Not an answer, but I think you overestimate the space of possibilities. If you restrict to the set of stock abbreviations you believe you have, then the range to consider consists of all cases where one of the three letters is altered. So 75 possible such alterations per stock owned. This assumes I understand correctly the problem you need to solve.

You can try a simulation. Here are 45 given combinations:

givenSample = RandomChoice[Alphabet[], {45, 3}]

Here I assume that you are interested in the matching of 2 consecutive letters:

subSequencesOf2LettersConsecutive = 
 Union[Map[Rest, givenSample], Map[Most, givenSample]]
shares2lettersQ[{a_, b_, c_}] := 
 MemberQ[subSequencesOf2LettersConsecutive, {a, b} | {b, c}]

Here I run the simulation:

sampleSize = 10000;
randomTriplets = RandomChoice[Alphabet[], {sampleSize, 3}];
results = Map[shares2lettersQ, randomTriplets];
Count[results, True]/sampleSize // N

I get a probability of 23%. You may have to adapt the code to your own assumptions.

Be careful that my simulation is made with some assumptions that may or may not acceptable in your situation. For example, I assumed that abc matches abx and xbc and xab, but not axc, and not bac. Otherwise the code must be modified. Also, you must provide your own 45 combinations and start the simulation with them.