Theoretical probability question

[littlebabyjesus]

I'm now still lost as to why my descriptions were wrong - but for the love of dog please no one try to explain it to me. I'd rather leave it on a high.

I'm not letting you off. You now need to grapple with irrational numbers, which the set of real numbers includes

 

[kabbes]

But you can count the whole numbers between and 10.
And you can count the number of x.x numbers between 1 and 10.
And the second set is bigger than the first set.
So no matter how far you extend it, even to infinity, the infinite set of x.x numbers will be bigger than the infinite set of whole numbers.

No?

There are certainly more numbers between 1 and 10 than there are numbers divisible by 10 between 1 and 10.

But when you extend infinitely, both sets have infinitely many members, so you can’t make that comparison in that way. All you can do is count the members of both sets and see what you find.

If you count them you get:

1 — 10
2 — 20
...

100 — 1000

...

And so on for ever. You never get to a whole number that you can’t match up with a brand new multiple of 10. So they have the same size.

That is totally different to the number of real (irrational) numbers between 0 and 1, which can’t be counted by maching them against whole numbers.

 

[mrs quoad]

Could you edit the thread title so it just says “probability question,” Corax?

Otherwise this will bug me every time I see it.

Tah.

 

[littlebabyjesus]

Come back corax! I was just starting to have fun.

I have lots of questions about the nature of irrational numbers and whether or not they can be taken to represent anything real in the universe. Whole new discussion.

 

[littlebabyjesus]

Could you edit the thread title so it just says “probability question,” Corax?

Otherwise this will bug me every time I see it.

Tah.

No, especially not now you've said that.

I don't quite accept that all probability is theoretical, for the reason kabbes gave really. The maths is precise enough. What it maps on to in the real world may be theoretical, but then you can say that about lots of things. I'd say that about π!

 

[kabbes]

And what if there was suddenly a batch of measurements that didn’t match the theory at all.

That could happen.

Couldn’t it?

how would you interpret such a thing?

 

[kabbes]

There is exactly a 1 in 6 chance that a perfect die thrown perfectly randomly will give you a 6.
In reality, there is no such thing as perfect, which means we have additional uncertainty. That doesn’t invalidate the theoretical construct, however.

 

[littlebabyjesus]

Well. Yeah.

Edit: otherwise, it’s all 0s and 1s.

Ah, just saw your edit. Maybe it is all 0s and 1s.

It is an open question as to whether or not the universe is continuous or discrete. My money (probability based on incomplete knowledge) is on discrete. If that is the case, then computers have it right, really, and irrational numbers, for example, have no exact correlation in the universe, are not actually needed for a precise description of the universe (which might actually just be the universe itself). But, as quantum mechanics tells us, complex numbers do appear to map onto something fundamental in the universe. We just don't yet know what it is.

 

[mrs quoad]

No, especially not now you've said that.

I don't quite accept that all probability is theoretical, for the reason kabbes gave really. The maths is precise enough. What it maps on to in the real world may be theoretical, but then you can say that about lots of things. I'd say that about π!

Bollocks, IMO.

 

[littlebabyjesus]

Bollocks, IMO.

Which bit?

 

[mrs quoad]

Ah, just saw your edit. Maybe it is all 0s and 1s.

Probably is. IMO.

 

[mrs quoad]

Which bit?

The bit that was bollocks.

Edit: for clarity, all of it.

 

[littlebabyjesus]

The bit that was bollocks.

Edit: for clarity, all of it.

Even the bit about π?

 

[littlebabyjesus]

There is exactly a 1 in 6 chance that a perfect die thrown perfectly randomly will give you a 6.
In reality, there is no such thing as perfect, which means we have additional uncertainty. That doesn’t invalidate the theoretical construct, however.

Problem of transferring maths over to the real world (always a mistake - poor maths). Course once you've done that you're in messy bayesian territory.

 

[mrs quoad]

There is exactly a 1 in 6 chance that a perfect die thrown perfectly randomly will give you a 6.

D4.

#owned

 

[littlebabyjesus]

D4.

#owned

Lovely platonic solid. hashtag not owned.

See now there are five platonic solids, and it's quite easy to prove that there can't be any more than that. There are 16 regular four-dimensional shapes, it appears, 6 'convex', others 'star'. Are the convex ones the only ones that are equivalent to the platonic solids? If so, why is it that there is one more than in 3D? Is there a pattern, and if so, why?

I don't have a deep feel for why there should be just these numbers of these things, even though, in the case of the 3D solids, I can give a proof of it.

Mapping that on to a real world problem, I also don't have a good feel for why there are three spatial dimensions. The complex numbers of quantum physics hint that there is at least one more that we're missing. Why do we perceive fewer dimensions than there actually are?

 

[squirrelp]

This might entertain you Corax - Hilbert's Hotel

 

[Fez909]

How can walking into a wall not result in bumping into a wall? That's what walking into a wall means - 100%. Answered using words, not numbers.

"How can walking into a door not result in bumping into a door? That's what walking into a door means."

Except when the door is in the opened state.

Now imagine instead of doors you've got walls. And instead of open/closed you've got "are my atoms lined up in a very specifc way that they won't interact with the wall, allowing me to pass through it?"

