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This problem which I've been working on is a variant of the Broken Stick problem and it is as follows:

-A spaghetti stick of length one is dropped on the floor and has $n$ breaks. What is the probability that one of the breaks is longer in length than $\frac{1}{2}$ and what is the general formula for determining the probability that at least one break is longer than length $x$.

Anything helps, thanks.

Vivek Kaushik
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  • For a related question see https://math.stackexchange.com/questions/262388/broken-stick-probability-problem – mlc May 10 '17 at 05:37
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    @mlc This is not a related question. The another algorithm of breaks is provided in the link, and another joint distribution of breaks arises. – NCh May 10 '17 at 16:16
  • @NCh: not to press my point, but "related" is not "duplicate". Sometimes one can gets ideas from related problems. – mlc May 10 '17 at 16:49

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The experiment to break a stick of unit length into $n$ pieces at random can be considered as follows.

Let us consider a circle of unit length and choose $n$ points on the circle at random and independently one from another. We obtained $n$ arcs between neighboring points. By symmetry, all arc lengths are equally distributed. The lengths are mesured in some predefined direction - say. clockwise. After that let us break the circle in the last chosen point. We get a stick of unit length and rest $n-1$ points that break this stick into $n$ equally distributed parts.

The probability that the largest part is greater than $1/2$ is the probability that all poins in the circle lie in the same semicircle. This probability can be found by splitting this event into $n$ disjoint events.

Let $A_i$ be the event that all points lie clockwise from the $i$th point, and all arc lengths are less then $1/2$.

The events $A_1,A_2,\ldots, A_n$ are disjoint and have equal probabilities $$ \mathbb P(A_i) = \frac{1}{2^{n-1}}. $$ The probability that all poins in the circle lie in the same semicircle (event $A$) is then $$ \mathbb P(A)=\mathbb P(A_1\cup \ldots\cup A_n)=\frac{n}{2^{n-1}}. $$ This is the probability that one of the breaks is longer in length than $\frac12$.

The general formula for determining the probability that at least one break is longer than length $x$ can be found in W.Feller book "An Introduction to Probability Theory and Its Applications". Vol.2, Chapter I, paragraph 9, Theorem 3. It provides the probability of opposite event "all breaks are less than smth".

Theorem 3. Let the interval $\overline {0,t}$ be partitioned into $n$ subintervals by choosing independently at random $n-1$ points $X_1,\ldots,X_{n-1}$ of division. The probability $\varphi_n(t)$ that none of this subintervals is of length exceeding $a$ equals $$ \varphi_n(t)= \sum_{v=0}^n (-1)^v {n\choose v}\left(1-v\frac{a}{t}\right)^{n-1}_{+}. $$

Here $$(x)_+^n=\begin{cases}x^n, & x>0 \\ 0, & x \leq 0\end{cases}.$$

The theorem is proved in this book by reсurrence relations.

If $t=1$, the probability that at least one break is longer than length $x$ is equal to $$ 1-\sum_{v=0}^n (-1)^v {n\choose v}\left(1-v\frac{x}{t}\right)^{n-1}_{+}=\sum_{v=1}^n (-1)^{v+1} {n\choose v}\left(1-v\frac{a}{t}\right)^{n-1}_{+}. $$

NCh
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  • Thanks man, that helps so much. –  May 11 '17 at 00:29
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    Note that the result for $a=\frac12$ generalizes to $a\ge\frac12$. The same argument as for $a=\frac12$ goes through, and in Feller's Theorem $3$, only the terms for $\nu=0,1$ contribute, so the probability for one of the intervals to exceed $a$ is $na^{n-1}$. (For $a\lt\frac12$, the argument for $a=\frac12$ doesn't go through because the events are no longer disjoint.) – joriki Jul 07 '18 at 08:32