Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
@Waldemar I have changed to a strict inequality also rephrased the definiition –  ARi Oct 2 '13 at 17:46
    
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 2 '13 at 20:40
1  
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 2 '13 at 21:51
3  
Indeed, oeis.org/A005044 gives the "number of triangles with integer sides and perimeter $n$." –  Gerry Myerson Oct 3 '13 at 0:20
1  
@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 3 '13 at 8:28

1 Answer 1

up vote 5 down vote accepted

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).

share|improve this answer
    
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$) –  ARi Oct 3 '13 at 3:38
4  
Doesn't Pietro's answer give you everything you need to calculate that? –  Gerry Myerson Oct 3 '13 at 3:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.