Not for a useful meaning of "potential customer". If I publish a Mario clone for iPhone my potential customers are iPhone owners who would consider playing a Mario clone or who fail to understand my description of the game and think they're buying something else. In talking about potential customers you have to think about the market of people who want your product.
So how do you get an accurate count of the people who want your product? You can't. Also, what do you mean by "want"? Are willing to try it for free? Are willing to pay for it?
It's rather obvious that the number of people who are willing to try something for free is far bigger than the number of people willing to pay money for it. Yet the industry's estimate compares the number of non-pirates who were willing to pay the asking price to the number of pirates who were willing to try it for free. It's comparing apples to oranges. It's a completely meaningless statistic.
But that's the entire point! The number of pirated copies is completely irrelevant when estimating the number of lost sales.
No, it isn't. If 80.000 people pirate my Mario clone then, ignoring publicity effects, the number of lost sales is somewhere between 0 and 80.000. So at least it gives an upper bound, which is more than knowing the number of people with an iPhone does.
But a hard upper bound is not the same as a meaningful estimate. The number of iPhone owners also gives an upper bound to the number of lost sales.
But more than that: the industry's logic that "anywhere between 0 and 80,000 == 80,000" is completely wrong, yet that is what they do: they don't claim that this is a (very loose) upper bound, they claim it's the number of lost sales. And that's a lie.
There are 6 relevant figures:
A. The number of people with jailed phones who bought the product
B. The number with jailed phones who didn't buy
C. The number with jail-broken phones who bought
D. The number with jail-broken phones who pirated it but would buy it if they couldn't pirate (~= lost sales)
E. The number who pirated it but would never buy it
F. The number with jail-broken phones who neither bought nor pirated it.
We are given two figures: (A + B) / (C + D + E + F) ~= 9 ; (A + C) / (D + E) ~= 0.25. That is simply insufficient information to even estimate D.
True, but which of the following two claims would you say is most likely:
A/B =~ (C+D)/(E+F)
E =~ 0
The industry is claiming the second one is true. TFA says the first is a more reasonable assumption. I'm inclined to agree with TFA.