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What is the probability that a particular set of integer edge lengths selected from an interval $[1, N]$, can form a triangle? How might this extend to the case where one selects real number edge lengths from the unit interval?

I'd also be very interested in any comments on what the probability distribution would look like for the distances between the triangles vertices.

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I don't understand what permutations have to do with it? –  Raskolnikov Dec 29 '11 at 16:57
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@Raskolnikov They have to do with a mental lapse on my part, since the problem I was originally thinking about involved assemblies with more than three edges. My apologies. –  Steve Dec 29 '11 at 17:03
    
@Raskolnikov Looking about the site, I was unable to find an equivalent version of this question, though there was a somewhat related and interesting question on the probability that a stick broken in two places will form a triangle (math.stackexchange.com/questions/676/…). –  Steve Dec 29 '11 at 17:10
    
You're right, I had that question in mind. But it's definitely different. –  Raskolnikov Dec 29 '11 at 17:37
    
great question I must say! –  Nikhil Bellarykar Dec 29 '11 at 17:53
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2 Answers 2

up vote 6 down vote accepted

To form a triangle you need each of the three sides to be less than the sum of the other two.

So for the unit interval case you might consider a unit cube and exclude three corner pyramids each with volume $1/6$ making the probability $1/2$.

For the discrete case you have $N^3$ points and again have to exclude three pyramids, each with $(N^3-N)/6$ points, which will give a probability of $\frac{1}{2}+\frac{1}{2N^2}$.

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But it is easy to see that the probability for $n=2$ equals 1. Please see my post. –  Sasha Dec 29 '11 at 18:26
    
@Sasha: For $N=2$, I think $(2,1,1)$, $(1,2,1)$ and $(1,1,2)$ are not triangles so the probability is $5/8$. –  Henry Dec 29 '11 at 18:42
    
Yes, as seen in my post, I was using $\geqslant$ instead of $>$ in the triangular inequalities. I have deleted my post. +1 –  Sasha Dec 29 '11 at 20:37
    
@Sasha In other words, you solved for the probability that a set of rigid edges, with lengths chosen in the described manner, can be cyclized in R^3, which I think is interesting. –  Steve Dec 29 '11 at 21:35
    
@Sasha, it seems like the probability using $>=$ instead of $>$ is: (2 + 5*(n - 1) + (n - 1)^2)/(2*n^2), which holds up to {n, 1, 200}. Is there an obvious reason why this should be true? –  Steve Dec 29 '11 at 21:40
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This answers a slightly different question. The original question seeks, instead, to determine $$ \mathbb{P}\left( \left[ \begin{array}{c} X_1+X_2 > X_3 \\ X_1+X_3 > X_2 \\ X_2+X_3 > X_1 \end{array}\right] \right). $$


Let $X_1$, $X_2$, $X_3$ be uniform discrete random variables over interval $[1,n]$. We are seeking to determine: $$ \begin{eqnarray} p = \mathbb{P}\left( \left[ \begin{array}{c} X_1+X_2 \geqslant X_3 \\ X_1+X_3 \geqslant X_2 \\ X_2+X_3 \geqslant X_1 \end{array}\right] \right) &=& \sum_{k_3=1}^n \sum_{k_2=1}^{n} \sum_{k_1=1}^{n} \frac{1}{n^3} \left[ \begin{array}{c} k_1+k_2 \geqslant k_3 \\ k_1+k_3 \geqslant k_2 \\ k_2+k_3 \geqslant k_1 \end{array}\right] \\ &=& \sum_{k_3=1}^n \sum_{k_2=1}^{n} \sum_{k_1=\max(1,|k_3-k_2|)}^{\min(n, k_2+k_3)} \frac{1}{n^3} \\ &=& \sum_{k_3=1}^n \sum_{k_2=1}^{n} \frac{ \min(n,k_2+k_3) - \max(1,|k_3-k_2|) + 1}{n^3} \\ \end{eqnarray} $$ Evaluating the above sum requires careful consideration of three cases, $k_2=k_3$, $k_2< k_3$ and $k_2 > k_3$, each of which needs to be further split to resolve $\min(n,k_2+k_3)$. The end result is bound to be rational function of $n$, so can as well "guess" it: $$ p = \frac{1}{2} + \frac{3n-2}{2 n^2} $$

In[33]:= Table[
  Sum[1/n^3, {k3, 1, n}, {k2, 1, n}, {k1, Max[1, Abs[k3 - k2]], 
    Min[n, k2 + k3]}], {n, 1, 12}] // FindSequenceFunction[#, n] &

Out[33]= (-2 + 3 n + n^2)/(2 n^2)
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Running your Mathematica script for {n, 1, 100} yields: (2 + 5*(n-1)+(n-1)^2)/(2*n^2) –  Steve Dec 29 '11 at 19:47
    
The scaling behavior seems right, but I'm hoping for an exact result! –  Steve Dec 29 '11 at 19:55
    
@Steve But $2+5(n-1)+(n-1)^2$ is identically $n^2 + 3 n -2$, which coincides with the result I stated. –  Sasha Dec 29 '11 at 20:31
    
@Steve By the exact result, do you mean step-by-step evaluation of the sum ? If so, I can add one to the post. –  Sasha Dec 29 '11 at 20:33
    
Are we sure that the FindSequenceFunction result is correct for all values of N? –  Steve Dec 29 '11 at 22:10
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