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/sci/ - Science & Math

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File: 47 KB, 603x950, Problem.png [View same] [iqdb] [saucenao] [google]
5427204 No.5427204 [Reply] [Original]

since /sci/ is so good with probability, here is a puzzle for you, i even included 3 solutions so all you have to do it pick one and motivate why (or come up with a 4rth).

lets see if this goes better than the boy-girl problem.

>> No.5427218

That's not three possible solutions.

That's an ill-defined problem, with solutions based on three interpretations of the problem. Namely, you don't say how the chord is constructed, so there's no way to answer it.

>> No.5427222

>>5427218
so you cant solve it?

>> No.5427228

>>5427222
Well yes, but neither can you.

>> No.5427231

>>5427218
>you don't say how the chord is constructed
its random

>> No.5427233

>>5427218

lel

Recognizing and exploiting ambiguities is a characteristic of what it means to be a good problem solver. The solutions are valid. Strang does a demonstration asking whether or not a randomly constructed triangle is equilateral, obtuse or acute, and there are different answers depending on the method.

3/10 is the best I can give you.

>> No.5427239

obviously you have to choose the correct sample space, just like how with the boy-girl problem you could get different answers depending on which sample space you chose.

I believe that solution number 3 is the correct sample space.

And testing by simulation would yield a ratio closest to 1/3.

>> No.5427241

solution number one is correct. assuming symmetry under rotation is not valid. it leads to the same cords being counted multiple times, leading to higher answers.

>> No.5427243

>>5427239
>completely missing the point
The simulation results depend on how you build your simulation, hence which sample space you choose.

>> No.5427246

>>5427241
I like you.

>> No.5427250

>>5427233
>Strang does a demonstration asking whether or not a randomly constructed triangle is equilateral, obtuse or acute, and there are different answers depending on the method.

Yes, exactly. So it's not a valid question unless you specify the method.

>> No.5427251

>>5427231

0/10

>> No.5427255

>>5427241
but what about all the chords passing through the centre?

Solution number one does not count them enough.

>> No.5427260

>>5427243
choose the simulation with the least loss of generality.

one way that would come time mind would be to choose a random variable 0=<x<360 to choose the the starting point of the chord, then another random variable 0<y<180 to choose the angle which the chord progresses from to reach the next point on the circle.

This would lead to every chord being as likely to be chosen as every other chord.

It's objectively the best sample space and would provide the best simulation to reveal teh best result.
FACT

>> No.5427265

>>5427260
>somehow still missing the point

>> No.5427270

>>5427255
It's a null-measure set.

>> No.5427272

>>5427260
>build his simulation following proof 3 sample space
>get the result from proof 3
What a tweest!

>> No.5427278

>>5427265
I'm not missing the point.

my proposed way of simulating chords randomly is perfect.

It would lead to one number and that number would objectively be the correct answer.

prove me wrong.

>> No.5427280

The first one is correct. Like in internet arguing, assumptions apply various arbitrary restrictions that result in everything becoming wrong.

If we think of the first example and its unique midpoint and apply it to the other two examples.

In example two, we then "restrict the chords to midpoints lying on a horizontal line through the center"
And in three, we "restrict the chords to those having a midpoint tracing a tear shape."

There's probably other assumptions/restrictions one could make, but if the possible location of chord midpoints does not 'shade' the entire circle area, it's invalid.

>> No.5427287

>>5427272
if that's the case then proof 3 uses the correct sample space and the others use incorrect sample spaces.

my method of simulation has no loss of generality and would lead to every possible chord being chosen as much as every other possible chord. Therefore my proposed method of simulation is objectively ocrrect and the best.

Mathematics: 1
wooly faggots who should be studying philosophy: 0

>> No.5427296

>>5427287
It's as if we couldn't see you coming.
I bet you're OP too.

>> No.5427304

>>5427296
Am I to assume that you realise that your position has no mathematical basis and so have resorted to substance-less, smug replies ?

What a typically philosopher thing to do.

>> No.5427309

>>5427304
Why don't you choke on my fat substance?

>> No.5427306

>>5427270
Are you sure, there are more than one chord going through the center

>> No.5427308

I think this question is too intellectually taxing for /sci/.

You should have just used the baby problem, OP.

>> No.5427313

>>5427306
It's a 1-parameter subset in a 2-parameter set, it's like worrying about the area of a line.

>> No.5427311

Well, my GCSE maths I used came back with the answer of 1/3.
Yet I am less than qualified to give a proper proof that uses any logic.

