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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?

N. F. Taussig
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Jack
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  • 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? – Mason Apr 30 '18 at 01:32
  • Yes exactly, and yes we can select the same length twice – Jack Apr 30 '18 at 01:48
  • 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. – Mason Apr 30 '18 at 01:59
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    Duplicate of [Probability that a triangle can be formed from a permutation of three edges of random length](https://math.stackexchange.com/questions/94965/probability-that-a-triangle-can-be-formed-from-a-permutation-of-three-edges-of-r) – dxiv Apr 30 '18 at 02:22

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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.

poetasis
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