The smaller your bet, the less often they have to fold for it to be worth it. For example, if you bet half the pot and you have NO chance of winning if called, they only have to fold 1/3 of the time for it to be a +EV play. If you have a 50% chance to win, then it is a +EV play unless they NEVER fold.
Example (some math involved)
More realistically, say you have a 20% chance to win and you bet 400$ in to a 400$ pot. Then when they call you, you have an EV of -160$ (1 of 5 times you win 800 and 4 of 5 you lose 400). When you fold, you have an EV of 400$ (you always win 400$).
Let x be the probability they call, and y be the probability they fold. Therefore, y*400 + x*-160 is your EV. Seeing where this equation is 0 tells
you when you break even, and thus the largest amount of times they can fold before the play becomes -EV. So we can set the equation equal to 0 and do some basic algebraic manipulation:
y*400 + x*-160 = 0
y*400 = 160*x
x = 2.5*y
So when they call 2.5 times for every 1 fold you break even. Any more and you are making money. In terms of percentages, they must be folding at least 28.57% of the time for it to be +EV. [Note: 28.57% = 100%/(2.5+1)]
Obviously, in a real poker situation you don't know your odds
of winning, but you can make educated guesses (this can only really come from experience). This is also an oversimplification because it neglects any betting on future streets or a possible raise - this only accounts for them deciding to call or fold on one street.