The man in the middle

Let Fargo's number be $x$ , Margo's be $y$ , and Kent's be $z$ . Then their numbers can be plotted as a point on an $xyz$ -coordinate system inside a unit cube. Divide the unit cube into $27$ congruent $\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}$ smaller cubes (like a Rubik's cube). Then the following green cubes have a middle coordinate between $\frac{1}{3}$ and $\frac{2}{3}$ , and the following red cubes do not.

Since $13$ out of $27$ cubes are green, the probability is $\frac{13}{27} \approx \boxed{0.48}$ .

If $M$ is the middle number then, if $0 < x < 1$ and $\delta x > 0$ is an infinitesimal, then $P[x < M < x + \delta x] \; = \; 3! \times x \times \delta x \times (1-x - \delta x) \; \approx \; 6x(1-x)\delta x$ (one of the values has to be less than $x$ , one has to be greater than $x+\delta x$ , while the third has to lie between $x$ and $x + \delta x$ . Thus the probability density function of $M$ is $f_M(x) \; = \; \left\{ \begin{array}{lll} 6x(1-x) & \hspace{1cm} & 0 < x < 1 \\ 0 & & \mathrm{o.w.} \end{array}\right.$ and hence $P[\tfrac13 < M < \tfrac23] \; = \; \int_{\frac13}^{\frac23} 6x(1-x)\,dx \; = \; \boxed{\tfrac{13}{27}}$

$\frac{1}{27}+\frac{3}{27}+\frac{3}{27}+\frac{3}{27}+\frac{3}{27}=\frac{13}{27}$