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]$)**
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I am looking for a result based on the parameter 't'.


* Now P(t= 1)= 1;
* 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.
* 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](https://mathoverflow.net/q/2014/34859)", 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](https://mathoverflow.net/q/142983/34859) which may be solved by this discrete approach.



Number of 3-tuple partitions of a multiple of three which follow the triangle inequality
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](http://oeis.org/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](http://oeis.org/A002620)). 


