What is the probability that a particular set of integer edge lengths selected from an interval $[1,n]$, can form a triangle? That is, let $a,b,c \in \{1,2, \dots n \}$. What is the probability that $a$, $b$ and $c$ are the side lengths to a triangle?

How might this extend to the case where one selects real number edge lengths from the unit interval? Can I look for a cube then exceed the pyramids from it?

  • $\begingroup$ I think your question is: What is the probability that 3 lengths selected randomly from a set satisfy the triangle inequality? Are we allowed to select the same length twice? $\endgroup$ – Mason Apr 30 '18 at 1:32
  • $\begingroup$ Yes exactly, and yes we can select the same length twice $\endgroup$ – Jack Apr 30 '18 at 1:48
  • $\begingroup$ Well... Let's start small... $n=1$. Then $a,b,c \in \{1\}$ which means there's an 100% chance that they satisfy the triangle inequality. What about $n=2$? Note that all lengths will satisfy the triangle inequality except when we select exactly one length equal to $2$. That is, $a=b=1, c=2$ fails the triangle inequality. It fails $3$ out of the $8$ cases. $\endgroup$ – Mason Apr 30 '18 at 1:59
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    $\begingroup$ Duplicate of Probability that a triangle can be formed from a permutation of three edges of random length $\endgroup$ – dxiv Apr 30 '18 at 2:22

Here is an except from my answer to a similar question:

If A+B > C, you have 100% chance of making a triangle. If A+B=C, the triangle is 2d and does not qualify. If A+B < C, you have three joined line segments that do not meet at two ends. You will have to figure the probabilities of each of these equalities/inequalities and go from there.

  • $\begingroup$ If A+B > C, you have 100% chance Not so, try $A=1, B=100, C=10$ for example. $\endgroup$ – dxiv Apr 30 '18 at 2:24
  • $\begingroup$ It should be $A + B > C$ and $|A - B| < C$ $\endgroup$ – Phil H Apr 30 '18 at 4:13

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