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/sci/ - Science & Math


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9271013 No.9271013 [Reply] [Original] [archived.moe]

Can /sci solve this?

>> No.9271015

No they can't.

>> No.9271053

Ill-formed question. Won't bother.

>> No.9271143

>>9271013
~31,6%

>> No.9271165

>>9271013

((1/16)^m)*16

>> No.9271171

>>9271165
oh I missed this
((1/m)^n)*m)

>> No.9271186

>>9271165
>>9271171
read correctly will you?

>> No.9271200

So, exactly one instance of exactly three darts in the same square? That's a lot of cases.

>> No.9271207

>>9271013
no

>> No.9271220

>>9271013

Does more than 2 darts in a square count? Is it ok if we have multiple squares with 3 or more darts?

We need more information here.

>> No.9271243

[math]{n \choose m} {\frac{1}{d}}^{m-1} [/math]

>> No.9271259

>>9271013
How can you throw the darts both uniformly and randomly?

>> No.9271286

>>9271259
The (X, Y) of where each dart lands has a Uniform([0,1]x[0,1]) distribution, brainlet.

>> No.9271320

>>9271013
Surely it's just a binomial thing?

>> No.9271366

>>9271320
Depends on >>9271200
If it means exactly one instance, then I don't think it's binomial. If it doesn't, then it seems like it'd be P(X = 2), X ~ Binomial(7, 1/16) since the first one goes wherever it goes and the other two must match it.

>> No.9271372

>>9271013

counting problems are kind of boring

>> No.9271389

>>9271013
0.00024%

>> No.9271855
File: 58 KB, 931x777, Screenshot from 2017-11-02 16-55-59.png [View same] [iqdb] [saucenao] [google]
9271855

Bumping with simulations.

>> No.9271960

>>9271855
I don't understand this.

>> No.9271978

>>9271960

brainlet

>> No.9271995
File: 64 KB, 889x848, Screenshot from 2017-11-02 18-25-31.png [View same] [iqdb] [saucenao] [google]
9271995

>>9271960
Do you mean the code or the results. Here's a different version with some expository examples.

>> No.9272042

>>9271013
d choices for the square.
[math] \binom{n}{m} [/math] choices for which darts you'll pick so that you can put in the square.
[math] d \binom{n}{m} [/math] acceptable configurations in total.
[math] \binom{n+d-1}{d-1} [/math] possible configurations in total (just put the squares in an array and think of it as the classic problem with the cells and the balls.
Therefore, the probability is [math] d \binom{n}{m} / \binom{n+d-1}{d-1} [/math] .

>> No.9272050

18.274004% for the first one

>> No.9272063

>>9272042
Oh shit forgot to put the rest n-m darts in other squares .
Multiply by [math] \binom{(n-m)+(d-1)-1}{(d-1)-1} [/math] if you want exactly m balls and by [math] \binom{(n-m)+d-1}{d-1} [/math] if you want at least m balls.

>> No.9272068

>>9272063
actually nvm, ignore that, it give probabilities greater than 1

>> No.9272130

>>9272042
>>9272063
>>9272068
Oh I figured it out. I cosidered the darts being different. Remove [math] \binom{n}{m} [/math] and everything is fine.
You choose a square (d ways). Put m darts in it. Ignore them. Put the rest of the darts in (if you want exactly m, you ignore the square you put the m darts in, otherwise you include it) .

Probability of Exactly m dots in one square is [math] d \binom{n-m+d-2}{d-2} / \binom{n+d-1}{d-1} [/math] .

Probability of m or More dots in one square is [math] d \binom{n-m+d-1}{d-1} / \binom{n+d-1}{d-1} [/math] .

For d=16, n=8 , m=3 the probabilities are 96/253 ~ 38% and 128/253 ~ 50.6 %

>> No.9272315

>>9272130
Plug 16 darts into the first one and you get something greater than 1.

Seems like you're just pulling stuff out of your ass.

>> No.9272444

>>9272315
Yeah I noticed that too, but I can't see where is the damn flaw. It's driving me mad.

>> No.9272744

>>9271013
100%

>> No.9272750

>>9272315
Any attempt to isolate any SINGULAR variable will always result in the following constant: 0.716531310573789250425604096925379667453112059821479157140...

Derived by this formula: [math]\frac{1}{e^{1/3}}[/math]

Because you are allow for 'square' then the probability will be 100%, if you never allow one variable to submit to the above constant (to your desired level of machine precision).

