Can /sci solve this?

Can /sci solve this?

No they can't.

Ill-formed question. Won't bother.

~31,6%

((1/16)^m)*16

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

read correctly will you?

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

no

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.

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

How can you throw the darts both uniformly and randomly?

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

Surely it's just a binomial thing?

Depends on 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.

counting problems are kind of boring

0.00024%

Bumping with simulations.

I don't understand this.

brainlet

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

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] .

18.274004% for the first one

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.

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

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 %

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.

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

100%

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).

bump

100%

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

bumping

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} - ...

>thread is up for 2 days
>still no correct solution
Veeky Forums is really filled with brainlets

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

Wrong brainlet.

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

Tohohohoho, hohoho, hohohohoho.

Wrong brainlet, right answer?

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

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)

Ok nigger.

and with this

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

for the second problem i see is the same way

obviously n-m

Sorry I shouldn't have teased you. I actually normally don't on Veeky Forums, my apologies.

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]

For m = 1 this gives negative probabilities.

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

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.

Someone follows nn taleb

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

Solution

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?

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

autist detected

I don't understand it.

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.

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

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

nvm I do now

xd fuck you nigger