>> No.5427315

>>5427304
The best way to spot a troll or an idiot is to see if he mocks philosophy.

>> No.5427318

>>5427306
there are more higher infinite chords in the rest of the circle than teh infinite chords going through the centre of hte circle.

so it makes infinitessimal difference that we're under counting them.

>> No.5427323

>>5427315
The best way to spot a butthurt loser is to see if he fails to provide any valid rebuttal but replies anyway.

Why don't you go write an essay about the meaning of truth or something?

Let the adults talk about mathematics.

>> No.5427326

>shit question is asked
>first reply explains exactly why it's shit
>OP continues to selfbump with manufactured controversy anyway

typical /sci/ thread

>> No.5427329

>>5427318
>teh infinite chords going through the centre of hte circle.

There are not a single chord going through the center.

>Chords can not go through the center of a circle.
>http://en.wikipedia.org/wiki/Chord_(geometry)

>> No.5427331

>>5427313
Ok. But don't the chords that are counted multiple times in the second argument also form a null-measure set?

>> No.5427334

>>5427323
>implying I was him

>> No.5427335

>>5427329
That's wrong you retard.

diameters are just a subset of chords which pass through the centre.

How does it feel to be a retard man?

>> No.5427337

1/3 is correct.

>> No.5427338

>>5427329
So you wouldn't count the diameter as a special case of chord?

>> No.5427339

>>5427331
Nah, it's different, it makes the distribution denser near the center of the circle.

Keep in mind >>5427218 is still right though, there is not a distribution more correct than the others.

I can see what "picking a segment at random" means without much equivocacy, but picking a chord at random? Doesn't mean much.

>> No.5427342

>>5427337
>did not read thread

>> No.5427343

we already know that 1/4 is correct.

The only question that remains is whether we could have determined that by simulation.

And the answer is yes.

OP is simply refusing to acknowledged that his "ambiguous" problem isn't ambiguous at all out of asspain.

>> No.5427344

>>5427323
im writing an essay abut ur mum lel xD

>> No.5427349

>>5427339
>picking a chord at random?

>a chord is a line segment joining two points on any curve

So why not pick two points at random and draw a line between them?

>> No.5427350

>>5427344
>>5427334
>proving my point further
Thanks.
Looks like I win.

>> No.5427347

>>5427339
Yes there is, retard.

obviously only the first answer is correct.

>> No.5427352

>>5427339
>it makes the distribution denser near the center of the circle
Oh yes, I didn't realize this.

>but picking a chord at random? Doesn't mean much.
Wouldn't picking two points on the circle at random angles be the best way? it makes the distribution uniform.

>> No.5427353

>>5427349
That's another valid way of simulating this problem that would lead to the correct, true answer.

>> No.5427357

>>5427350
Congratulations, now go to bed and let grown ups discuss about philosophy, will you?

>> No.5427356

>>5427349
Yeah why not?
Why not also choose a midpoint and pass through it?

There are many different distributions to pick from. There is no "correct" one.

>> No.5427358
File: 30 KB, 313x464, 1238782717002.jpg [View same] [iqdb] [saucenao] [google]
5427358

>>5427350
> I win

>> No.5427359

>>5427352
>Wouldn't picking two points on the circle at random angles be the best way? it makes the distribution uniform.

"Uniform" doesn't mean anything in that context.

>> No.5427363

>>5427356
how would you choose the midpoint randomly?

As long as you truly are choosing the midpoint randomly and every chord is as likely to be chosen as every other chord then it is a correct distribution and would lead to the same answer as any other correct distribution.

What you meant to say is that there are many ways of creating the correct distribution and htere are many incorrect distributions and many ways of creating them.
solutions 2 and 3 are incorrect distributions

>> No.5427364
File: 43 KB, 911x531, some chords.jpg [View same] [iqdb] [saucenao] [google]
5427364

>>5427241
Yup, a chord tangent to the circle, for example,

>> No.5427361

>>5427353
> the correct, true answer.
Which is 1/4

>> No.5427368

>>5427363

1/10

>> No.5427371

>>5427363
>how would you choose the midpoint randomly?

Uniformly distributed random variable over the area of the circle.

>> No.5427372

>>5427363
Are you OP trying a new approach.
If not, I will say again that there is no correct distribution.
"every chord is as likely to be chosen as every other chord" has no meaning when talking about infinite set. It all depends on how you perform your integration.

I will not discuss this further.

