OK, Hearthstone. Lets say my deck has all 30 different cards.

>>7905264
cmon how old are you ?

the chance is 1 / n
n = cards in your deck

>>7905274
No.

>>7905274
assuming he only runs 1 of the card

>>7905279
he says "all 30 different cards."
i know that you can have 2 same cards (or more in arena)

>>7905277
i should write
n = cards left in your deck

>>7905282
This is still wrong.

Are you telling me that it is same same chance to draw My desired card out of a deck of 15 cards as it is to draw my desired card out of a deck of 15 cards after trying to draw it 14 times from the previously bigger decks?

>>7905287
14=15*

>>7905287
this is not conditional probability. the chance at any given time remains 1/n

>>7905293
>>7905292
no i understand what he is talking about, give me 5 minutes to check my notes from the stochastics course

>>7905287
>Are you telling me that it is same same chance to draw My desired card out of a deck of 15 cards as it is to draw my desired card out of a deck of 15 cards after trying to draw it 14 times from the previously bigger decks?
This tells me you probably phrased the question wrong.

Chance of drawing the card GIVEN you already drew 16 cards that were not it = 1/14

Chance of drawing 16 cards that are not it and drawing the card = (29/30)(28/29)...(14/15)(1/14) = 1/30

>>7905305
this

>>7905305
i think you can use
https://en.wikipedia.org/wiki/Hypergeometric_distribution
to find the chance to draw the card in a specific amount of draws

>>7905318
this is to find the chance of having a success withing a sample size of a pool. for example if i pull 5 cards out of a 30 card deck, and want to find the chance of opening at least 1 success out of x successes, i would use hypergeometric. when it comes to drawing a success from the top card, its simply x/n

>>7905318
No, that won't really help you and is unnecessary. Remember that (Bayesian) probability is a measure of how much information you have. Before you draw any cards, the chance of drawing the card you want after any particular number of turns is 1/30. After you draw n cards that aren't that card, you now have more information about where the card is in the deck. The chance of drawing the card on any particular turn after n turns is now 1/(30-n)

>>7905324
you are right, this is what i mean to say

>>7905305
Ill simplify my question.

We can say all other cards that are not my card are the same card C.
So now we have 29 C cards and 1 my card.
I am interested in the chances of me pulling my card on the second draw.

So the first draw i do not pull my card and on the second i do.
That translates to (29/30)(1/29) and that is not equal to 1/29.

>>7905333
>I am interested in the chances of me pulling my card on the second draw.
Before you draw any cards the chance of this occurring is 1/30. If you draw a card and it's not the card you want, then the chance of drawing it is 1/29.

>>7905337
he is interested in drawing the second card specifically.

and you answered your own question. its 29/30)(1/29

>>7905345
>he is interested in drawing the second card specifically.
Yes, it's 1/30.

>>7905337
Ill change my quesiton.
Lets say that i have two coins.
First coin has 30 sides and second has 29.
The first coin has 29 sides that are all tails and one that is heads.
The second coin si the same but with one less tails side.

so much like the (1/2)*(1/2) when we want to calculate my chances of getting tails first and heads next,
in this case it translates to (29/30)*(1/29).

>>7905355
Which means that my chances of getting my card on the second draw is (29/30)*(1/29), not 1/29.

>>7905355
yes, so what now ?

>>7905345
The question is what is the equation that gives me the chance per chosen turn of when i finally manage to draw my card.

>>7905360

>>7905363

its is not 1/n, so what is it?

>>7905365
Anyone?

>>7905365
>>7905378
i dont know how to explain it well in english but you have to difference your questions

if you want to know what the chance is to draw 1 specific card when the you have "n" cards left it is 1/n

but if you want to know what the chance is to not to draw the card for x rounds and then to draw it you have to do something like ("29/30)(28/29)...(14/15)(1/14) = 1/30
i can make a equation for it

i'm not 100% sure about it, stochastics confuses me everytime when i start it again and its 11pm here

>>7905363
you either are wording your question poorly or dont know what you are asking

are you asking on turn 1, what is the chance you draw it exactly on any given turn? or on any given turn, you draw it as your next card. the answers to both have been given

>>7905392
I want a general equation that describes this:

>>7905359

(29/30)*(1/29) is the chances of me drawing my card out of a deck of 29 cards after i tried, failed and discarded a card from the previously 30 card deck.

