I don't see a list of security team members on that page you linked to, which is what I believe Boris was referring to when he mentioned they have "all the security response team's addresses" for Microsoft

Another fallacy of equivocation. The set I am talking about is the set of digits that is drawn from for this infinite sequence. The set itself is not infinite. Set A has 10 members {0,9}. Set B has 9 members {0,2-9}.

You are arguing based on a different understanding of what the sets are. Even still, your argument does not hold up.

In your case, let set X be the set of all infinitely long number randomly generated. Pick any number from this set, and remove all of the ones. The resulting number exists in a set that is not a subset of set X. Why? Because set X is the set of all numbers that are both infinitely long and random and it is impossible for a number without 1s to exist in set X. Both conditions are important.

If the number is truly randomly generated, each digit (0-9) have an equal probability of appearing. For a finite length of digits (even if said length is large), it is possible (though improbable) that the number wouldn't contain a 1. If we were arguing for finite length numbers, your argument would be fine. However, for infinite length, it is impossible because as length approaches infinity, the probability of a digit not appearing approaches 0. At infinity, it would be 0. This is where your argument breaks down. The number with all 1s removed is not a part of set X so your argument above as a counter example is invalid. This number would be a part of set Y where Y is the set of all infinitely long numbers randomly generated from set B.

These sets are not equivalent, and it is important to keep this in mind when constructing a proof.

Author also forgets to take into account that the number of options available for streaming generally suck. I gave up on Netflix for movies when 85% of what I wanted to watch wasn't available. I'll use it for TV shows, but that's it.

The first set is a subset of the second set.

Yes, but it is not a super set which is the issue. A subset can have additional properties that the super set doesn't have.

-Square is a proper subset of rectangle (all 4 sides are equal)
-Natural numbers are a proper subset of integers (includes negatives of the non-zero natural numbers)
-{0,2-9} is a proper subset of {0,9}

My point still stands. Removing the ones from a random sequence of numbers from the set {0,9} means the resulting sequence of numbers is no longer random by that set. It is random by the set {0,2-9}. Your sequence of numbers cannot be random with respect to the set {0,9} because you know with absolute certainty that a 1 is not in the sequence. There is no equal probability for all the numbers in {0,9} to be chosen.

Note, I'm not disagreeing the possibility that Pi doesn't include every finite pattern within it (though I'm inclined to think it does). I'm disagreeing with your proof for why its not.

Yes, I agree with that. I realized as soon as I stepped away from my computer I may not have been clear. I don't disagree with Hatta's point that it's possible Pi might not contain all possible finite patterns. I merely disagreed with his proof.

My intuition says that Pi does, but that is just my opinion and I have no mathematical proof to back that up.

I also know that there is a 0% chance there will be a 1 in the sequence.

My issue with your analogy is that you allow the the possible subsets (finite patterns within the infinite sequence) to pull from a different set of numbers then the original sequence itself. The original assertion phrased another way would be given a set of 10 digits {0-9}, an infinite and random ordering of those digits will contain every possible finite pattern of 10 digits (0-9).

When you remove the digit 1 from the sequence above, you are left with a random infinite sequence of numbers from the set {0,2-9}, not a random sequence of numbers from the original set. You've done nothing to prove that an infinite random sequence of numbers does not contain every possible finite pattern because said finite pattern must be drawn from the same set of numbers {0,2-9}. I can argue that this new set of numbers contains every possible finite pattern for numbers from the set {0,2-9}

Your argument suffers the fallacy of equivocation. The resulting sequence is random by the set {0,2-9} but not by the set {0,9}.

Yes, that's exactly what I'm doing. This proves that random sequences don't necessarily contain all finite sequences.

Not at all. The sequence is no longer random when you remove all of the 1s from it. It's akin to flipping a coin an infinite number of times, and removing all of the times heads came up leaving you with an infinite sequence of tails. That is not random.

My example proves your intuition wrong. It doesn't prove that pi fails to contain the encoding for every song. But it does prove that irrationality is not sufficient to support that claim.

Your example is flawed.

I don't think your analogy works the way you think it does. When you drop all the 1's from the sequence, you are limiting in scope (for lack of a better term) the subset of possible sequences so that they no longer have 1 in them. This doesn't prove the impossibility of containing every possible pattern when you similarly apply the same condition (ie, every pattern that doesn't contain a 1). Because Pi is irrational, my intuition tells me it would contain the encoding for every song.

Long time lurker, had to create an account to post on this one. Wasn't this the whole premise behind the space shuttle...a reusable craft to ferry people to/from the ISS? And didn't this fail because of the extreme abuse the shuttle suffered upon re-entering the atmosphere? And unless he's planning on mining for fuel on Mars, there is going to be the cost of ferrying the fuel to Mars in the first place, regardless of whether or not you are on that ferry...