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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.

Emperor Concerto
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  • Here's one way to make this precise: for any $a>0$, the area of ${(b,c)\in [0,a]\times [0,a]: b+c>a}$ is exactly ${a^2\over 2}$ (= exactly half the area of the whole square $[0,a]\times[0,a]$). – Noah Schweber Sep 10 '21 at 00:22
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    The problem is, you can't randomly select a number, at least without specifying the distrubution. There are no uniform distribution (which is usually implied) available. – Trebor Sep 10 '21 at 00:23
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    “Theee numbers chosen at random” is not well-spwcified – Thomas Andrews Sep 10 '21 at 00:31
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    Roughly speaking, the triple $(a,b,c)$ is "good" if and only if the triple $(a,a-c,a-b)$ is "bad". – Alexander Burstein Sep 10 '21 at 01:15
  • For the positive real number case, if you choose A,B,C such that A=B=C, then you have an equilateral triangle. If A<(B=C), or B<(A=C), or C<(A=C), you have an isosceles triangle; this inequality holds 100% of the time for "random distributions" producing such results. Find a random distribution for choices A, B, C that creates a triangle exactly half the time. Next, if the integer must be positive, and the three people select integers with equal likelihood, n=2 gives 50%. – Gwendolyn Anderson Sep 10 '21 at 04:35
  • I think the limit works (yes, it's a late response!!) , I would assume that in these books, they usually allow stuff to happen over a finite measure space , on which one can discuss the uniform probability measure, and then take the limit as $n \to \infty$. I'm a little embarrassed to submit this as an answer, I believe it's incomplete since I haven't substantiated it. Can you let me know the name of your probability textbook, @EmperorConcerto? – Sarvesh Ravichandran Iyer Sep 16 '21 at 19:00

1 Answers1

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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.

David E Speyer
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