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# /sci/ - Science & Math

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Can /sci solve this?

 >> Anonymous Thu Nov 2 08:28:56 2017 No.9271015 No they can't.
 >> Anonymous Thu Nov 2 08:57:00 2017 No.9271053 Ill-formed question. Won't bother.
 >> Anonymous Thu Nov 2 09:52:46 2017 No.9271143 >>9271013~31,6%
 >> Anonymous Thu Nov 2 10:04:55 2017 No.9271165 >>9271013((1/16)^m)*16
 >> Anonymous Thu Nov 2 10:08:29 2017 No.9271171 >>9271165oh I missed this((1/m)^n)*m)
 >> Anonymous Thu Nov 2 10:18:18 2017 No.9271186 >>9271165>>9271171read correctly will you?
 >> Anonymous Thu Nov 2 10:27:03 2017 No.9271200 So, exactly one instance of exactly three darts in the same square? That's a lot of cases.
 >> Anonymous Thu Nov 2 10:30:21 2017 No.9271207
 >> Anonymous Thu Nov 2 10:41:58 2017 No.9271220 >>9271013Does 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.
 >> Anonymous Thu Nov 2 10:57:03 2017 No.9271243 ${n \choose m} {\frac{1}{d}}^{m-1}$
 >> Anonymous Thu Nov 2 11:02:45 2017 No.9271259 >>9271013How can you throw the darts both uniformly and randomly?
 >> Anonymous Thu Nov 2 11:15:34 2017 No.9271286 >>9271259The (X, Y) of where each dart lands has a Uniform([0,1]x[0,1]) distribution, brainlet.
 >> Anonymous Thu Nov 2 11:28:14 2017 No.9271320 >>9271013Surely it's just a binomial thing?
 >> Anonymous Thu Nov 2 11:48:01 2017 No.9271366 >>9271320Depends on >>9271200If 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.
 >> Anonymous Thu Nov 2 11:50:13 2017 No.9271372 >>9271013counting problems are kind of boring
 >> Anonymous Thu Nov 2 12:00:28 2017 No.9271389 >>92710130.00024%
 >> Anonymous Thu Nov 2 16:56:17 2017 No.9271855 File: 58 KB, 931x777, Screenshot from 2017-11-02 16-55-59.png [View same] [iqdb] [saucenao] [google] Bumping with simulations.
 >> Anonymous Thu Nov 2 18:10:26 2017 No.9271960 >>9271855I don't understand this.
 >> Anonymous Thu Nov 2 18:20:54 2017 No.9271978 >>9271960brainlet
 >> Anonymous Thu Nov 2 18:26:53 2017 No.9271995 File: 64 KB, 889x848, Screenshot from 2017-11-02 18-25-31.png [View same] [iqdb] [saucenao] [google] >>9271960Do you mean the code or the results. Here's a different version with some expository examples.
 >> Anonymous Thu Nov 2 18:50:52 2017 No.9272042 >>9271013d choices for the square.$\binom{n}{m}$ choices for which darts you'll pick so that you can put in the square.$d \binom{n}{m}$ acceptable configurations in total. $\binom{n+d-1}{d-1}$ 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 $d \binom{n}{m} / \binom{n+d-1}{d-1}$ .
 >> Anonymous Thu Nov 2 18:56:58 2017 No.9272050 18.274004% for the first one
 >> Anonymous Thu Nov 2 19:06:07 2017 No.9272063 >>9272042Oh shit forgot to put the rest n-m darts in other squares .Multiply by $\binom{(n-m)+(d-1)-1}{(d-1)-1}$ if you want exactly m balls and by $\binom{(n-m)+d-1}{d-1}$ if you want at least m balls.
 >> Anonymous Thu Nov 2 19:08:59 2017 No.9272068 >>9272063actually nvm, ignore that, it give probabilities greater than 1
 >> Anonymous Thu Nov 2 19:50:05 2017 No.9272130 >>9272042>>9272063>>9272068Oh I figured it out. I cosidered the darts being different. Remove $\binom{n}{m}$ 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 $d \binom{n-m+d-2}{d-2} / \binom{n+d-1}{d-1}$ .Probability of m or More dots in one square is $d \binom{n-m+d-1}{d-1} / \binom{n+d-1}{d-1}$ .For d=16, n=8 , m=3 the probabilities are 96/253 ~ 38% and 128/253 ~ 50.6 %
 >> Anonymous Thu Nov 2 21:53:38 2017 No.9272315 >>9272130Plug 16 darts into the first one and you get something greater than 1.Seems like you're just pulling stuff out of your ass.
 >> Anonymous Thu Nov 2 23:15:49 2017 No.9272444 >>9272315Yeah I noticed that too, but I can't see where is the damn flaw. It's driving me mad.
 >> Anonymous Fri Nov 3 02:19:44 2017 No.9272744 >>9271013100%
 >> /psy/mon|0-X's|/sci/mon Fri Nov 3 02:28:01 2017 No.9272750 >>9272315Any attempt to isolate any SINGULAR variable will always result in the following constant: 0.716531310573789250425604096925379667453112059821479157140...Derived by this formula: $\frac{1}{e^{1/3}}$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).
