Posts for arflech

arflech
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I actually had to look through the Riddler's Twitter feed for hints he has given to other people for this one, because the riddle as written is ambiguous: http://fivethirtyeight.com/features/how-many-bananas-does-it-take-to-lead-a-camel-to-market/
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arflech
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Samsara wrote:
Nahoc wrote:
WHY DO PEOPLE KEEP DOING THIS
I know the answer. I'm going to reveal it on stream on October 14th (date subject to delays).
It's because all the kids are still into the dubs tep amirite?
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arflech
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The probability that nobody gets a 6 is (5/6)^n, and that one person gets a 6 is n*(1/6)*(5/6)^(n-1); therefore the probability of a tie at 6 is 1-(n/6)*(5/6)^(n-1)-(5/6)^n. Then the probability that nobody gets a 6 or a 5 is (2/3)^n, and that one person gets a 5 and everybody else gets less is n*(1/6)*(2/3)^(n-1); therefore the probability of a tie at 5 is (5/6)^n-(n/6)*(2/3)^(n-1)-(2/3)^n. The pattern is apparent, ending in the case of a tie at 2: The probability that nobody gets above 2 is (1/3)^n, that nobody gets above a 1 is (1/6)^n, and that one person gets a 2 and everybody else gets 1 is n*(1/6)*(1/6)^(n-1); therefore, the probability of a tie at 2 is (1/3)^n-(n/6)*(1/6)^(n-1)-(1/6)^n. The final case of a tie at 1 includes a couple of degenerate cases: (1/6)^n-(n/6)*0^(n-1)-0^n=(1/6)^n. Adding up these cases yields 1-(n/6)((5/6)^(n-1)+(2/3)^(n-1)+(1/2)^(n-1)+(1/3)^(n-1)+(1/6)^(n-1)) For the case of the median of three rolls of a d6, the following probabilities for one roll hold: 1: (1/6)^3+3*(1/6)^2*(5/6)=2/27 2: (1/6)^3+3*(1/6)^2*(5/6)+6*(1/6)*(1/6)*(2/3)=5/27 3: (1/6)^3+3*(1/6)^2*(5/6)+6*(1/6)*(1/3)*(1/2)=13/54 4: (1/6)^3+3*(1/6)^2*(5/6)+6*(1/6)*(1/3)*(1/2)=13/54 5: (1/6)^3+3*(1/6)^2*(5/6)+6*(1/6)*(1/6)*(2/3)=5/27 6: (1/6)^3+3*(1/6)^2*(5/6)=2/27 Then the probability that nobody gets a 6 is (25/27)^n, and the probability that one person gets a 6 is n*(2/27)*(25/27)^(n-1), so the probability of a tie at 6 is 1-n*(2/27)*(25/27)^(n-1)-(25/27)^n. The probability that nobody gets 6 or 5 is (20/27)^n, the probability that one person gets 5 and nobody else gets above 5 is n*(5/27)*(20/27)^(n-1), so the probability of a tie at 5 is (25/27)^n-n*(5/27)*(20/27)^(n-1)-(20/27)^n. A similar telescoping argument yields the following probability of a tie: 1-n*((2/27)*(25/27)^(n-1)+(5/27)*(20/27)^(n-1)+(13/54)*(1/2)^(n-1)+(13/54)*(7/27)^(n-1)+(5/27)*(2/27)^(n-1)) I seem to get the chance of no tie for 3d6 median as usually higher than for 1d6. ___ Now on to the FiveThirtyEight Riddler... The basic problem for this one is straightforward, but the Extra Credit is tough for those of us who like closed-form solutions: http://fivethirtyeight.com/features/can-you-bake-the-optimal-cake/
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arflech
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HHS wrote:
I got an answer of phi. My solution: Let t be the man's position along the east side of the paddock, with the length of the sides being 1. Let x be the ram's position along the same side, let v be the ratio of the ram's speed to the man's speed, and let s be the distance between the man and the ram. Then, dx/dt = v(t - x)/s, and ds/dt = (t - x)/s - v = (1/v)dx/dt - v. Since x = 0 and s = 1 at t = 0, we get s = 1 + x/v - v. At t = 1, x = 1 and s = 0, which means that 1 + 1/v - v = 0. This has two solutions. One of them is negative and therefore unphysical, and the other one is v = phi.
