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.
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我认为我可以重新整理温克勒的论点，使其更有说服力。
桌子上有一些点的排列。您还有一个由硬币制成的大模板，以蜂窝状图案粘在一起。现在，您可以进行蒙特卡洛模拟，将蜂窝重复地扔到桌子上的随机位置（但始终具有相同的方向），并计算被覆盖的点的数量。如果您获得足够的样本，您最终会发现每次投掷覆盖的预期平均点数为 9.069 个。
关键的见解是，如果平均值为 9.069，则一定有 10 个点被覆盖的投掷。因为如果您从未覆盖 10 个点，则平均值将为 9 或更少。
现在您知道至少有一次投掷覆盖了 10 个点。您重复该投掷，并移除所有未覆盖点的硬币。剩余硬币数量将少于 10 个。
一个小题外话：对于一些巧妙的点排列，覆盖点的长期平均值是否有可能不是 9.069？答案是否定的，因为每个点都可以单独考虑。换句话说，在蜂巢抛出 10000 次时，每个点预计被覆盖的次数是 9069 次。因此，我们预计总共覆盖 90690 个点，平均每次抛出 9.069 个点。
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.