 >> Anonymous Fri Nov 3 11:07:23 2017 No.9273292 bump
 >> Anonymous Fri Nov 3 11:10:10 2017 No.9273296 100%
 >> Anonymous Fri Nov 3 13:22:43 2017 No.9273568 >>9271013dCm*(1/d)^m*(1-(1/d))^(n-m)Probability function
 >> Anonymous Sat Nov 4 02:14:57 2017 No.9274696 bumping
 >> Anonymous Sat Nov 4 02:37:20 2017 No.9274731 >>92710131)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} - ...
 >> Anonymous Sat Nov 4 02:41:05 2017 No.9274736 >thread is up for 2 days>still no correct solution/sci/ is really filled with brainlets
 >> Anonymous Sat Nov 4 11:37:04 2017 No.9275248 >>9274736Nigga 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
 >> Anonymous Sat Nov 4 12:45:59 2017 No.9275333 >>9275248Wrong brainlet.
 >> Anonymous Sat Nov 4 12:48:34 2017 No.9275336 >>9271013The real problem is whether or not to do what you asked. I have chosen not to do it. Problem solved. Tohohohoho, hohoho, hohohohoho.
 >> Anonymous Sat Nov 4 12:58:17 2017 No.9275352 >>9275333Wrong brainlet, right answer?
 >> Anonymous Sat Nov 4 13:04:21 2017 No.9275362 >>9271013At least correct the formulation of the problem as "at least 3 darts" t. Taleb
 >> Anonymous Sat Nov 4 13:06:10 2017 No.9275368 How about the old $\frac{1}{16}^3 \times \frac{15}{16}^5$ (if they're talking about exactly 3 hits in one square)
 >> Anonymous Sat Nov 4 13:08:02 2017 No.9275370 >>9275333Ok nigger.
 >> Anonymous Sat Nov 4 13:10:44 2017 No.9275372 >>9275368and with this(1/d)^m * (d-1/d)^(m-n)for the second problem i see is the same way
 >> Anonymous Sat Nov 4 13:14:24 2017 No.9275377 >>9275372obviously n-m
 >> Anonymous Sat Nov 4 13:22:16 2017 No.9275389 >>9275333>>9275352Sorry I shouldn't have teased you. I actually normally don't on 4chan, my apologies.
 >> Anonymous Sat Nov 4 14:02:01 2017 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 $n \choose m$ sets of m darts.So the probability there is at least one square with exactly m darts in it is:${n \choose m} d ({\frac{1}{d}})^{m}={n \choose m} ({\frac{1}{d}})^{m-1}$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:${n \choose m} ({\frac{1}{d}})^{m-1}(1-{n-m \choose m} ({\frac{1}{d-1}})^{m-1})$
 >> Anonymous Sat Nov 4 14:25:57 2017 No.9275440 >>9275417For m = 1 this gives negative probabilities.
 >> Anonymous Sat Nov 4 14:30:08 2017 No.9275446 >>9275440Think about what m=1 would mean in terms of there being exactly 1 square with m darts.
 >> Anonymous Sat Nov 4 14:33:50 2017 No.9275452 >>9275446It 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.
 >> Anonymous Sat Nov 4 14:37:06 2017 No.9275457 >>9271013Someone follows nn taleb
 >> Anonymous Sat Nov 4 14:39:50 2017 No.9275460 >>9275452Hmm. Good point. Where's the flaw in my argument?
 >> Anonymous Sat Nov 4 14:41:48 2017 No.9275464 File: 127 KB, 1190x808, IMG_0053.jpg [View same] [iqdb] [saucenao] [google] >>9271013Solution
 >> Anonymous Sat Nov 4 15:00:21 2017 No.9275490 File: 75 KB, 980x812, Screenshot from 2017-11-04 14-58-01.png [View same] [iqdb] [saucenao] [google] >>9275464Agrees 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?
 >> Anonymous Sat Nov 4 15:24:12 2017 No.9275519 File: 52 KB, 935x656, Screenshot from 2017-11-04 15-23-28.png [View same] [iqdb] [saucenao] [google] Actually I think these are the predicates you'd want to use.
 >> Anonymous Sat Nov 4 15:28:40 2017 No.9275521 >>9271259autist detected
 >> Anonymous Sun Nov 5 00:30:00 2017 No.9276476 >>9275464I don't understand it.
 >> Anonymous Sun Nov 5 00:36:21 2017 No.9276482 File: 4 KB, 117x125, 1433132237110s.jpg [View same] [iqdb] [saucenao] [google] 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.
 >> Anonymous Sun Nov 5 02:05:02 2017 No.9276644 File: 2 KB, 464x134, Solved.png [View same] [iqdb] [saucenao] [google] >>9271013An 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 ~ 0.16905667632) Pic
 >> Anonymous Sun Nov 5 02:11:35 2017 No.9276651 >>9276644This is for at least 1 box with at least m btw.
 >> Anonymous Sun Nov 5 03:40:46 2017 No.9276730 >>9275464>>9276476nvm I do now
 >> Anonymous Sun Nov 5 05:56:35 2017 No.9276867 >>9276482xd fuck you nigger
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