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'.

  • 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", 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.

  • $\begingroup$ I’d like to point out one of the problems when trying to relate discretized results to the original (real-valued) problem. In the discretized version it does matter if you use strict or non-strict inequality while of course in the real-valued version it does not. It seems that with a strict inequality you can try to approximate the lower bound of the probability while with a non-strict one – the upper bound. $\endgroup$ – Waldemar Oct 2 '13 at 20:40
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    $\begingroup$ I'm not sure why it's "obvious" that there isn't a closed form expression. I expect the opposite, I would guess that this is not a hard exercise, and that if you compute the first few values you can find a related sequence in the OEIS. But I don't see why this problem is interesting, unlike the continuous problem. $\endgroup$ – Douglas Zare Oct 2 '13 at 21:51
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    $\begingroup$ Indeed, oeis.org/A005044 gives the "number of triangles with integer sides and perimeter $n$." $\endgroup$ – Gerry Myerson Oct 3 '13 at 0:20
  • $\begingroup$ (I meant to put the easy computation in the comments, as people do in similar cases, but the comment was a bit too long, so I put the long comment in the answer, as people also use to do) $\endgroup$ – Pietro Majer Oct 3 '13 at 0:30
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    $\begingroup$ @Waldemar: It looks like you are going backwards. Nothing ARi or you have said indicates otherwise. I see nothing that is simpler about the discrete versions than the continuous ones. I see no extra tools you bring to bear on the problem, only difficulties and ARi's bad guess that there is no closed form. Can you do the simplification that I suggested in that thread, of marked edges, ignoring the nonlinear constraint? The continuous version is a simple computation with qhull. The discrete version looks like a mess. I don't understand why your intuition says otherwise, but good luck. $\endgroup$ – Douglas Zare Oct 3 '13 at 8:28

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).

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    $\begingroup$ Doesn't Pietro's answer give you everything you need to calculate that? $\endgroup$ – Gerry Myerson Oct 3 '13 at 3:47

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