Your conclusion that the EV is -100% is not right. What you did was separate out the calc of the frequency with which you are a winner, which is 0% and concluded that the EV was -100%. With more care, you would have realized that this approach actually results in an indeterminate form of 0*infinity e.g.
EV = lim ((3/4)*(2))^n - 1, as n goes to infinity
= (lim (3/4)^n)*(lim (2)^n) - 1, "
= 0 * infinity - 1, "
= 0 *** Not Kosher ***
Of course this is not correct. This approach does not work. Better is
EV -> ((3/4)*(2))^n - 1, as n -> infinity
-> lim 1.5^n - 1, as n -> infinity
-> infinity, as n -> infinity
So the conclusion is that the EV goes to infinity as n goes to infinity. The frequency calc portion you made is okay. This means that you wind up a winner overall, but you are a winner a vanishingly small percent of the time. A curious result, but correct.
Analogously, you can also devise a scenario where it is -EV, but you are nevertheless a winner virtually 100% of the time.
Reference:
http://en.wikipedia.org/wiki/Indeterminate_forms