Spaceship Number Plates

ray giraffe

In space there are spaceships, each with a number plate, with numbers going from 1 up to n.

The only spaceship I have seen so far has number 5.

What is the best guess for n?



The next license plate I see is number 8.

What is the best guess for n now?

CJC86

Is this a puzzle that we are allowed to answer? Or a homework question that you need guidance on?

I would recommend thinking about the probability of seeing #5 given that there are n spaceships, then using Bayes' Theorem to give you an approximation of the probability that there are n spaceships, given that you see #5. Same logic follows for the second part as far as I can see.

http://en.wikipedia.org/wiki/Bayes%27_theorem

R0UF

5 and 8.

CJC86

R0UF wrote: »

5 and 8.

Come on, if you edit your post you should mention why you edit it.

You are right, of course, but you should give us some of the reasoning too. I assume the OP was unsure about it, hence wanted more than just "the correct answer".

R0UF

Sorry I edited as I misread the question.

I'm a little puzzled by the whole thing to be honest. How could anyone reach any answers other than 5 and 8?

Yakuza

Something about Maximum Likelihood Estimators springs to mind.

See here:
http://en.wikipedia.org/wiki/Maximum_likelihood#Discrete_uniform_distribution
and here:
http://en.wikipedia.org/wiki/German_tank_problem

im invisible

just having a quick browse through those links, does the fact that there are two numbers known in the second half of the question change things at all?

MathsManiac

It seems to me that you can't make progress on this without making some kind of "a priori" assumption about the relative likelihood of various values of n before any sightings.

Are you assuming that, if you never saw any number plates, then all values of n are equally likely? This seems a strange assumption. The existence of one spaceship makes it unlikely that there is only one, or indeed that there are only a few.

Imagine there are no cars. Then someone invents a car and it works. Surely, in a relatively short period of time, lots of cars would exist.

Also, if some type of (interplanetary?) agency has decided that spaceships need to be registered and assigned a number, then that indicates that there must be quite a few of them. I would tentatively suggest, for example, that spaceships wouldn't need registration numbers if there were only five or eight of them.

And finally, how many spaceships would need to be knocking about the galaxy in order that Ray Girraffe, going about his daily business on an insignificant little planet on the outskirts of the galaxy, would glance up and happen to spot one of them, close enough to read its number plate, not to mind seeing another one!

The more I think about it the more scary it is.
Flee, I say, flee for your lives!
:eek:

Yakuza

Start with the Drake Equation, and multiply the number of planetary civilisations by an arbitrary number of space ships

Just to add, in my initial post I was referring to ships within a given solar system (i.e. non warp-capable).

Then MM had to make it all complicated 'n' stuff

ray giraffe

Thanks ppl!


I have a question based on the German tank problem wikipedia article:

If there are 3 tanks observed (the maximum number observed being m), the minimum variance unbiased estimator for the number of tanks is

(4m/3)-1

However using Bayesian Analysis, the expected number of tanks is

2m-2



I'm confused how you can have two "best" estimates of the same thing

Can someone please explain in what sense each of these estimates is best?

Thank you!


[I know a good bit about pure maths, but not much about statistics. I read this problem in a puzzle book ages ago and never looked up the solution at the time.]

Yakuza

Two statisticians will analyse the same sequence of numbers differently.;)

The first method above is roughly based on the highest number seen plus the average gap between observed numbers as its guess for the highest.

The second method uses conditional probabilities (i.e. the probability that the highest observed is the highest one in the sequence), and using this information to derive an estimate for N

Fremen

ray giraffe wrote: »

I'm confused how you can have two "best" estimates of the same thing

In much the same way that if you have metric space M, a set A in M and a point x in M, the "closest" point in A to x can depend on your choice of metric (think Euclidean vs. Manhattan distance).

I'd need to think more about it, but that's probably exactly what's going on here. Instead of minimising a distance, you're minimising a so-called 'loss function'. Exactly which loss functions correspond to min. unbiased variance and posterior mean, I can't say off the top of my head.