Vtyjkyuk

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!

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 beginalign*P(M>frac12mid X=x)&=P(Y<x-frac12)+P(frac12<Yleq x)\&=2fracx-frac12x\&=2-frac1xendalign*

So that beginalign*P(M>frac12)&=int_frac12^1P(M>frac12mid X=x)cdot 2dx\&=int_frac12^1 (4-frac2x )dx\&=left[4x-2ln xright]_frac12^1\&=(4-2)+2ln(frac12)\&=2(1-ln 2)endalign*
This is the probability that the three segments do not form a triangle.

You do not need a normalising constant

• If $X=x < frac12$ you have $Y$ conditionally distributed uniformly on $[x,1]$ with density $frac11-x$ and you have a triangle if $Y in left(frac 12, x+frac12right)$ which has conditional probability $fracx1-x$




• If $X=x > frac12$ you have $Y$ conditionally distributed uniformly on $[0,x]$ with density $frac1x$ and you have a triangle if $Y in left(x-frac12, frac12right)$ which has conditional probability $frac1-xx$




• So integrating over $x$ uniform on $[0,1]$, you get the overall probability of a triangle being $$int_0^1/2 fracx1-x , dx+ int_1/2^1 frac1-xx , dx = 2log_e(2)-1 approx 0.386$$

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