A curious inconsistency.
But still... ok, the bullshit is elsewhere.
For a distributed (i.e., multi-node) system to not require partition-tolerance it would have to run on a network which is guaranteed to never drop messages (or even deliver them late) and whose nodes are guaranteed to never die. You and I do not work with these types of systems because they don’t exist.
This is bullshit; it assumes that the only options are complete reliance on absence of failures, and tolerance of arbitrary partitions. Specifically, it claims that P(system fails) = 1 - (1 - P(any particular node fails)) ^ (number of nodes), and that therefore "the question you should be asking yourself is: In the event of failures, which will this system sacrifice? Consistency or availability?". There are plenty of real-life counter-examples to this; given a real-world (ie, at least partially synchronous) network, it is possible to maintain consistency and availability in the face of partition/failure of up to half-minus-one of your nodes. This blows the probability calculations completely out of the water.
You cannot, however, choose both consistency and availability in a distributed system. ... As a thought experiment, imagine a distributed system which keeps track of a single piece of data using three nodes—A, B, and C—and which claims to be both consistent and available in the face of network partitions.
Hey, see how that "in the face of network partitions" snuck in there? It's bullshit, you want "in the face of these specific kinds of network partitions", things like crashed nodes. Just enough to invalidate that abuse of statistics used to claim that Availability is impossible.