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In like a beholder, out like a halfling.
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# ? Mar 1, 2015 18:12 |
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# ? May 13, 2024 10:34 |
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in like a djinn
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# ? Mar 1, 2015 18:12 |
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After the leak, it had to go out of layout and back into editing for being too long, but apparently it went back out of editing shortly after that so who knows.
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# ? Mar 1, 2015 18:17 |
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Is there any way to pronounce Cuthbert that doesn't make you sound like you have a lisp?
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# ? Mar 1, 2015 18:30 |
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whydirt posted:Is there any way to pronounce Cuthbert that doesn't make you sound like you have a lisp?
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# ? Mar 1, 2015 18:37 |
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I would love to play in an Exalted game but I seem to have the same problem with Exalted that I have with a lot of L5R groups: an absurd amount of metaplot I don't have any particular interest in reading or being beholden to. It sounds like it could be an interesting game with a cool setting, but every group I find expects me to know a million fiddly little details scattered throughout a hundred different books and I just don't have any desire to do more research for an RPG character than I did for grad school.
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# ? Mar 1, 2015 18:39 |
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Adept Nightingale posted:
Galaga Galaxian posted:My RPG mood is way too easily influenced by whatever video games or movies I've been watching lately. MadScientistWorking fucked around with this message at 18:44 on Mar 1, 2015 |
# ? Mar 1, 2015 18:40 |
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whydirt posted:Is there any way to pronounce Cuthbert that doesn't make you sound like you have a lisp?
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# ? Mar 1, 2015 18:40 |
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Q-Bert
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# ? Mar 1, 2015 18:42 |
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I used to like Exalted a lot, then I started getting wise to the creepier aspects of it. As soon as I discovered Legends of the Wulin had all the things I wanted in Exalted without the skeevy poo poo I never looked back. Exalted still has some cool ideas that I like, but 2E was a mechanical mess and I really doubt 3E is gonna fix it enough for me to care.
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# ? Mar 1, 2015 18:44 |
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Splicer posted:My new job just wrecked my usual RPG schedule, but I'll hopefully be playing Some Fallout Thing and running the occasional Danger Patrol. LOL last month a friend asked me: I have a deck of cards. 2 are worth 10 points, 10 are worth 3, 20 are worth 1. What's the average value if I draw 2 cards from a full deck? And I was like oh yeah super easy to build in Excel. Don't even need vba... hubris!
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# ? Mar 1, 2015 18:47 |
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FactsAreUseless posted:Q-Bert @#$&!
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# ? Mar 1, 2015 18:50 |
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Exalted is pretty much Feat Tax: The RPG only with about 50 times more mechanics than you need to do loving anything, with a side of WW's usual "how dare you expect to be a unambiguous good guy, here's some mechanics to make you do evil poo poo if you want to or not".
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# ? Mar 1, 2015 18:51 |
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FactsAreUseless posted:Q-Bert
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# ? Mar 1, 2015 19:00 |
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The Racist Q*Bert meme.
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# ? Mar 1, 2015 19:02 |
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Quarex posted:Uhh, it is pronouncd "Q*Bert"
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# ? Mar 1, 2015 19:02 |
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fosborb posted:LOL last month a friend asked me: I have a deck of cards. 2 are worth 10 points, 10 are worth 3, 20 are worth 1. What's the average value if I draw 2 cards from a full deck? 38.31% chance you get 2 points 40.32% chance you get 4 points 9.07% chance you get 6 points 8.06% chance you get 11 points 4.03% chance you get 13 points 0.20% chance you get 20 points
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# ? Mar 1, 2015 19:15 |
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Mark Rein "Q*Bert" Hagen.
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# ? Mar 1, 2015 19:16 |
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FactsAreUseless posted:The Racist Q*Bert meme. A @#$%&! A @#$%&! Look there's a @#$%&!
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# ? Mar 1, 2015 19:28 |
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K*Ramer
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# ? Mar 1, 2015 19:29 |
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FMguru posted:4.375 points
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# ? Mar 1, 2015 19:47 |
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FMguru posted:4.375 points Mind running through the math? I ended up estimating it with a markov chain but it occurs to me I may have been over thinking it. I was calculating it in a way that recalculated the total population each draw, but the exact order of draws doesn't matter if it's just "draw 3 at once." Whoops.
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# ? Mar 1, 2015 19:47 |
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I'm not super great at math, but this sounds like conditional probability problem.
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# ? Mar 1, 2015 19:50 |
It doesn't seem that rough, I think I could put together the math in a notepad right now.