I got the same answer, but by looking at both the x position and y position simultaneously, which probably complicated things.
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arflech
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arflech
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Toothache wrote:
Am I crazy for wanting to see some of the SMT series games and spinoffs TASed?
I have just discovered how awesome the SMT series is, and I would like to see a TAS of Persona 4; then again, for that and Persona 3 FES, all of the unskippable cutscenes would probably trigger YouTube copyright strikes for the encodes, which would be more than 10 hours each. I'm not sure whether Nocturne is like that too, because I've only seen a little bit of gameplay for that one.
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arflech
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I hope you saved that pastebin, because it has been removed.
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arflech
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The extension to arbitrary regular polygons is obvious, and the limit as the polygons approach circles is also obvious. What is a little less obvious is the question of which shape is most efficient: Something to note, however, is that if, for any point on the edge of the eaten portion, the segment from the center to the edge of the sandwich passing through this point is not bisected by this point, that means the sandwich is not at the most efficient shape. For example, in the polygonal case, the corners could be rounded off with circular arcs so that the eaten region is the same, but the uneaten region is smaller. Also, it seems like you don't need to have the initial requirement that distances are measured with respect to the centroid, to end up with the most efficient shape as if you do have this initial requirement. I think a rigorous proof of this would require a similar strategy to the isoperimetric problem.
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arflech
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Invariel wrote:
4. Game developers have to approach each new game as a new game, even if you're talking about ShootyGame 9, which builds right out of ShootyGame 8. There is always going to be a portion of the audience who hasn't played ShootyGame 8, and who needs that tutorial section. So, in mixing gameplay and story, you, the experienced gamer, have to suffer through a boring tutorial session filled with story you don't necessarily need or care about so that all new players can have a baseline of skill to move forward with.
The problem is that even in comparison to older games of their types, recent games rely too much on tutorials; even comparing Picross-e with the new Zelda-themed Picross is instructive.
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Post subject: Re: I need some life advice...
arflech
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Ready Steady Yeti wrote:
here in the state of SC
Flee once you're financially able; even IN is a step up.
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arflech
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The SMB3 thread links to this disassembly: http://sonicepoch.com/sm3mix/ (The main page is about a hack, but it links to both a disassembly and its documentation; the documentation does not link to the disassembly.) I also just found this disassembly for Mario 1: https://gist.github.com/1wErt3r/4048722 SonicRetro is disassembling the classic Sonic games: http://info.sonicretro.org/Disassemblies An older group called Hacking CulT has also released disassemblies: http://www.hacking-cult.org/?x/2
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arflech
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This Riddler entry reminds me of the isoperimetric problem, and as with that famous mathematical problem, I think I found out how to prove it without the calculus of variations: http://fivethirtyeight.com/features/can-you-solve-the-puzzle-of-the-picky-eater/
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arflech
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Bobo the King wrote:
FractalFusion wrote:
