the disks cover
90.69% of the infinite plane? The easiest way to answer is to say,
perhaps, that the percentage of any
large square covered by the disks
approaches this value as the square
expands. What is "random" about the
placement of the disks? One way to
think it through is to fix any packing
and any disk within it, then pick a
point uniformly at random from the
honeycomb hexagon containing the disk
and move the disk so its center is at
the chosen point.
I don't understand. Doesn't the probabilistic nature of this proof simply mean that in the majority of configurations, all 10 dots can be covered. Can't we still come up with a configuration involving 10 (or less) dots where one of the dots can't be covered?

Answer 1

I think that I can re-arrange Winkler's argument to make it a little more convincing.
You're given an arrangement of dots on a table. You also have a big template made of coins glued together in a honeycomb pattern. You now do a Monte Carlo simulation, repeatedly throwing the honeycomb on the table at a random location (but always with the same orientation), and counting the number of covered dots. If you get enough samples you will eventually find that the expected average number of dots covered is 9.069 per throw.
The key insight is that if the average is 9.069, there must have been a throw where 10 dots were covered. Because if you never covered 10 dots, the average would be 9 or less.
So now you know that there was at least one throw that covered 10 dots. You duplicate that throw, and remove all the coins that aren't covering a dot. There will be 10 or fewer coins remaining.
A small digression: Is it possible that for some clever arrangement of dots the long range average of covered dots is something other than 9.069? The answer is no, because each of the dots can be considered separately. In other words, in 10000 throws of the honeycomb, the expected number each dot will be covered is 9069 times. So we expect a total of 90690 dots to be covered, for an average of 9.069 per throw.