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01212013, 05:24 AM  #21  
I heard you like lasers?
Join Date: Feb 2011

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I have no idea what you just said, but I already know how to use that formula. It's used to figure out the pitch in Hz (x) of a certain amount of semitones (y) higher or lower than A4. For example, if y was three C5 (x) = 523.25... Hz. I'll have to admit defeat on this one; I don't have the knowledge required to figure out the probability with more than one draw. Looks like I have some studying to do.
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Last edited by JimDawson : 01212013 at 05:41 AM. 

01212013, 05:41 AM  #22  
Feminist
Join Date: Feb 2006

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Why do you always ruin our ignorant fun? You're like the Grinch who stole Ignorance.
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01212013, 05:42 AM  #23  
Banned
Join Date: Jul 2009

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He's a menace really. Needs to be shackled in to the R+P thread. 

01212013, 05:44 AM  #24  
Feminist
Join Date: Feb 2006

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Which Yugioh card will help us with that? Calculate me the odds of drawing that, NOW!
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01212013, 05:48 AM  #25  
Est. 1966.
Join Date: Apr 2007
Location: Burnley, UK

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sorry.
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“Our life is what our thoughts make it.” ― Marcus Aurelius Slacker's Art Website. 

01212013, 05:51 AM  #26  
Feminist
Join Date: Feb 2006

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I shackle you to the live free or die erect thread.
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01212013, 05:58 AM  #27  
Est. 1966.
Join Date: Apr 2007
Location: Burnley, UK

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Didn't work.... I ducked.
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“Our life is what our thoughts make it.” ― Marcus Aurelius Slacker's Art Website. 

01212013, 06:01 AM  #28  
Feminist
Join Date: Feb 2006

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01212013, 06:07 AM  #29  
I heard you like lasers?
Join Date: Feb 2011

I don't know how TC is going to get practical use out of this, but I'm just having fun figuring out some math.
Actually, if you get a statistic in your favour (>50%) all you need to do is repeat your number of tries over and over for it to work out for you. Kind of creepy, but it actually works with less tries than you would think. Just get a die, pick a number, and start rolling. With a sixsided die you should quickly see your actual statistic of getting your number hovers around the 16.666...% mark and generally gets more accurate with more rolls. EDIT: Works better the higher your probability is; like if you picked four numbers on a sixsided die you should quickly see you statistic of "winning" hovers around 66.666...% and gets more accurate with more tries.
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Last edited by JimDawson : 01212013 at 06:10 AM. 

01212013, 06:08 AM  #30  
Feminist
Join Date: Feb 2006

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Aren't you now falling for the gambler's fallacy?
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01212013, 06:11 AM  #31  
I heard you like lasers?
Join Date: Feb 2011

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Nope, just real math. It's silly to gamble with math like this. EDIT: Because before you get enough tries in for it to start levelling out you could be in the hole. If your chances are higher, it works out much faster than betting small amounts over and over again on 16 at the roulette wheel. In roulette, your payout isn't even equal to the actual odds anyway.
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Last edited by JimDawson : 01212013 at 06:13 AM. 

01212013, 06:21 AM  #32  
Feminist
Join Date: Feb 2006

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Sorry, it sounded more like you were saying that with more tries the outcome you want becomes more likely.
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01212013, 06:23 AM  #33  
I heard you like lasers?
Join Date: Feb 2011

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Not at all, you just need to read closer. It's less erratic if you use a range of numbers and dice with more sides; like using a percentile and selecting numbers 150. (two 0's is 100 if you are using two ten sided dice.) With more rolls, your statistics should get closer and closer to 50%.
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Last edited by JimDawson : 01212013 at 06:28 AM. 

01212013, 06:58 AM  #34 
Tab Contributor
Join Date: Feb 2009
Location: strawberry fields

i got 50% chance
(8/40)/(32/40)/3 x 6? 
01212013, 07:01 AM  #35  
Tab Contributor
Join Date: Feb 2009
Location: strawberry fields

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thats why we play 12 rounds of swiss matches. consistency is everything. youre only allowed 3 of most cards some 2 some none, so iv managed to use psuedo versions to make my deck only consist of 5 different sets of 8 cards. basic combos need 2 pairs, game winning ones 3. but thats without ersponse from the opponent, so i have 2/5 used for killing their shit. i want 1 of each in every opening hand and id be happy all day. 

01212013, 07:40 AM  #36  
Feminist
Join Date: Feb 2006

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But how does it help him to know that throwing a dice the probability of throwing a 6 gets close to 1/6 in the long run? All very true, but how does it help in a card game? There are much more different groups he has to draw from, and I doubt he can create a deck that gives him a win with >50% preset probability.
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01212013, 07:54 AM  #37 
Tab Contributor
Join Date: Feb 2009
Location: strawberry fields

its not auto win, its basically if my opponent has nothing to stop me win. but i have imput on whether or not they have impact on my play. not much but enough to turn that 50% into a best 2/3 match win at least.

01212013, 07:58 AM  #38  
Feminist
Join Date: Feb 2006

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But do you know their deck? Otherwise it will be a very complicated procedure to calculate it for all the cards available and all the possible combinations. I don't think you can have 1 deck per se with a very high probability of winning.
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01212013, 08:14 AM  #39  
up the hoods
Join Date: Jun 2007
Location: Bangor, Norn Iron/Manchester

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Probability of drawing no reds is (32/40)*(31/39)*(30/38)*etc So 1 minus that is the probability of drawing at least 1 red You can use similar logic for the others 

01212013, 08:18 AM  #40 
Registered User
Join Date: Dec 2011

Edit: Basically ^that, but in a more convoluted way. If you understood that, there's no reason to read this, if not this might help (I'm not guaranteeing anything, though )
/edit Well, unless I'm wrong, the chance should be 76.4%, 96.5% and 97.9%. Basically, since the events are so likely to take place it's easier to find the likelihood that they don't take place and subtract that from one (or 100% in this case). So you want to find the odds of not getting any reds in the first example, which is 32/40 * 31/39 * 30/38 * 29/37 * 28/36 * 27/35: The chance of the first card not being red is 32/40, however, for the second card you're assuming you've already drawn a nonred card, so the chance of it being nonred is (321)/(401), and so forth. What you get in the end is the chance of not getting any reds, thus all other events must contain at least one red, so you just subtract this chance of not getting any reds from the total chance of any outcome (1/100%, whatever) and get the chance of drawing at least one red card. You repeat the process in the other scenarios, but replacing 32 with 24 and 16. There's probably a much nicer way to express this mathematically, but I think it should be correct. Last edited by lncognito : 01212013 at 08:19 AM. 
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