All this problem is, is what Bobo the King said above with the ratio tending to 3:2:1, and the negative binomial distribution applied to the expected number of total drops required to get one of each gem. If probability of success is p and probability of failure is q=1-p, then the expected number of trials required to obtain success is 1(p)+2(qp)+3(q2p)+... = p(1+2q+3q2+...) . Using the identity 1/(1-x)2 = 1+2x+3x2+... , the expected number is p(1/(1-q)2)=p/p2=1/p. Ex. The expected number of rolls of a 6-sided die required to roll a 1 is 1/(1/6)=6. To calculate the expected number of total drops required to get one of each gem, condition on the order in which the types of gems are first encountered. That is, we calculate: 1 + P(Common)*{E(# of drops to get Uncommon or Rare) + P(Uncommon|Uncommon or Rare)*E(# of drops to get Rare) + P(Rare|Uncommon or Rare)*E(# of drops to get Uncommon)} + P(Uncommon)*{E(# of drops to get Common or Rare) + P(Common|Common or Rare)*E(# of drops to get Rare) + P(Rare|Common or Rare)*E(# of drops to get Common)} + P(Rare)*{E(# of drops to get Common or Uncommon) + P(Common|Common or Uncommon)*E(# of drops to get Uncommon) + P(Uncommon|Common or Uncommon)*E(# of drops to get Common)} = 1 + (1/2)*{2 + (2/3)*6 + (1/3)*3} + (1/3)*{3/2 + (3/4)*6 + (1/4)*2} + (1/6)*{6/5 + (3/5)*3 + (2/5)*2} = 1 + (1/2)*7 + (1/3)*(13/2) + (1/6)*(19/5) = 1 + 7/2 + 13/6 + 19/30 = 30/30 + 105/30 + 65/30 + 19/30 = 219/30 = 7.3 Since the ratio of the types of gems tends to 3:2:1, the expected number of common, uncommon, and rare gems, respectively is 7.3/2 = 3.65, 7.3/3 = 2.43333... , and 7.3/6 = 1.21666... . This is consistent with the numbers given by OmnipotentEntity's program.
Bravo, FractalFusion! I knew it had to do with the negative binomial distribution but I couldn't quite put all the pieces together. Your explanation is clear and it appears to be correct.
I used a very complicated tree-branching model, looking at the chances that a sequence of a certain type of gem would be terminated by each of the other gems, and then I got 4.7 common gems, 103/36 (about 2.861) uncommon gems, and 299/225 (about 1.329) rare gems; I wasn't confident in the model, and I had tinkered with it from the original one, which kept saying less than one rare gem on average. The idea I had was to look at the average length of a sequence of one type of gem, then add on the average length of a sequence of that first type and a second type, which begins with that second type, and look at the average proportions of the gems in each two-type sequence, and I ended up using six cases: CU, CR, UC, UR, RC, and RU.
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arflech
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This puzzle is relevant for RTAs, and I think I found an answer: http://fivethirtyeight.com/features/can-you-slay-the-puzzle-of-the-monsters-gems/ I have rephrased it below: If a certain type of monster drops common, uncommon, and rare gems in the respective ratio 3:2:1, and you slay the monsters until you have at least one of each type of gem, how many gems of each type will you have, on average?
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arflech
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Warp wrote:
The most common consensus is that with proper names, 's should be used regardless of whether the name ends in S or not. With common nouns only the apostrophe ought to be used.
The actual consensus is that 's should be used for all singular nouns, whether they end in s or not (like kudos's or Lewis's or OS's), and also for plural nouns that don't end in s (like children's); with plural nouns that end in s, only the apostrophe should be used (like systems' or the Lewises'). Another common consensus is that the apostrophe should not be used to pluralize a word, not even an initialism (CDs), nor to refer to a time period (like the 1900s, but it may indicate omission of a leading part of a date, like the '90s).
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arflech
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WST wrote:
According to the scientists, Sonic causes cancer.
The name was coined about 20 years ago by a postdoc whose daughter liked the Sonic comics: https://en.wikipedia.org/wiki/Sonic_hedgehog
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arflech
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^I believe that is covered in typical classes on real analysis and follows from the fact that all pairs of infinite digit strings representing the same number represent rational numbers. ^^I think you meant (ry)^t in that last line.
