Given n=3t, t$\in \mathbb N$; let $\mathbb L_3$ be set of all distinct integer partitions of n having 3 parts; say $\lambda_1,\lambda_2,\lambda_3$ .
If I chose any one partition randomly from $\mathbb L_3$ what is the probability of the parts following the triangle inequality.
Given any $\lambda\in\mathbb L_3$, what is P($\lambda|\lambda_i<\lambda_j+\lambda_k \;;[0<i,j,k\leq 3 ] $ $ \wedge [ i\neq j \neq k]$)
I am looking for a result based on the parameter 't'.
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Now P(t= 1)= 1;
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P(t=2)= 1/3 as there are only three distinct 3-tuple partitions of n= 3.2 ie 6 viz (2,2,2) and (1,2,3) (4,1,1) and triangle inequality holds for only one case.
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for t= 3 , n= 9; number of 3-tuple partitions of 9 are 6 with 2 ie (4,3,2),(4,4,1) following the triangle inequality so P(3) = 1/3 and so on.
Obviously a closed form expression for P(t) will not be there but we can look for bounds.
Motivation
I am looking for a solution to type of geometrical problems such as "Find the probability of the parts of a stick forming a triangle when broken in three parts", using a discreet approach for which I give my treatment above.The bounds for large t will be valid for a general case of uniformly breaking a stick in two parts; and if possible asymptote of P(t) or its bound as t tends to infinity can be considered.
Related problems would be "Probability that a stick randomly broken at five places can form a tetrahedron which may be solved by this discrete approach.
Best Answer
So your finite probability space is the set of all $(a,b,c)$ satisfying $1\le a \le b\le c$ and $a + b+c =n$ (counted by OEIS A069905); no need that $n$ be a multiple of $3$. Lucky triples are those also satisfying $a < b < c$ and $a+b > c$. The complement is easier, as it reduces to: $a < b$ and $ a +b \le c= n -(a+b)$, that is $ a +b \le \lfloor n/2\rfloor $. Since $c= n -(a+b)$ the number of the wrong triples is the same as the number of integer pairs $(a,b)$ with $1\le a < b$ and $ a +b \le m $ with $m:=\lfloor n/2\rfloor$ (OEIS A002620).