Consider this as all the possible permutations of n numbers. Letter to the right envelope then matches the case of some number k is at the k'th place.
n = 1: only sequence is 1. 1 / 1! = 1, so 100% chance.
n = 2: new sequences can be generated by replacing any number in any above sequence with a 2, and moving the replaced number to the end of the sequence. If it was formerly in the right place, this reduces the # of letters to the right place by one; if not, it the number of letters to the right place stays the same. OR, for any sequence above, 2 can be added to the end, increasing the # of letters to the right place by one.
Replace: 1 --> 21. reduces the # of hits to zero. "zero hits, two misses." (0,2)
Add to the end: 1 --> 12. adds the # of hits to two. "zero misses, two hits." (2,0)
n = 3:
(0,2) sequence; replace either with a three for 2x(0,3) or add 3 to the end (1,2).
(2,0) sequence; replace either with a three for 2x(1,2) or add 3 t the end (3,0)
(3,0) x1 four involve hits. 4/3! = 67%
n = 4:
(0,3)x2: replace any for 6x(0,4) or add 4 to the for 2x(1,3)
(1,2)x3: replace the only hit for 3x(0,4), replace either miss for 6x(1,3) or add the end for 3x(2,2)
(3,0): replace any hit for 3x(2,2) or add 4 to the end for (4,0).
(4,0) x1 all but 9 include at least one hit, so 15/4! gives the likelihood.
And so on and and so forth. There is a non-trivial limit for the chance of at least one hit as n grows without limit, and the sequences approaches that value fast. Sadly, so far I haven't figured out a pretty way to get the likelihood for k envelopes in a pretty, closed form.