This article reveals a fundamental misunderstanding of Bayesian statistics by the author, when they say "for the sake of simplicity, let's call it a wash and assume the odds are the same". Because the odds ratio is key in statistics.
You cannot just go "this chance is very small and so is this chance therefore we can assume them to be similar". That's just wrong. The chance that the data we see happens if there are aliens is a lot smaller than the chance of the data given that are none. Yes, both are very small but that does not mean the odds ratio can be assumed to be 1. As the author illustrates, this incorrect reasoning breaks the usefulness of Bayesian statistics.
As for an example let's say that you claim to be using magic to win the lottery, which I don't believe. Now, the lottery happens and the winning number is 4529640, which is not yours. The probability of that number winning is small regardless of these initial hypotheses. If we follow the reasoning in the article we may say that that means because both chances are small this gives us no information on these hypotheses, which is clearly wrong.
I might have a proof that this list is complete (I am very tired though and should be sleeping instead of doing this, so my apologies if I'm wrong):
Because we can only get one extra by carrying, each digit of 2^(k - 1) is at most 4 (otherwise the next digit in 2^k will be odd).
Assume this list is complete up to 10^n. We find the biggest l such that 2^(5^(l - 1)*4) < 10^n. Let us consider the 10^(n+1) > 2^k > 10^n such that 2^k has all even digits.
By cyclicity of powers of 2 mod 10^l (that's why we chose this l), this means that 2^(k - 1) = a*10^l + b, where a is some integer and b is 1,2,4,32 or 1024 (because those are the only options with digits less than 5 mod 10^l). If l > 10,that means that we can divide by b to get 2^(k-1)/b = c*10^d + 1 where c and d are nonzero integers. But this is a contradiction.
Now we only need to show up to 2^(5^10 * 4) to allow l > 10, which has already been done by other comments.
> By cyclicity of powers of 2 mod 10^l (that's why we chose this l), this means that 2^(k - 1) = a*10^l + b, where a is some integer and b is 1,2,4,32 or 1024 (because those are the only options with digits less than 5 mod 10^l).
I'm pretty sure this is the part where the argument breaks down. Just because 2^(k-1) mod 10^l only has small digits doesn't mean that it corresponds to a lesser power of 2 with small digits. E.g., 2^18 ends in 2144, which is not one of 1, 2, 4, 32, or 1024. (And for that matter, 1024 ends in 24.)
The hard part is showing that eventually you must hit a digit greater than 4 if you look at a long-enough suffix.
You cannot just go "this chance is very small and so is this chance therefore we can assume them to be similar". That's just wrong. The chance that the data we see happens if there are aliens is a lot smaller than the chance of the data given that are none. Yes, both are very small but that does not mean the odds ratio can be assumed to be 1. As the author illustrates, this incorrect reasoning breaks the usefulness of Bayesian statistics.
As for an example let's say that you claim to be using magic to win the lottery, which I don't believe. Now, the lottery happens and the winning number is 4529640, which is not yours. The probability of that number winning is small regardless of these initial hypotheses. If we follow the reasoning in the article we may say that that means because both chances are small this gives us no information on these hypotheses, which is clearly wrong.