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# ? Mar 1, 2015 19:52 |
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fosborb posted:Mind running through the math? 4 points = (10/32) * (20/31) + (20/32) * (10/31) 6 points = (10/32) * (9/31) 11 points = (20/32) * (2/31) + (2/32) * (20/31) 13 points = (10/32) * (2/31) + (2/32) * (10/31) 20 points = (2/32) * (1/31)
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# ? Mar 1, 2015 19:55 |
The average is approximately 4.58. First card 2/32 results in 10 10/32 results in 3 20/32 results in 1 First + Second Card if First is 10 2/32 * 1/31 results in 10 + 10 2/32 * 10/31 results in 10 + 3 2/32 * 20/31 results in 10 + 1 First + Second Card if First is 3 10/32 * 2/31 results in 3 + 10 10/32 * 9/31 results in 3 + 3 10/32 * 20/31 results in 3 + 1 First + Second Card if First is 1 20/32 * 2/31 results in 1 + 10 20/32 * 10/31 results in 1 + 3 20/32 * 19/31 results in 1 + 1 Probability Distribution 2/992 results in 20 20/992 results in 13 40/992 results in 11 20/992 results in 13 90/992 results in 6 200/992 results in 4 40/992 results in 11 200/992 results in 4 380/992 results in 2 Weighted Averages 2/992 * 20 = 40/992 40/992 * 13 = 520/992 80/992 * 11 = 880/992 90/992 * 6 = 540/992 400/992 * 4 = 1600/992 380/992 * 2 = 960/992
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# ? Mar 1, 2015 20:03 |
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Meinberg posted:It doesn't seem that rough, I think I could put together the math in a notepad right now. Splicer posted:Where's your general solution
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# ? Mar 1, 2015 20:04 |
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Splicer posted:2 points = (20/32) * (19/31) Thanks, yeah I was definitely over thinking it. At the same time, the actual request was something like "what's the distribution if I draw 4 cards? or 5? or 6?" and while the underlying math remains the same, I think I prefer the "iterate 100,000 tests" method to more quickly play with the number and value of cards.
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# ? Mar 1, 2015 20:06 |
Splicer posted:It's really easy to do intuitively, but: Just use a probability tree! And then if I wind in a situation with way too many probabilities to track by hand, then I'll look into using a program.
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# ? Mar 1, 2015 20:07 |
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Was going to run Dungeon World today, but two of my three players couldn't make it, so I had to cancel. Bummed.
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# ? Mar 1, 2015 22:24 |
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This month I'll be writing and writing all the stretch goals for Strike! Speaking of which, if you haven't backed Strike yet, there are less than 48 hours left on the Kickstarter. The Warlord says "Get to Kickstarter and pledge right now!" Jimbozig fucked around with this message at 01:50 on Mar 2, 2015 |
# ? Mar 1, 2015 22:41 |
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Can you timg that? That is breaking some tables.
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# ? Mar 1, 2015 22:42 |
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It's a stretch goal.
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# ? Mar 1, 2015 23:03 |
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Splicer posted:It's a stretch goal. At $20,000 no more tables broken ever!
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# ? Mar 2, 2015 00:01 |
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Quarex posted:At $20,000 no more tables broken ever! Hello and welcome to the Something Awful Forums. Would you like to hear the good news about our fabulous new implementation of VBulletin, TITAN?
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# ? Mar 2, 2015 00:35 |
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Reene posted:Wait, are you in PDX? I am indeed! Also, a cat was sitting on my character sheet for the first half hour or so of the night.
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# ? Mar 2, 2015 00:36 |
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Ettin posted:The completed core book of Breakfast Cult, hopefully.
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# ? Mar 2, 2015 00:44 |
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Mors Rattus posted:Can you timg that? That is breaking some tables. Hah! Sorry, it looked fine on my screen. Fixed it now.
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# ? Mar 2, 2015 01:51 |
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I'm running TOR again today at a friend's place. Hope the cats won't mess up all the dice.
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# ? Mar 2, 2015 09:49 |
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# ? May 13, 2024 10:34 |
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fosborb posted:Thanks, yeah I was definitely over thinking it. At the same time, the actual request was something like "what's the distribution if I draw 4 cards? or 5? or 6?" and while the underlying math remains the same, I think I prefer the "iterate 100,000 tests" method to more quickly play with the number and value of cards. If you want to compute it exactly for 4 or 5 or 6 cards your best bet is polynomials. If you have a set of cards, each with value mi, then the elementary symmetric polynomial ek(Xm1,Xm2,...,Xmn-1,Xmn) will be the sum of Xt for every possible choice of t, and then you can just divide by the value at X=1(which is just a binomial coefficient C(n,k)) and get probabilities. This seems like it's just a restatement of the problem, but Newton figured out how to go between that form and the power sums, which are very straightforward to compute in the given case; you wouldn't necessarily want to do this by hand but there are computer libraries like numpy that should have polynomial multiplication algorithms, etc.
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# ? Mar 2, 2015 11:05 |