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.
Best Answer
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.