What is the general equation that describes this chances based on the turn on which i finally draw my card.

File: 1444495632258.jpg (104KB, 640x427px) Image search: [Google]

104KB, 640x427px

>>7905355
>Ill change my quesiton.
>Lets say that i have two coins.
>First coin has 30 sides

>>7905430
Point was to make an equivalent question that is identical.

>>7905446
It's funny because it was your 4th attempt to ask something, no offense.

What you want is: 1-(1-p)^n
Where p is the chance of your event, n is the number of tries.

>>7905455
but with p=1/30 and n=1/30 you get 0.638, shouldnt it be 1.0 ?

>>7905462
ups n =30 of course

>>7905462
I was skimming the thread and didn't see your final post, 1-(1-p)^n is the chance of drawing a desired card at least once in any of the n turns.

>>7905473
ye but should it equal 1 when you do 30 turns ?

>>7905484
That's why I gave up. I can only answer for endless decks.

>>7905264
1/30 if you have 30 cards
if you take out 14 of them without finding it, then the chances of finding it will become bigger at every next drawing, for 14 cards it's 30-14= 16, so 1/16

am I retarded ?

chance to draw it in any turn n is 1/30, but the chance to draw the card at turn n is 1/(30-n)

my brain hurts

hope that the thread is still alive in 12 hours when I wake up

good night

>>7905264
Say you have a full deck, and you've got reno in there, and that's who you're drawing for.

Turn 1, assuming you go first, you have 27 cards in your deck and 3 in your hand, before the first draw.
If none of them are reno, you have a 1/27 chance of drawing reno on the first draw.
Your second draw is a 1/26 chance of being reno.
Continue until you draw reno.
Then your next draw has a 0 chance to be reno.

>>7905264

You, and everyone else who's post I've read are retarded

CHANCE YOU WILL DRAW ANY SPECIFIC CARD: 1/(NUMBER OF TOTAL CARDS - NUMBER OF CARDS YOU'VE DRAWN ALREADY)

It does NOT matter how many cards you've drawn. The chance is the SAME

>>7906643
>calls everyone retarded
>answers question with literal 3rd grade math

>>7905423
>(29/30)*(1/29) is the chances of me drawing my card out of a deck of 29 cards after i tried, failed and discarded a card from the previously 30 card deck.
No, that's 1/29. The chance of this occurring before you draw any cards is 1/30. I don't know why you are having such trouble understanding this.

>>7906643
>gives equation based on how many cards you've drawn
>then says how many cards you've drawn doesn't matter
>doesn't realize it's already been explained much better than this retarded explanation
OK retard.

>>7907313
he is calculating the success of drawing the card ONLY on his second draw. meaning both his first draw is not the card (29/30), and the second card is the card (1/29). multiply them for the answer

>>7907327
Fuck, you people are dense. I'll explain this one more time:

The chance he will draw the card on his second turn *before he draws any cards* is 1/30

The chance he will draw the card on his second turn *after he didn't draw that card on his first turn* is 1/29

How the fuck do you idiots not understand this?

>>7907331
how did we ever get to space

>>7905295
It's just a uniform distribution where N = 30 and find p at exactly 15

This is trivial. Or op was asking a far stupider question and assuming that the events aren't independent and just wanted to know what 1/16 was

Kind of hard to tell

>>7907432
Meant hypergeometric and after readinf the full thread, that is what you want Op

Your sample population is 30
Your sample size is n (however many cards have been drawn)
Your success in population is 1

http://stattrek.com/m/online-calculator/hypergeometric.aspx
The outcome of it NOT being in the first 16 cards is .46667, the it being the 17th card is just 1/14, ie there's a 46.66% chance you havent drawn it and a 7.14% you'll draw it next given you not having already drawn it

You also need to know where to set the result to 1-p for specific outcomes

Have fun

>>7907480
And if those calculations are slightly off, I'm typing from a phone while walking around, so fix them yourself

Yo guys
>>7905581

1/2, you either draw it out not

>>7905355
That's is its exactly on the second draw.

If you want the odds for the first n draws, it's additive