The triangle inequality holds... $50\%$ of the time? Or is this problem asking for something else? Here's a question from my probability textbook:

If three numbers be named at random it is just as likely as not that every two of them will be greater than the third.

I don't understand what this problem is even asking, it's not clear what this means. Could anyone help?
One interpretation is just if I randomly select three numbers $a > b > c$, we want to show that the probability that $b + c > a$ is ${1\over2}$. I guess that's true, but I'm not sure since infinities are weird.
EDIT: Using Mike Earnest's answer of ${1\over2} + {1\over{2n^2}}$ to my previous question here:

Three different persons have each to name an integer not greater than $n$. Find the chance that the integers named will be such that every two are together greater than the third.

We take the limit as $n \to \infty$, and get ${1\over2}$. Does that work?
EDIT 2: This textbook was written in the 19th century, maybe explaining the lack of rigorous formulation on its part.
 A: This problem is badly written. One needs to specify how the three positive real numbers are chosen. Here are three interpretations, all of which I would consider reasonable, and which give very different answers to this question:
Interpretation 1 Choose $a$, $b$ and $c$ uniformly and independently in $[0,1]$. Then the probability that this condition holds is $1/2$.
Proof: We can think of the point $(a,b,c)$ as uniformly chosen in the cube $[0,1]^3$. The subset of $[0,1]^3$ where $a>b+c$ is a triangular pyramid with height $1$ and base of area $1/2$, so its volume is $1/6$; the same is true for $b>a+c$ and $c>a+b$. So the probability that, instead $a+b>c$, $a+c>b$ and $b+c>a$ is $1-1/6-1/6-1/6 = 1/2$.
Presumably, this is the interpretation the textbook intends. But I think that, at least equally reasonable, are the following:
Interpretation 2 Fix $a+b+c = N$. Sample $(a,b,c)$ uniformly at random in the triangle $\{ (a,b,c) : a,b,c \geq 0,\ a+b+c=N \}$. Then the region where $a+b>c$, $a+c>b$ and $b+c>a$ is a smaller triangles whose vertices are the midpoints of the edges of the original triangle, and thus has area $1/4$. So the probability is $1/4$.
You will get the same $1/4$ if you sample $(a,b,c)$ uniformly in the pyramidal region $\{ (a,b,c) : a,b,c \geq 0,\ a+b+c \leq N \}$, or if you sample $a$, $b$, $c$ independently at random for an exponential probability distribution.
Interpretation 3 Fix $a^2+b^2+c^2 = R^2$. Sample $(a,b,c)$ uniformly at random in the spherical triangle $\{ (a,b,c) : a,b,c \geq 0,\ a^2+b^2+c^2=R^2 \}$. This spherical triangle is $1/8$ of a sphere, so it has area $\tfrac{4 \pi}{8}R^2 = \tfrac{\pi}{2}R^2$. The region where $a+b>c$, $a+c>b$ and $b+c>a$ is a smaller spherical triangle all of whose angles are $\cos^{-1} (1/3)$, so it's area is $(3 \cos^{-1}(1/3)-\pi) R^2$ and the probability is $\tfrac{3 \cos^{-1}(1/3)-\pi}{\pi/2} \approx 0.35$.
You get this same $\tfrac{3 \cos^{-1}(1/3)-\pi}{\pi/2}$ if you sample from uniformly the region $\{ (a,b,c) : a,b,c \geq 0,\ a^2+b^2+c^2 \leq R^2 \}$, or if you sample $a$, $b$, $c$ independently at random from one-sided Gaussian distribution.
This could be a very nice textbook example to illustrate how answers to problems in probability depend on how they are formulated, and could lead to good class discussion of the best formulation, but I can't endorse your textbook singling out Interpretation 1 as the right one without discussion.
