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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.

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ARi
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  • 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. – Waldemar Oct 02 '13 at 20:40
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    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. – Douglas Zare Oct 02 '13 at 21:51
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    Indeed, http://oeis.org/A005044 gives the "number of triangles with integer sides and perimeter $n$." – Gerry Myerson Oct 03 '13 at 00:20
  • (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) – Pietro Majer Oct 03 '13 at 00:30
  • @DouglasZare We can approach a continuous problem through this discrete approach. Finding a parametric expression and taking the value as it tends to infinity. – ARi Oct 03 '13 at 03:43
  • @ARi: The continuous problem is easy and well-known. In the discrete version you have to choose whether you want the inequality to be strict or not, you have to decide whether to include perimeters not a multiple of $3$, and simple approaches run into tedious cases by parity. Then you incorrectly conjectured that there should be no closed form. Why would this look more attractive to you than the continuous version? – Douglas Zare Oct 03 '13 at 07:36
  • @DouglasZare If I can add something… The real issue here is how to attack the tetrahedron problem. First, I suggested (http://mathoverflow.net/questions/142983/probability-that-a-stick-randomly-broken-in-five-places-can-form-a-tetrahedron#comment369669_142983) that by discretization maybe we can get some nonprobabilistic (strict) bounds to the continuous problem. Then ARi presented his own idea of discretizatization. It seemed reasonable to apply such an idea first to the triangle problem and not directly to the tetrahedron one. – Waldemar Oct 03 '13 at 08:01
  • @DouglasZare (continued) Knowing the answer to ARi’s post I’m not sure how to translate the results into the bounds of the original continuous problem. I applied a different version of a discretization to the triangle problem and got promising results. However, it seems that this approach do not necessarily extends easily to the tetrahedron case. – Waldemar Oct 03 '13 at 08:02
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    @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. – Douglas Zare Oct 03 '13 at 08:28
  • (@Waldemar: compare e.g. the difficulty of evaluating a discrete summation $\sum_{j=0}^n j^r$ wrto the easy computation of the integral $\int_0^t x^r dx$). – Pietro Majer Nov 24 '13 at 08:11

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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).

Pietro Majer
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