I'm not that good wit the vagaries of the market, but I do loves me some math.
Assuming all outcomes are equally likely, if you have a set with 6 enhancements in it and you want to draw 1 specific one, it'll take on average 5 conversions at 3 converters per try.
Every time, you need 1 converter, 80% of the time, you need a second, 64% of the time you need a third and so on, so the average number of conversions is
1 + 1*.8 + 1*.64 + 1*.512 + ...
or
1 + .8 + .8^2 + .8^3 + ...
The answer to how many conversions you can, on average, expect to need becomes a geometric sum, in which case the answer is:
1/(1-f) where f is your chance of failure, in this case 80%.
1/(1-.8) = 5
This generalizes out to:
If I'm trying to draw 1 IO from a set with N pieces, my average cost will be N-1 conversions.
If I'm OK with any M out of the N pieces available, that average cost drops to (N-1/M) conversions.
That said, there is no top end to how many tries an individual conversion may take, so for smaller sample sizes you'll need to be prepared for really lousy RNG. About 5% of the time, it'll take more than 13 conversions (or 39 converters) to hit a single IO target in a 6 piece set and we all know how often we manage to land that kind of critical fail.
Remember that all events are INDEPENDENT. Just because you've failed 4 times in a row doesn't make try number 5 any more likely to succeed. Your chance to hit a winner with that next conversion is the exact same as it was when you started, no matter how good or bad your prior luck has been.