h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

To begin with, the chance that the 2-point breaker will _not_ be possible to assemble into a triangle is equal to the chance that one of the sections formed is larger than half the rod.

-------|-----|------
A B C

Now, this breaks down into two cases:
1) A or C is larger than half the rod (if C, turn the rod around!)
2) B is larger than half the rod.

For 1) to occur, we need that both breaks be in the same half-rod, the probability of which is 1/2 per break, equaling 1/4 for this situation.

For 2) to occur, we need that the first break is 'lower than 0.25' and the second one 'higher than 0.75', the probability of which is the product of the two individual probabilities, that is 1/16.

For a total, the rod is unusable iff on of these two breaks down, the probability of which is the sum of the probabilities: 1/4+1/16=4/16+1/16= 5 / 16



When it comes to the rectangles, it is necessary that the pieces be pairwise of equal length -- in the figure below:

------|----|------|----
A B C D

where ---- is the rod, and | are the breaks, we need that A=C and B=D for it to form a rectangle.

I have a feeling that the nitpickers can break this probability down to be equal to zero, so I'll lighten the requirements, and only require the ends to be able to meet. Perhaps some sort of insight will come out of this.

The ends will meet iff none of the pieces is larger than 1/2 rod. Thus, we can distinguish two cases again:

1) A or D is larger,
2) B or C is larger.

If 1) occurs, then all three breaks are in the same halfrod. This has the probability of 1/2*1/2*1/2=1/8.

If 2) occurs, then one break is in the first (or last) quarterrod, and the two others in the other quarterrod. Thus 1/4*1/4*1/4=1/64,

giving a total of 1/8+1/64=9/64.

(I probably screw up some part of identity between cases argument, but it's an idea to build on)

I probably will need some help with the conclusion from someone who has heir combinatorics fresh in memory.... Mine are swapped out on slow media for the moment...

// Mikael Johansson

In my opinion, your solution needs a few corrections. You have already structured the problem, though, so it was easy for me to concentrate just on the calculations.


First about triangles.

In your case 1), we need to multiply the probability by 2 because there are 2 half-rods. This is 1/4 * 2 = 1/2.

In your case 2) things are little complicated, because the 0.25/0.75 situation you described is _not_ the only one when (middle) section B is larger than half the rod. In fact, when the first break is at length X < R/2 (where R stands for rod length), then the second break anywhere between X+R/2 and R makes the middle section large enough to prohibit triangle construction. Thus, for each point X < R/2 the probability of section B being longer than R/2 equals 1-(X+1/2) = 1/2-X. (I assume that the length of a rod is 1. In this situation the probability of having a break in some part of the rod equals the length of this part, which will simplify writing.)

To calculate the total probability, we need to integrate this over the interval (0,1/2). The calculations are fairly simple and give the result of 1/8. Again, we have to multiply this by 2, as we have covered just half of the chances for the first break. This is 1/8 * 2 = 1/4.

So, the total probability of getting _unusable_ rod equals 1/2 + 1/4 = 3/4. This means that the chance to assemble a triangle from a randomly broken rod is just 1/4.



Now, let's talk about the tetragons (that's how my dictionary calls four-sided shapes - can you confirm ? I have no experience writing mathematics in English...)

This is actually very similar to the one in the triangles part, except that one more break makes expressions a little more complex in case 2).

Again, case 1) should be considered twice for the two half-rods. This equals 1/8 * 2 = 1/4.

In case 2) we have to deal with three breaks. The basic argument is mostly the same, but now when the first break is at X < R/2, then the next section will be longer than R/2 in two situations:
a) the second and third breaks are somewhere between X+R/2 to R, or
b) one of the other two breaks is somewhere between X+R/2 to R and the second somewhere between 0 to X.

Situation a) has the probability of (1/2-X)**2, b) has the probability of (1/2-X)*X.

The above expressions have to be integrated over (0,1/2) and the results multiplied by 2 to cover the second half-rod.

The calculations are still not too difficult and the result is 1/24 * 2 = 1/12 for a),
plus 1/48 * 2 = 1/24 for b), giving a total of 1/12 + 1/24 = 3/24 = 1/8.

To conclude the case, we have determined the total probability of _not_ getting a tetragon to equal 1/4 + 1/8 = 3/8. This means that the chance to assemble a tetragon from a randomly broken rod is 5/8.

I leave this for verification and to give myself some time for reading about notation before starting combinatorial part...coming soon.

Amator (Slawek Tarlecki)

I agree with Amator about the triangle problem. The correct answer is a probability of 1/4 that the three pieces will form a triangle. There are two points worth noting here:

1. In questions of probability, it is very important that you state exactly how the random elements are chosen. In this case, I assume that the two points are chosen along the length of the rod before the rod is broken. Each can range along the entire length of the rod. An alternative way to pick would be to pick the first point anywhere along the rod, then pick the remaining point anywhere along one of the two pieces left. This looks the same at a casual glance but will produce different results.

2. A graphical solution can very easily show the right answer, but is difficult to present here in a text-based forum:

Draw a square.
Divide it in half twice to make four smaller squares.
Divide the top left square in half using a diagonal line, going from bottom left to top right, and shade the bottom right half of this square.
Divide the bottom right square in half using a diagonal line, going from bottom left to top right, and shade the top left half of this square.

This diagram is a graph of the two points at which the rod breaks. The horizontal axis represents the position along the rod at which the first break occurs. The vertical axis represents the position along the rod at which the second break occurs. A quick check will persuade you that the shaded areas represent the combinations of positions which allow all three rod sections to be less than 0.5 in length. It is easy to see that the shaded areas amount to one quarter of the total area, so the probability is one quarter.

The four-sided figure, which is normally called a "quadrilateral" in English, rather than a "tetragon", is harder to demonstrate graphically, because three dimensions are needed. I haven't figured it out yet.

Thanks for help in naming the figure. I found 'quadrilateral', too, but it looked like less used for me. Obviously words of Latin origin are more natural in English than in Polish. We feel them somehow ultra-scientific and tend to avoid them, even when talking about science.

I like your graphical solution - it says basically the same with no need for integrals (which I don't like personally).

Anyway, I can not agree with you about getting different results depending on the method of choosing the points. This is an issue in problems of another type, where the order of events influences the outcome. In our problem the order has no meaning, and both approaches will produce the same results, if we explore it carefully.

Ok, when we pick the first point and break the rod, we get two pieces. Now we pick the remaining point 'anywhere along one of the two pieces' - but the probability of 'anywhere' being in any one of the pieces is proportional to the length of that piece. This means that the probabilities are still based on the total length of the rod (even though it is broken). So, the result _will_ be the same.

I will eventually post some calculations to prove it, but not now

Amator

I see you are just as careful in specifying how the points are chosen as I was asking you to be. The danger is when someone says: pick the first point, break the rod, then pick a point along the right-hand piece and break it. This looks the same to a casual observer, but will give different results.

Of course I am careful - after all, this is mathematics and not something to play with

too deep for me man I will just lurk for the moment, until something makes sense!

::Feeling , not the dwarf, but blame it on enness::