Covering ten dots on a table with ten equal-sized coins: explanation of proof

Nice! The above proof proves that any configuration of 10 dots can be covered. What you have here is an example of the probabilistic method, which uses probability but gives a certain (not a probabilistic) conclusion (an example of probabilistic proofs of non-probabilistic theorems). This proof also implicitly uses the linearity of expectation, a fact that seem counter-intuitive in some cases until you get used to it.

To clarify the proof: given any configuration of 10 dots, fix the configuration, and consider placing honeycomb-pattern disks randomly. Now, what is the expected number $X$ of dots covered? Let $X_i$ be 1 if dot $i$ is covered, and $0$ otherwise. We know that $E[X] = E[X_1] + \dots + E[X_{10}]$, and also that $E[X_i] = \Pr(X_i = 1) \approx 0.9069$ as explained above, for all $i$. So $E[X] = 9.069$. (Note that we have obtained this result using linearity of expectation, even though it would be hard to argue about the events of covering the dots being independent.)

Now, since the average over placements of the disks (for the fixed configuration of points!) is 9.069, not all placements can cover ≤9 dots — at least one placement must cover all 10 dots.

The key point is that the 90.69% probability is with respect to "the disks [being] placed randomly on the plane", not the points being placed randomly on the plane. That is, the set of points on the plane is fixed, but the honeycomb arrangement of the disks is placed over it at a random displacement. Since the probability that any such placement covers a given point is 0.9069, a random placement of the honeycomb will cover, on average, 9.069 points (this follows from linearity of expectation; I can expand on this if you like). Now the only way random placements can cover 9.069 points on average is if some of these placements cover 10 points -- if all placements covered 9 points or less, the average number of points covered would be at most 9. Therefore, there exists a placement of the honeycomb arrangement that covers 10 points (though this proof doesn't tell you what it is, or how to find it).

If you read carefully, this proof is for an arbitrary placement of dots. So given any dot arrangement, if we just place the coins randomly (in the honeycomb arrangement,) then on average we will cover slightly more than 9 of the dots. But since we can't cover "part" of a dot (in this problem) then that means that there exists a random placement of the coins that covers all 10 dots. So no matter the configuration of dots, we know that there is always a way to cover the dots with at most 10 coins :)