To understand this, one must think about the process of digestion itself. Trypsin cleaves proteins through hydrolysis. The newly-created C-terminus carboxyl acquires one of its oxygen atoms from the water. So, at this point, half of the C-termini will be singly labeled, and half will be unlabeled. (I'm ignoring the low probability of O18 atoms already being present at the site of cleavage).
However, in most cases, trypsin will continue to interact with the C-termini, swapping out C-terminal oxygens to create new water molecules while at the same time swapping in oxygens from existing water molecules to replace them. After this process has proceeded for some time, the chance of a given C-terminal oxygen atom being O16 approaches the level of its presence in the water: i.e. 50%. The same is true for its chance of being an O18 atom.
So, a simple binomial distribution applies. Since the C-terminal carboxyl has two oxygen atoms:
Chance of both being unlabeled = 0.5 times 0.5 = 0.25.
Chance of both being labeled = 0.5 times 0.5 = 0.25
Chance of one being labeled and the other being unlabeled = the remainder of the probability = 1.0 - 0.25 - 0.25 = 0.50.
So, one quarter of the peptides are unlabeled, one half singly-labeled, and one quarter doubly-labeled. A 1:2:1 ratio should be evident.
That being said, different peptides show different rates of back exchange, with a few being totally resistant to it. So while the majority of peptides should display a 1:2:1 ratio pattern, some exceptions might be noted."
From internal communications.