*The following might look like a simple problem - but the question has been unanswered for more than a week on math.stackexchange.com, and I have asked quite a few of the Ph.d. students at our department - hopefully someone will come up with a shamefully simple solution. The problem originates from analyzing a specific attack in relation to a specific cryptographical protocol.* First, a $k \in {0,\ldots\}$ is fixed. I am analyzing the following experiment: 1. Pick an $x \in \{0,\ldots,2k\}$ uniformly at random 2. Pick $(2k+1)$-bit bitstring $b_1=(u,v_1) \in \{0,1\}^x \times \{0,1\}^{2k+1-x}$ uniformly at random 3. Pick a $(2k+1-x)$-bit bitstring $v_2 \in \{0,1\}^{2k+1-x}$ uniformly at random What is the probability that the majority function of $b_2 = (u,v_2)$ is bigger than the majority function of $b_1 = (u,v_1)$? **Remark:** The reason for picking a bit string of length $2k+1$ is for the majority function to be well-defined. It can be analyzed as follows. Define the random variables: - $X \sim Uniform(\{0,\ldots,2k\})$ - $Y(x) \sim Binom(x,\frac{1}{2})$ - $Z_1(x),Z_2(x) \sim Binom(2k+1-x,\frac{1}{2})$ What is: $\Pr[Y(X) + Z_1(X) \leq k \wedge Y(X) + Z_2(X) \geq k+1]$? The challenge of the problem is easiest shown by fixing a specific $x$, and calculating: $\Pr[Y(x) + Z_1(x) \leq k \wedge Y(x) + Z_2(x) \geq k+1]$ $= \sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y] \Pr[Z(x) \geq k+1-y]$ $= \sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y] (1 - \Pr[Z(x) \leq k-y])$ $= \Pr[Y(x) + Z(x) \leq k] -\sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y]^2$ $= \frac{1}{2} -\sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y]^2$ where we just let $Z_1=Z_2=Z$ after the dependence has been removed. But how to go on from here? If we let * $f(k,x) = \sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y]^2$ * $g(k,x) = \frac{1}{4}(1 + \frac{x}{2k+1})$ Then a plot from maple suggests that $g(k,x) \geq f(k,x)$ for all values that we consider. **[A plot can be seen here. As a new user, I am not allowed to embed it.][1]** **How can I show that $g$ is an upper bound to $f$?** I tried all kinds of things - everything from rewriting to expressions about the variance of some complicated variable, to trying out different induction strategies. I also looked into the theory of moment generating functions. Maybe I was just not creative enough. If successfully proven, it will result in the lower bound $\Pr[Y(x) + Z_1(x) \leq k \wedge Y(x) + Z_2(x) \geq k+1] \geq \frac{1}{2} - \frac{1}{4}(1 + \frac{x}{2k+1})$ Taking the average over all $x \in \{0,\ldots, 2k\}$, we end up with a lower bound on the expectation of $\frac{1}{2k+1} \sum_{x=0}^{2k} (\frac{1}{2} - \frac{1}{4}(1 + \frac{x}{2k+1})) = \frac{1}{4} \frac{k+1}{2k+1} \geq \frac{1}{8}$. [1]: http://sorenhaagerup.dk/files/f_and_g.jpg