>> No.9273292

bump

>> No.9273296

100%

>> No.9273568

>>9271013
dCm*(1/d)^m*(1-(1/d))^(n-m)
Probability function

>> No.9274696

bumping

>> No.9274731

>>9271013
1)
16 * {Probability of 3 darts in the first square} - 120 * {Probability of 3 darts in the first and second square}
2)
d * {Probability of m darts in the first square} - {d choose 2} * {Probability of m darts in the first and second square} + {d choose 3} * {Probability of m darts in the first, second and third square} - ...

>> No.9274736

>thread is up for 2 days
>still no correct solution
/sci/ is really filled with brainlets

>> No.9275248

>>9274736
Nigga it's x^n where x is the chance of a specific square with one try and n is the amount of times you throw

(1/16)^3 is the solution

>> No.9275333

>>9275248
Wrong brainlet.

>> No.9275336

>>9271013
The real problem is whether or not to do what you asked. I have chosen not to do it. Problem solved.

Tohohohoho, hohoho, hohohohoho.

>> No.9275352

>>9275333
Wrong brainlet, right answer?

>> No.9275362

>>9271013
At least correct the formulation of the problem as "at least 3 darts"
t. Taleb

>> No.9275368

How about the old [math]\frac{1}{16}^3 \times \frac{15}{16}^5[/math] (if they're talking about exactly 3 hits in one square)

>> No.9275370

>>9275333
Ok nigger.

>> No.9275372

>>9275368
and with this

(1/d)^m * (d-1/d)^(m-n)

for the second problem i see is the same way

>> No.9275377

>>9275372
obviously n-m

>> No.9275389

>>9275333
>>9275352
Sorry I shouldn't have teased you. I actually normally don't on 4chan, my apologies.

>> No.9275417

Pick a square, and pick m darts then the probability those m darts land in that square is (1/d)^m.
There are d squares and [math]n \choose m[/math] sets of m darts.

So the probability there is at least one square with exactly m darts in it is:
[math]{n \choose m} d ({\frac{1}{d}})^{m}={n \choose m} ({\frac{1}{d}})^{m-1}[/math]

To get exactly one square we need to look at the remaining d-1 squares and the remaining n-m darts.
The probability that there are no squares with m darts is 1-P(at least one square with m darts).
We already know the formula for that probability since it's the same one as previously, but now with d-1 squares and n-m darts.

The final probability is therefore:
[math]{n \choose m} ({\frac{1}{d}})^{m-1}(1-{n-m \choose m} ({\frac{1}{d-1}})^{m-1})[/math]

>> No.9275440

>>9275417
For m = 1 this gives negative probabilities.

>> No.9275446

>>9275440
Think about what m=1 would mean in terms of there being exactly 1 square with m darts.

>> No.9275452

>>9275446
It means there's one square with one dart, a perfectly reasonable scenario, yet for e.g. n = 3, m = 1, d = 3 you get a "probability" of -3.

>> No.9275457

>>9271013
Someone follows nn taleb

>> No.9275460

>>9275452
Hmm. Good point. Where's the flaw in my argument?

>> No.9275464
File: 127 KB, 1190x808, IMG_0053.jpg [View same] [iqdb] [saucenao] [google]
9275464

>>9271013
Solution

>> No.9275490
File: 75 KB, 980x812, Screenshot from 2017-11-04 14-58-01.png [View same] [iqdb] [saucenao] [google]
9275490

>>9275464
Agrees with simulations but a little stupid that his solution is to brute force the problem with a method that won't scale to large n, m, d. What's the point of posting something like this online?

>> No.9275519
File: 52 KB, 935x656, Screenshot from 2017-11-04 15-23-28.png [View same] [iqdb] [saucenao] [google]
9275519

Actually I think these are the predicates you'd want to use.

>> No.9275521

>>9271259
autist detected

>> No.9276476

>>9275464
I don't understand it.

>> No.9276482
File: 4 KB, 117x125, 1433132237110s.jpg [View same] [iqdb] [saucenao] [google]
9276482

Who gives a shit, how does this help me figure out how to pay for my rent AND get the brakes fixed on my car?

If there was less maths bullshit, there would be more people working as mechanics and it would cost less to get my brakes fixed. That's real math in the real word.

>> No.9276644
File: 2 KB, 464x134, Solved.png [View same] [iqdb] [saucenao] [google]
9276644

>>9271013
An exact symbolic answer is difficult.
I just do 1 minus the probability of getting less than m in each box.
Generating functions ftw.
1) 371759562752/2199023255552 ~ 0.1690566763
2) Pic

>> No.9276651

>>9276644
This is for at least 1 box with at least m btw.

>> No.9276730

>>9275464
>>9276476
nvm I do now

>> No.9276867

>>9276482
xd fuck you nigger

>>
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