I don't think it's a bad question really but I don't know how good an answer I can arrive at..
You flip a coin between one an 10.000 times (let's say 9999 times just because). If you flip until you've flipped 9999 times the probability that you're ahead is P(profit | 9999 flips)=50% and the probability that you're down is also 50% so if we flip it out it's not +EV as expected.
Furthermore if we repeat the 9999 flip set infinate times and plot a graph with the net gain/loss on the x axis and the percentage we got that net gain/loss on the y axis we would be plotting a standard normal distrubution curve.
(only with different x and y scales as well as the top of that bell shaped thing intersecting the y axis)
If we cut the flippage as soon as we get ahead the first half of the curve will remain close to the same except we move the vertex quite a bit down the y axis but we basically let the y axis cut the shape in half and replace the right hand side of it with a single point on the graph located at P(X=0.5, Y=a big %) since we're gonna profit most of the times.
Now this is a part I can't prove but I suspect that the integral of the positive x values will be equal in magnitude to the integral of the negative x values meaning that even this will be breakeven. So to answer the original question I'd have to say no, it's not +EV but it's gonna be really low varience until the system valuerapes you (much like the bet one unit, try to double up, if fail bet two units, if fail bet 4 units until win)
Edit: Feel free to correct me, I've been wrong before