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01212013, 03:47 AM  #1 
Registered User
Join Date: Feb 2009
Location: strawberry fields

help me with some probability
40 cards
8 red 8 blue 8 green 8 yellow 8 purple i get to draw 6 cards and keep them all in my hand what are the odds of drawing AT LEAST 1 red? odds of drawing at least 1 red and 1 blue odds of drawing at least 1 red 1 blue and 1 green thats about as much as i need to know. thanks if anyone shows me how its done. stats are harder thani thought.
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01212013, 03:51 AM  #2 
Organ Donor
Join Date: Jan 2005
Location: Madison, Wisconsin

I'd say your odds are pretty good.
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01212013, 03:54 AM  #3 
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Join Date: Feb 2009
Location: strawberry fields

id like to think that too, but im sitting here thinking, whats the best way to beat all these spoiled rich kids in yugioh after their parents buy them boxes of the new pack series that just came out, and i thought, lets use math! then i remembered i never took stats
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01212013, 04:02 AM  #4  
I heard you like lasers?
Join Date: Feb 2011

Well, I didn't take this in school but...
8 is one fifth of 40. Since you're finding the odds of two types of cards with the same odds, I think you would figure out what one fifth of one fifth is, making it 1/25. Then you multipy it by six because that's how many times you're drawing cards. Pretty sure your odds of getting at least one red and one blue are six out of twenty five, or 24%. The second one would be 6/125, or 4.88%. Did the percentage in my head, and it's wrong. It's actually 4.8%. Damn, I've done more difficult things...
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Last edited by JimDawson : 01212013 at 04:06 AM. 

01212013, 04:03 AM  #5 
yoloswag420
Join Date: Nov 2007
Location: Vancouver, BC

are you drawing with replacement or no? makes sort of a big difference. if yes, use permutations. if no, use combinations

01212013, 04:03 AM  #6 
Banned
Join Date: Jul 2009

Help you with probabilities?
No chance mate. 
01212013, 04:07 AM  #7  
Registered User
Join Date: Feb 2009
Location: strawberry fields

Quote:
no, i pick up 6 cards and keep them all in my hand, no replacement. how do i use combinations? is the above sir correct though?
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01212013, 04:10 AM  #8  
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Join Date: Feb 2009
Location: strawberry fields

Quote:
it cant be though... because i draw one of each very often. it would be hard to imagine having 5 sets of 8 cards and shuffling them then drawing 6 and not getting a card from 3/5 that i want. the other 2/5 is filler.
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01212013, 04:13 AM  #9  
I heard you like lasers?
Join Date: Feb 2011

Quote:
Yeah, I think the number of varieties comes into play here somehow... something seems a bit screwy to me too. This intrigues me, so I'll stick around and try some more things.
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01212013, 04:18 AM  #10 
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Join Date: Feb 2009
Location: strawberry fields

i feel silly putting this much thought into a childrens card game but you guys have no idea how competitive yugioh gets. to make the question easier:
40 cards 8 red 8 blue 8 green 8 yellow 8 purple i get to draw 6 cards and keep them all in my hand what are the odds of drawing AT LEAST 1 red? odds of drawing at least 1 red and 1 blue odds of drawing at least 1 red 1 blue and 1 green thats about as much as i need to know.
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01212013, 04:31 AM  #11  
I heard you like lasers?
Join Date: Feb 2011

On each individual draw, for the first question, you would have a clear 1/5 chance of getting a red card. The hard part is that you have six draws I know that it doesn't make sense to have your odds go above 100% in this kind of math, so 120% (6/5) is definitely wrong. I suspect there's some kind of calculus for this, and the odds are in the >80% range. I'm going to try messing with the fractions some more to get some screwy number with lots of decimal places.
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01212013, 04:46 AM  #12  
yoloswag420
Join Date: Nov 2007
Location: Vancouver, BC

Quote:
didn't really look over his math but i don't think he's right, just from his approach. if you're doing probability you need to read and understand this or you're boned, honestly. don't worry about pascal's triangle. i'm pretty rusty on the mechanics so i'd probably make a stupid mistake if i tried. 

01212013, 04:55 AM  #13  
I heard you like lasers?
Join Date: Feb 2011

^ Yeah, it's wrong. I didn't take into account that as you're drawing cards, the overall number of cards changes. That makes it more complicated. That's only one of the things which don't add up. It has to be some kind of calculation which can't exceed 1 or 100%, and the only basis I have to go on from experience is an equation to figure out note pitches if A4=440Hz.
Right now, I am trying to mutilate this formula: X= 440*2^(y/12) I'm still going hard at it, but I am pretty sure you need some kind of fractional exponent to make it work.
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Last edited by JimDawson : 01212013 at 05:03 AM. 

01212013, 04:57 AM  #14 
Registered User
Join Date: Feb 2009
Location: strawberry fields

explain this? i think its the formula i have to use
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01212013, 04:58 AM  #15 
Banned
Join Date: Jul 2009

It's not. In that, n is the number of terms and r is the ratio. That's for either geometric are arithmetic progressions, I forget which. Probably arithmetic.

01212013, 04:59 AM  #16  
yoloswag420
Join Date: Nov 2007
Location: Vancouver, BC

Quote:
it's the equivalent of nCr or n choose r formula. think of it this way: if i have n cards in a deck, how many different ways can i choose a hand of r cards? so if you have a deck of 4 cards and you want to draw 2 cards, there are 6 different ways you can draw 2 cards. Quote:
it is the formula he needs. it's simplified into every calculator ever as nCr though 

01212013, 05:01 AM  #17 
Banned
Join Date: Jul 2009

Ooops, maybe I monged out. Christ, I got a B in ALevel Maths not 7 months ago...

01212013, 05:07 AM  #18 
yoloswag420
Join Date: Nov 2007
Location: Vancouver, BC

a bit of an example to help:
you have a deck of 40 yugioh cards. you want to draw exodia on your first turn and win the game. so on your first turn there are 5 specific cards you need to get. from 5 specific cards, you must choose all 5 of them. this divided by the total number of ways to choose 5 cards from a deck of 40 (5 choose 5)/(40 choose 5) = 1.5197x10^6 = 0.00015197% that's a low chance, by the way also remember 0! = 1 if you happen to need it 
01212013, 05:09 AM  #19  
Est. 1966.
Join Date: Apr 2007
Location: Burnley, UK

Quote:
Knowing the odds is one thing, being lucky enough that the odds work in your favour or knowing how to manipulate the odds are both something else entirely.
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01212013, 05:14 AM  #20  
yoloswag420
Join Date: Nov 2007
Location: Vancouver, BC

Quote:
if you want to solve for y you'll need to use logarithms, then it's quite simple. x = 440*2^(y/12) log x = log 440*2^(y/12) log x = log 440 + (y/12)log 2 (log x  log 440)*12/log 2 = y 

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