The Great Rod Puzzle...

[DavidHooper]

I'm gonna start a new thread on The Great Rod Puzzle. I'll post a question then wait for the answer, and then post an extension question &c...

So first up, really easy to start with:

1) Take a rod of length 2a. Break it into two pieces. What is the average (mean) length of each piece?

[Vlatko]

a

[Slaine]

Lets see now:

suppose the rod is broken into two lengths of xa and ya, then

xa+ya=2a

therefore:

average length = (xa+ya)/2

replacing xa+ya :

average length = (2a)/2
average length = a

QED.

[DavidHooper]

Well done.

Second question:

2)Take a rod of length 2a. Break it into two pieces. What is the average (mean) length of the shortest piece?

[HarryW]

a/2 ?

[DavidHooper]

Well done

3) Take 2 rods each of length 2a. Break both into two pieces. What is the average (mean) length of the shortest piece?

[HarryW]

a/4?

Not sure about that one, I'm half-guessing.

[Behemoth]

2a/4 [2 rods into 4 pieces]

=

a/2

[HarryW]

But if the mean length of the shortest piece of one rod is a/2, then that can't be right.

2 rods of length 2a and 2b (where a=b but different letters for clarity)

mean shortest of a is a/2
mean shortest of b is b/2

mean shortest of a/2 and b/2 will always be less than a but usually less than a/2 as well.

I don't know the maths exactly but I'm still guessing a/4.

[Behemoth]

damnit, I dodn't read the question properly!
I don't think I understand anything now...

[kedaman]

a/6

[DavidHooper]

Some good (intuitive!) guesses but no-one's come up with the right answer yet.

Keep trying and I'll give a clue tomorrow...

[kedaman]

a/3? i was assuming the rod was a, not 2a, so

[DavidHooper]

Well done keda. a/3 is the answer. Tell us how you did it?

I may write a easy (easier!) to understand explanation later but for now.

1)Break the rods. Discard the longer length of each. Keep the shortest lengths.
2)The shortest lengths can be any length between 0 and a. So, draw a graph with the x axis being one rod and the y axis being the other.
3)Put a point (call it X) at some distance between 0 and a on both axes.
4)Find the area where the shortest length is smaller than X.
5)Remember to scale by a^2 because probabilities are out of 1.
6)This is the cumulative distribution function.
7)Differentiate with respect to x to find the probability density function.
8)Multiply by x and integrate between 0 and a.

Volia, a/3.

If everyone is sick of this puzzle then kewl, but say if you want another extension.

[HarryW]

I always hated stats, and I still do.

[kedaman]

And i remembered the result was 1/6 for a rod with length 1 anyways what I did was drawing a figure instead of integrating, a tetraeder (or tetrahedron, dunno how what it is, you know a pyramid with triangle as base) where the rightangled triangle base consists of the probability scale 0-1 and the length factor involved. As you divide off the height anyway, i've set it 1. The volume of a tetraeder is height* (length*width/2)/3 length*width being the base area, You get the same formula when you integrate, for sure. So that results in a/3 for length=2a