Character-Limited Patterns?

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That's an interesting statistic. How many words were in the text used?  I guess you counted all overlapping sequences, so that the total nuber of 4-word sequences is N(word)-3.

The full text was used. The number of four-word sequences is somewhat less than n-3 because it excludes sequences that overlap paragraph boundaries. The total number of non-distinct four-word sequences is 33349. The number of distinct four-word sequences is 33319, so 30 are repeated. Most of these are in f57v, and are essentially sequences of four single-character words.

If we exclude sequences consisting of only single-character words, then there are 33283 four-word sequences, of which 33281 are unique. So there are really only two repeated four-word sequences in the VM, visible here:
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The next question would be what would be a 'normal' number for a known plain text.

A sample of Pliny of slightly smaller size has 110 repeated four-word sequences (none of which consist of single-character words). In order to maximize the number of repeated four-word sequences in Pliny, the best characters to discard are i and x. However, this only raises the number of repeated four-word sequences by 17.

And the next.... since such a text would be edited and spell-checked, what would happen if one intruduces arbitrary errors in this text, e.g. one arbitrary substitution every 80, 40 or 20 characters.....

Given that the number of repeated sequences is not extremely large even in a typical spellchecked text, my guess is that the number of errors would need to be fairly high to reduce it even further. For example, the Pliny sample i used has a total of 35125 words. The 110 repeated four-word sequences, along with the original 110, account for (110*2)*4 = 880 words. That's 2.5% of the text, assuming that each sequence is repeated only once.

In order to reduce the number of 110 repeated four-word sequences by half, we'd need errors to affect at least 55 words. If the errors are random then it's possible to compute the total number of errors needed to achieve this.

The expected number of errors necessary to eliminate any one of the 110 sequences is 1/(880/35125) = 39.9. In order to eliminate another sequence after that we'd need an additional 1/((880-8)/(35125-8)) = 40.27 errors.

For the full 55 sequences we'd need a total of 3002 word errors, which corresponds to a per-word error rate of about 8.5%. That's not unreasonable. However, in order to eliminate all but 2 sequences, we'd need 16298 errors, a 46% per-word error rate.

Let me know if you see an error in my thinking.