For an explicit derivation of why this is fair:

Say that you believe the event is likely with probability p, and your betting partner tells you that it will fail with probability q. Then I am going to modify my estimate by c before I tell it to the other person. So my expected value is:

p*(q^2-(1-(p+c))^2) - (1-p)((p+c)^2 - (1-q)^2)

Naturally I want to find the local maximum for variation in c, for a fixed value of p and assuming q is out of my control. So we take the derivative with respect to c. Using Wolfram Alpha shows this is -2c. So the only local maxima possible are telling someone 0, telling them 1, or telling them the true value of p.