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A stick is broken into two at random, then the longer half is broken again into two pieces at random. What is the probability that the three pieces make a triangle?

I've been stumped at this question for a while now, and cannot seem to find a convincing answer on the internet. The answers that I find are usually 0. 386 (which is 2ln2 - 1), but I have worked the answer out to be 0.27865.

I let X be the event of the position of the first cut of a length of unit 1. Then X ~ Unif(0,1).

Then I let Y be the event of the position of the second cut. Finding the PDF of Y is slightly more tricky.

Y is Unif (a/(1-x)) for 0 < x < 0.5 and Unif (a/x) for 0.5 < x < 1

where the normalising constant a, I worked out to be 0.5/ln2.

Now the answers I have seen did not work out this normalising constant, and you will get 0.386 as the probability to create a triangle, but with the normalising constant, I get an answer of 0.27865...

Now, as with these probability questions, you can always make an experiment, and estimate a probability. I was too lazy to do this and found a simulation here:

http://www.sineofthetimes.org/the-broken-stick-puzzle/

I found the simulation to support the 0.38 answer... but then again, I don't know how they created the simulation.

What have I done wrong here? (Or am I right?) Thanks so much!

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  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – José Carlos Santos Aug 9 '18 at 19:17
  • $\begingroup$ What does "Unif (a/(1-x)) for 0" mean? $\endgroup$ – saulspatz Aug 9 '18 at 19:19
  • $\begingroup$ Some formatting problem with the "less than" sign, sorry. $\endgroup$ – Snow-Covered Island Aug 9 '18 at 19:21
  • $\begingroup$ I think without loss of generality you can consider $X$~$U(\frac12, 1)$ and $Y$~$U(0,X)$. Then the question is whether $\max(Y,X-Y,1-X)\leq \frac12$. $\endgroup$ – Arnaud Mortier Aug 9 '18 at 19:22
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    $\begingroup$ I think you should spell out what you mean by the "normalising constant" $a$. I don't see how and why it should arise. Also, could you link to some of the places where you found the $2\ln2-1$ result, and point out what parts of their derivation you don't understand? Then we could avoid replicating the entire derivation here. $\endgroup$ – joriki Aug 9 '18 at 19:40
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I think without loss of generality you can consider $X$~$U(\frac12, 1)$ and $Y$~$U(0,X)$.

If you set $M=\max(Y,X-Y,1-X)$ (the longer side) then the three segments do not form a triangle if and only if $M$ is larger than the sum of the other two sides, that is $M> 1-M$, in other words whether $M> \frac12$.

Note that $P(1-X)>\frac 12=0$, so $$\{M>\frac 12\}=\{\max(Y,X-Y)>\frac 12\}$$

Now \begin{align*}P(M>\frac12\mid X=x)&=P(Y<x-\frac12)+P(\frac12<Y\leq x)\\&=2\frac{x-\frac12}{x}\\&=2-\frac1x\end{align*}

So that \begin{align*}P(M>\frac12)&=\int_{\frac12}^1P(M>\frac12\mid X=x)\cdot 2dx\\&=\int_{\frac12}^1 (4-\frac2x )dx\\&=\left[4x-2\ln x\right]_{\frac12}^1\\&=(4-2)+2\ln(\frac12)\\&=2(1-\ln 2)\end{align*} This is the probability that the three segments do not form a triangle.

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You do not need a normalising constant

  • If $X=x < \frac12$ you have $Y$ conditionally distributed uniformly on $[x,1]$ with density $\frac{1}{1-x}$ and you have a triangle if $Y \in \left(\frac 12, x+\frac12\right)$ which has conditional probability $\frac{x}{1-x}$

  • If $X=x > \frac12$ you have $Y$ conditionally distributed uniformly on $[0,x]$ with density $\frac{1}{x}$ and you have a triangle if $Y \in \left(x-\frac12, \frac12\right)$ which has conditional probability $\frac{1-x}{x}$

  • So integrating over $x$ uniform on $[0,1]$, you get the overall probability of a triangle being $$\int_0^{1/2} \frac{x}{1-x} \, dx+ \int_{1/2}^1 \frac{1-x}{x} \, dx = 2\log_e(2)-1 \approx 0.386$$

The following simulation in R gives much the same result

set.seed(1)
cases <- 10^6
X <- runif(cases, min=0, max=1)
Y <- ifelse(X > 1/2, runif(cases, min=0, max=X), runif(cases, min=X, max=1))
triangle <- ((X > 1/2 & Y < 1/2) | (X < 1/2 & Y > 1/2)) & abs(X-Y) < 1/2
mean(triangle)
[1] 0.386273
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