Of course the chances of the door being open are many, many times greater than the chance of your atoms being lined up in the correct way to pass through the wall. But they're roughly the same thing

 

[kabbes]

mrs quoad — when you ask if probability is “theoretical”, are you asking if probability theory itself is like a scientific theory (ie hypothesis that appears to be correct based on empirical evidence) as opposed to mathematical theory (ie tautologous statements derived logically from axioms)?

Because if so, no it is not like a scientific theory. It is a construct deriving from something called measure theory, which is a well-definited and rigorous branch of maths. It gives you the tools to define the measurement of sets in such a way that you can apply the same kind of approach to likelihood than you do to other topologies. From that Wiki article:

• Every probability space gives rise to a measure which takes the value 1 on the whole space (and therefore takes all its values in the unit interval [0, 1]). Such a measure is called a probability measure. See probability axioms.

If, on the other hand, you’re asking if the application of probability in the real world is theoretically valid, the answer is that it is as valid as any other application of a theoretically pure construct. You really need to differentiate risk (the known unknowns) from uncertainty (the unknown unknowns), or process error (shit happens) from parameter error (but maybe not in the way envisaged) from model error (and what was THAT shit about?). You also need to accept model limitations (ie we know it isn’t perfect but use it anyway because meh, close enough).

These issues, though, can also be tackled with the tools of probability. Parameter uncertainty is dealt with via Bayesian approaches. Model uncertainty has its own mathematical branch. The implications are understood within a rigorous theoretical framework, not just accepted through empirical evidence.

 

[mauvais]

Ah right, so, my padlock requiring a 5 digit combination, has 10000 possible combinations. (4^10 not 5^10 because excluding 0?)

I am a bit tired TBH.

If it's got 5 rollers and each goes 0-9, then it's 10^5.

If it's got 5 rollers and each goes 1-9, then it's 9^5.

 

[xenon]

If it's got 5 rollers and each goes 0-9, then it's 10^5.

If it's got 5 rollers and each goes 1-9, then it's 9^5.

Yes, I get that but this has pins not rollers.

10 pins. Either pressed or not. So you can’t for example have a combination of 11111. You can only use a given digit once, on or off. So the probability must be considerably less than 10,000.

But how do I work it out.

 

[mauvais]

Yes, I get that but this has pins not rollers.

10 pins. Either pressed or not. So you can’t for example have a combination of 11111. You can only use a given digit once, on or off. So the probability must be considerably less than 10,000.

But how do I work it out.

Same principle but 2^10.

 

[xenon]

Same principle but 2^10.

Yeah giving a probability of 1024. If you can have a combination of any length. However in my example you have to have five digits, no more no less. I am okay with binary maths, for relatively simple stuff. But this has thrown me.

 

[xenon]

I keep wanting to say 32. I know it’s more than that but can’t get beyond 5^2.

 

[xenon]

I’m going to say 512. Just flailing around.

 

[kabbes]

Yeah giving a probability of 1024. If you can have a combination of any length. However in my example you have to have five digits, no more no less. I am okay with binary maths, for relatively simple stuff. But this has thrown me.

Just as a matter of notation, a probability must fall in the range 0 to 1. It can't be 1024. I think you mean 1/1024.

I'm not sure I understand your combination lock at all. I think you're saying you have 10 switches, right? Like those switches could be:
00000 00000
00000 00001
00000 00010

etc?

So how then does this five digit thing come into it?

 

[mauvais]

Yeah, I don't understand the five digit thing either.

Can you provide a picture?

 

[xenon]

Yeah sorry have probably phrase the question Slightly confusingly. I will try attach a pic of the padlock in question.
It has 10 buttons. And a 5 button, digit, combination. I know if I could have a combination of between one and 10 digits, it would give me 1024 possible combinations. But what if it must be five?

I see what you mean about probability though. Yes that would be asking what are the chances of hitting the right combination. Probability between zero and one.

 

[Signal 11]

Erm. Sorry Corax about the slight hijack but thought about this earlier and doesn't seem worth a new thread...

Combination locks. The type you get on padlocks, with 10 switches. So, 10 to the power 2, gives us 1024 possibilities. 2^10.

But how do I work out how many combinations there are if, the code has to have at least n digits. That is, use at least n switches? Say, 5 for example?

No, I've not forgotten the combination but thought, what if I did...

Assuming you mean 10 switches that can be either on or off and you have to switch at least 5 of them on...
There's 1 way to switch all 10, 10 ways to switch 9 of 10, 100 ways to switch 8 of 10 etc.
So for at least 5 it will be 11111.
edit: don't think that's right - e.g. for 8 out of 10 it will be 10x9 ways.
so 1 + 10 + 10x9 + 10x9x8 + 10x9x8x7 + 10x9x8x7x6 = 36100?

 

[kabbes]

Yeah sorry have probably phrase the question Slightly confusingly. I will try attach a pic of the padlock in question.
It has 10 buttons. And a 5 button, digit, combination. I know if I could have a combination of between one and 10 digits, it would give me 1024 possible combinations. But what if it must be five?

I see what you mean about probability though. Yes that would be asking what are the chances of hitting the right combination. Probability between zero and one.

The number of ways to choose 5 items out of a pool of 10 is given by the formula 10!/5!x5! = 252

So the chance of getting it right at random is 1/252

So it's not a great lock. A bit better than a normal 3 digit combination, but not by much.