>> No.5427376

>>5427357
>>5427358
>continuing to prove me right further by refusing to engage with the mathematics at hand

thanks again.

Now will they decide to try and provide some mathematical arguments, or will they continue being butthurt philosophy kiddies who are too butthurt to admit that they're wrong and have no argument to rebutt with?

>> No.5427378

>>5427363
>how would you choose the midpoint randomly?
Not that guy, but enclose the circle in a box and pix random x,y coordinates, and discard them if they are outside the circle.

>> No.5427383

>>5427372
>"every chord is as likely to be chosen as every other chord" has no meaning when talking about infinite set.


Yes it does.
Do you mean to tell me that solutions 2 and 3 are drawn from distributions where every chord is as likely to be chosen as every other chord?

No?
Then "every chord is as likely to be chosen as every other chord" does have meaning.
You even acknowledged it yourself earlier >>5427339

Looks like you're starting to get confused.
Better take a break so you can start thinking about this problem logically again like you were earlier and stop this wooly philosophy nonsense.

>> No.5427387

>>5427378
that's a valid distribution/method of simulation where every chord is as likely to be chosen as every other chord and would lead to the correct answer.

well done.

>> No.5427388

>>5427368
>no argument/10

I think you'd fit in better at /lit/ , kid.

>> No.5427390

>>5427378
I remember my physics teacher doing that wrong in a computational calculus in first year.
Man it was hard to convince him that his algorithm was incorrect:
"pick x coordinate on [-1,1]"
"pick y corrdinate in [-sin(acos(x)) , sin(acos(x))]"

>> No.5427389

>>5427326
lol. i wasn't even online.
>>5427222
was me. after that sci was just being sci.

>> No.5427394

>>5427389
Hey, OP's back!
So what's the answer?

>> No.5427399

>>5427394
first poster got it correct.

>> No.5427402

>>5427399
lol philosophy fag

>> No.5427449

>>5427383
Not the guy you're arguing with.
>Then "every chord is as likely to be chosen as every other chord" does have meaning.
Do tell what this meaning is, then.

>> No.5427471

Essentially, the very first reply is correct. >>5427218

More specifically, it's not enough to say "choose a chord at random."

Knowing that, the first "solution" is entirely wrong, because there are clearly multiple points on each chord (i.e. the midpoint of one chord is a non-midpoint on another chord) so it's not a one-to-one mapping of {points inside the area} to {chords}.

The second solution is probably the best choice, given the ambiguous wording.

The third solution is acceptable if you allow the definition of "random chord" to be: "Given a point on the outer circle, choose a random angle between <span class="math">-\frac{\pi}{2}[/spoiler] and <span class="math">\frac{\pi}{2}[/spoiler] to define a chord."

>> No.5427480

>>5427449
it means that there is an even , random distribution of chords.

instead of distributions where not every chord is has teh same chance of being chosen, like in solutions 2 and 3, as explained in arguments provided by >>5427280


I'm sorry, was that question meant to show me up?
No , sorry. I'm afraid you are going to have to deal with , concrete, technical mathematics rather than philosophy rubbish.

>> No.5427477

>>5427471
>there are clearly multiple points on each chord (i.e. the midpoint of one chord is a non-midpoint on another chord)
wat
It doesn't matter, since you define a chord by its midpoint only.
It's a mapping just like the other two.

>> No.5427484

>>5427471
>Knowing that, the first "solution" is entirely wrong, because there are clearly multiple points on each chord


Durr hurr.

I'd recommend against posting in this thread if your iq is below 80.

>> No.5427491

This is a well known problem called the Bertrand paradox. No solution is more or less correct than any other.

>> No.5427494

>>5427480
>it means that there is an even , random distribution of chords.
Which means what, exactly? I know what it means for finite sets. What does it mean for infinite sets?

>instead of distributions where not every chord is has teh same chance of being chosen
Still waiting on you to define that.

>I'm sorry, was that question meant to show me up?
No. It was meant to have you elaborate on your claims.

>No , sorry. I'm afraid you are going to have to deal with , concrete, technical mathematics rather than philosophy rubbish.
The irony is delicious. Waiting on you to provide concrete, technical mathematics rather than philosophy rubbish.

>> No.5427497

>>5427494
>seriously arguing with a troll
Pls /sci/ stop being trollbait.

>> No.5427504

>>5427491
and i went through all the trouble of editing out the name...

>> No.5427509

>>5427494
>I know what it means for finite sets. What does it mean for infinite sets?