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arflech
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I've been thinking about a simplified problem involving farming: Imagine that you're growing a crop on an infinite amount of farmland, starting with a certain number of seeds A (natural number), with average yield per seed y>1 (that is, if you plant A seeds, you'll most likely get yA seeds for the next harvest). Each season, you save a certain portion of seeds r (0<r<1) for the next season and eat the rest. Somehow, the yield per seed does not change as a result of practices like overusing the same plot of soil or tilling the rest of the plant back in (because only the seeds have value), and the present value of the seeds depreciates by a factor of k (0<k<1) per season, so the average present value of a seed's yield is v=ky. If N is the number of seeds available per harvest, and t is the number of seasons since the beginning (whole number), then N=A*(yr)^t, so to keep this enterprise going, r>=1/y. If H is the total number of seeds harvested after T seasons (whole number), then H=sum(A*(1-r)(yr)^t,t,0,T)=A*(1-r)(1-(yr)^(1+T))/(1-yr) If r>=1/y then H grows without bound as T->infinity, otherwise H->(1-r)A/(1-yr). For this case, dH/dr=(y-1)A/(1-yr)^2, which is positive because y>1, so the total harvest grows without bound as r->1/y. Now, let V be the net present value of the harvests, then V=sum(A*(1-r)(vr)^t,t,0,T)=A*(1-r)(1-(vr)^(1+T))/(1-vr). What is important to note here is that although it is possible for v>=1, this is not guaranteed, but what is guaranteed is that 1/y<1/v; if v<1, then r<1<1/v, so V=(1-r)A/(1-vr). Then dV/dr=(v-1)A/(1-vr)^2<0, so the ideal strategy to maximize net present value is to save no seed (r=0) and consume it all without planting. If v>=1 then it is possible for r=1/v, and then V=(1+T)(1-1/v)A, which grows without bound as T->infinity. The curious thing happened when I tried to maximize the net present value over limited timescales; in this case, dV/dr=(v*(1-r)(vr-1)(T+1)(vr)^T-(1-v)(1-(vr)^(1+T)))/(1-vr)^2 so dV/dr=0 when v*(1-r)(vr-1)(T+1)(vr)^T=(1-v)(1-(vr)^(1+T)) and although I was able to see that this means r=1/v for the cases T=0, T=1, T=2, T=3, and T=4, (after eliminating the extraneous roots), and that r=1/v is a root in the general case, it's curious that WolframAlpha wasn't able to solve this equation for r or at least figure out that r=1/v is always a root, not even with the sophisticated non-elementary functions that it knows about.
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arflech
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> would like RAM and PCI Phone/Fax card > lives in USA > not a TASer, only encoded like one vid > cri evrtim
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arflech
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scrimpeh wrote:
Bonus points for github also supporting subversion
I was not aware of that.
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arflech
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I thought I remembered some recent game that requires Windows 7 or later (and therefore will not run under Hourglass) and yet is TASable because the developer implemented that sort of functionality; it's possible that the creator of Undertale would do so in the future, enlisting the help of people like packsciences who can be trusted to not leak the source code they're working on.
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arflech
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Samsara wrote:
to see other people accepting who they are without questioning them
/srg/ and Reddit are where the hate concentrates nowadays.
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arflech
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something-something "public performance"
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arflech
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I remember there was a time when the rules just didn't allow TASes for some games to be published at all, because there was no way the games could have interesting TASes (but of course, if and when interesting TASes for those games were made, they would be accepted); I haven't followed the site's history close enough to remember whether feos was behind that, but I thought one of the purposes of the Vault was to allow any game to have a published TAS, as long as it met the technical standards, but even if there's no way to make an interesting TAS for the game.
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arflech
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EgxHB wrote:
Tangent ... "[P]lagiarism is an ethical violation only." Try telling that to the uni professors after they kick you out for submitting some pre-written essay you bought for a hundred quid.
"an ethical violation only" is not the same as "not subject to sanctions"; if this were a video submitted for a university course, that probably would be grounds for failing the course or expulsion from university (a.k.a. "Plagiarism is not copyright infringement; just because it's one doesn't make it the other.")
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