#1

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.

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.

*Last edited by jrcsgtpeppers at Jan 21, 2013,*

#2

I'd say your odds are pretty good.

#3

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

#4

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...

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...

*Last edited by JimDawson at Jan 21, 2013,*

#5

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

#6

Help you with probabilities?

No chance mate.

No chance mate.

#7

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

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?

*Last edited by jrcsgtpeppers at Jan 21, 2013,*

#8

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...

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.

#9

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 at least 3 different ones....

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.

#10

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.

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.

#11

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.

*Last edited by JimDawson at Jan 21, 2013,*

#12

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?

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.

#13

^ 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.

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.

*Last edited by JimDawson at Jan 21, 2013,*

#14

explain this? i think its the formula i have to use

#15

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.

#16

explain this? i think its the formula i have to use

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.

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.

it is the formula he needs. it's simplified into every calculator ever as nCr though

#17

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

#18

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

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

#19

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

*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.

#20

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.

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

#21

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

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.

*Last edited by JimDawson at Jan 21, 2013,*

#22

Knowingthe 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.

Why do you always ruin our ignorant fun? You're like the Grinch who stole Ignorance.

#23

Why do you always ruin our ignorant fun? You're like the Grinch who stole Ignorance.

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

#24

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

Which Yu-gi-oh card will help us with that? Calculate me the odds of drawing that, NOW!

#25

Why do you always ruin our ignorant fun? You're like the Grinch who stole Ignorance.

sorry.

#26

sorry.

I shackle you to the live free or die erect thread.

#27

I shackle you to the live free or die erect thread.

Didn't work.... I ducked.

#28

Didn't work.... I ducked.

#29

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 six-sided 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 six-sided die you should quickly see you statistic of "winning" hovers around 66.666...% and gets more accurate with more tries.

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 six-sided 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 six-sided die you should quickly see you statistic of "winning" hovers around 66.666...% and gets more accurate with more tries.

*Last edited by JimDawson at Jan 21, 2013,*

#30

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 dice, pick a number, and start rolling. With a six-sided dice 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.

Aren't you now falling for the gambler's fallacy?

#31

Aren't you now falling for the gambler's fallacy?

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.

*Last edited by JimDawson at Jan 21, 2013,*

#32

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.

Sorry, it sounded more like you were saying that with more tries the outcome you want becomes more likely.

#33

Sorry, it sounded more like you were saying that with more tries the outcome you want becomes more likely.

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 1-50. (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%.

*Last edited by JimDawson at Jan 21, 2013,*

#34

i got 50% chance

(8/40)/(32/40)/3 x 6?

(8/40)/(32/40)/3 x 6?

#35

Sorry, it sounded more like you were saying that with more tries the outcome you want becomes more likely.

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.

#36

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 1-50. (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%.

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.

#37

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.

#38

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.

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.

#39

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.

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

#40

Edit: Basically ^that, but in a more convoluted way. If you understood that, there's no reason to read this, if not this

/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

So you want to find the odds of

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 non-red card, so the chance of it being non-red is (32-1)/(40-1), 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.

*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 is32/40 * 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 non-red card, so the chance of it being non-red is (32-1)/(40-1), 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 at Jan 21, 2013,*