So you're telling me you don't understand what a random variable X=x where x is a real number 0<=x<1 with uniform disitrbution means?
that's an infinite set.

or in this case, see any one of the multiple ways provided: >>5427352
>>5427260

I mean if you don't understand anything except discrete probability distributions then how am I meant to help you?
I can't .
You're too terminally retarded to understand.

As evidenced by the fact that you refuse to engage with the explanation already provided:
there is an even , random distribution of chords.

instead of distributions where not every chord is has teh same chance of being chosen, like in solutions 2 and 3,

>> No.5427512

>>5427491
wrong, retard. see>>5427241
>>5427280


When did /sci/ become so full of people with low intelligence?
What's worrying is that he seems to have some education in mathematics yet is too stupid to even understand this problem.

>> No.5427510

>>5427504
So, are you entertained by your little troll buttbuddy.

>> No.5427520

>>5427494
it means that solution number 3 has a uniform distribution of chords, as defined by their midpoints. while the other two do not.

>> No.5427532

>>5427509
>So you're telling me you don't understand what a random variable X=x where x is a real number 0<=x<1 with uniform disitrbution means?
No. I'm telling you that it doesn't have a well-defined meaning for infinite sets *in general*. There is a well-defined uniform distribution for the set [0, 1), but but for infinite sets in general; and this uniform distribution is NOT based on the intuition of "every point having the same chance of being chosen".

>> No.5427542
File: 261 KB, 1712x746, totally uniform.png [View same] [iqdb] [saucenao] [google]
5427542

>>5427520
>it means that solution number 3 has a uniform distribution of chords, as defined by their midpoints. while the other two do not.


You might have to recheck the definition of uniform(you're arguing for the #1 example as enumerated by wikipedia)

>> No.5427545

sorry I don't understand why peoe are saying the random chord is ill defined.

I just assumed "random chord" meant
a random element of the set of all line segments which are chords of that circle...

is this wrong?

>> No.5427546
File: 22 KB, 207x239, 1350601794297.png [View same] [iqdb] [saucenao] [google]
5427546

>>5427532
and this uniform distribution is NOT based on the intuition of "every point having the same chance of being chosen".

>> No.5427551

Bijections in general don't preserve uniform distributions.

Consider the function f:[0,1] -> [0,1] with f(x)=x^2. This function is a bijection but if X is an on [0,1] uniform distributed random variable then f(X) is not uniform distributed.

>> No.5427557

>>5427545
>a random element of the set of all line segments which are chords of that circle...
That's correct, but this too is ill-defined. The set of possible chords of that circle is infinite, and there is not (in general) such a thing as a uniform choice from an infinite set. Simple generalizations exist for *particular* infinite sets (concepts like "a uniformly random number from the range [0, 1)" are well-defined), but by no means all infinite sets, and this is not one of them.

>> No.5427563

>>5427546
>Do you have a single fact to back that up?
Why, yes. A uniform distribution over a continuous interval like [0, 1) is based on the different points having the same probability density -- which is not the same thing as having the same pointwise probability, and which doesn't generalize to arbitrary infinite sets like the set of chords of a circle.

>> No.5427627

Picking a "random chord" amounts to picking the length of the chord and then picking some angle of rotation. Assuming both of these are done uniformly at random, since the probability sought, call it P, is independent of the angle of rotation, P is just equal to the proportion of chords whose length is just that the intersect the inner circle. Since a chord of length greater than sqrt(2) will have to cross the inner circle, P = 1 - sqrt(2).

>> No.5427641

>>5427510
as OP, im highly entertained by the fact that sci lived up to my expectations of arguing about shit.

>> No.5427645

>>5427627
>Picking a "random chord" amounts to picking the length of the chord and then picking some angle of rotation.
That's one way out of many, and therefore one of the incorrect answers out of many.

>> No.5428216

1

>> No.5428242

>>5427627
>Picking a "random chord" amounts to picking the length of the chord and then picking some angle of rotation.

OR it means picking a random point within the circle and then a random angle through that point

OR it means picking a random midpoint

OR it means picking a random point on the circumference and then a random angle

OR it means picking two random points on the circumference and connecting them

OR it means picking two ransom points within the circle and extending a line through them in both directions

OR etc. etc.

In other words, it's not properly defined what it means, and the very first reply in this thread is correct.

>> No.5429875

Why are people even arguing about this? its 2. the other does not even make sense.

>> No.5429893

>>5429875
Congratulations on also missing the point.

>> No.5429912

Boy, what a great read.