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Selenbrant

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Determine the right amount of copies; Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg94522#msg94522
« on: June 17, 2010, 08:55:00 pm »
Global Moderator Comment These tables do not account for the automulligan, so don't use them for pillars, towers, novas and the like.

Its about the chance to draw X copies of a certain card depending on:
  • turns or rather cards drawn "n"
  • the number of copies "M"
  • decksize "N"
  • what exactly X is (X=1, means chance to draw exactly 1 card; X>0 means chance to draw at least 1 card and so on)
There is also a topic that explains hypergeometric probability or at least how to use an online calculator:
http://elementscommunity.org/forum/index.php/topic,7893.0.html for those who lack math skills.
Edit: Coinich added some nice 4D graphs there.

Or just look at the tables I already made.
I will regularly add new tables for different deck sizes and different probabilities.
If the table is slope just zoom out a bit

N = 30
X>0
n --->
M = 1
M = 2
M = 3
M = 4
M = 5
M = 6
123456789turns: coin toss won
12345678turns: coin toss lost
789101112131415cards drawed
23,3326,6730,0033,3336,6740,0043,3346,6750,00
41,8446,9051,7256,3260,6964,8368,7472,4175,86
56,3862,0767,2471,9276,1379,9083,2586,2188,79
67,6973,3178,1682,3285,8688,8391,3293,3695,02
76,3981,5285,7289,1291,8493,9995,6696,9397,89
83,0087,4390,8693,4795,4396,8797,9298,6599,16
N = 32
X>0
n --->
M = 1
M = 2
M = 3
M = 4
M = 5
M = 6
123456789turns: coin toss won
12345678turns: coin toss lost
789101112131415cards drawed
21,8825,0028,1331,2534,3837,5040,6343,7546,88
39,5244,3548,9953,4357,6661,6965,5269,1572,58
53,6359,1964,2968,9573,1977,0280,4683,5586,29
64,8270,4575,3879,6683,3686,5389,2291,4993,38
73,6278,8983,2986,9389,9092,3094,2395,7496,93
80,4685,1588,8691,7794,0195,7297,0197,9598,63
An own table for 32 cards just because I saw so much decks with exactly 32 cards.
N = 40
X>0
n --->
M = 1
M = 2
M = 3
M = 4
M = 5
M = 6
123456789turns: coin toss won
12345678turns: coin toss lost
789101112131415cards drawed
17,5020,0022,5025,0027,5030,0032,5035,0037,50
32,3036,4140,3844,2347,9551,5455,0058,3361,54
44,7849,8054,5058,9163,0266,8470,3973,6876,72
55,2260,6565,5770,0174,0177,6080,8083,6486,16
63,9369,4074,1878,3481,9585,0687,7390,0091,93
71,1476,3980,8284,5387,6290,1892,2994,0095,39
N = 50
X>0
n --->
M = 1
M = 2
M = 3
M = 4
M = 5
M = 6
123456789turns: coin toss won
12345678turns: coin toss lost
789101112131415cards drawed
14,0016,0018,0020,0022,0024,0026,0028,0030,00
26,2929,7133,0636,3339,5142,6145,6348,5751,43
37,0441,4345,6149,5953,3756,9660,3663,5766,61
46,4151,4056,0360,3264,2967,9571,3274,4277,26
54,5759,8564,6368,9472,8376,3179,4382,2184,68
61,6466,9971,7075,8579,4782,6385,3787,7489,79
N = 60
X>0
n --->
M = 1
M = 2
M = 3
M = 4
M = 5
M = 6
123456789turns: coin toss won
12345678turns: coin toss lost
789101112131415cards drawed
11,6713,3315,0016,6718,3320,0021,6723,3325,00
22,1525,0827,9730,7933,5636,2738,9341,5344,07
31,5435,4239,1442,7246,1649,4652,6255,6458,53
39,9544,4848,7552,7756,5560,1063,4266,5469,45
47,4652,4156,9961,2165,0968,6571,9174,9077,63
54,1459,3364,0368,2672,0775,4978,5581,2983,73
I think thess tables could be used to find the right amount of copies you need depending on how necessary you need a card. Especially for cards you actually only need once. For this cards you have to decide between readiness of the card and getting this card more then once(=dead card).
For example(first table) with 3 Hopes you will have a chance of 67,24% to get at least 1 hope with your 9th card (or 2/3 turn). With 4 Hopes you would have a chance of 78,16%.
If you think you had a bad draw you can also use them to see whether you were really unlucky or not.  ;)

I hope the table is clearly arranged and useful. You can also tell me which table you need and I will edit them.
« Last Edit: December 27, 2013, 12:43:04 pm by Higurashi »

PuppyChow

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Re: Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg94609#msg94609
« Reply #1 on: June 17, 2010, 10:09:52 pm »
If you want the equation (anyone)

I have an equation I use that's a bit more limited, but simpler to use :) . All it will tell you is the probability of drawing at least one of a certain card which you have a certain amount of in a certain number of draws,

p = ( ( (t-c)! / (t-c-d)! ) / ( t! / (t-d)! ) )
p = Probability
c = Number of the type of card you're testing for in the deck
d = Number of cards you're drawing
t = Number of cards in the deck

It looks more complicated than it really is, since I can't use fraction bars :P .

So if I had six precognitions in a 30 card deck, and I wanted one in my hand of seven, then the probability would be:

p = 1-((30-6)!/(30-6-7)!) / ((30)!/(30-7)!)
p = 1-.17
p = .83
p = 83%

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Re: Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg94636#msg94636
« Reply #2 on: June 17, 2010, 10:39:49 pm »
http://elementscommunity.org/forum/index.php/topic,7893.0.html

I actually think we're probably all using the same formula to get the same answers, but that table puts it nicely too.  I was almost envisioning a 3D graph with the x axis being the number of cards wanted included in the deck, the y axis being the number of cards in the deck, and the z axis being the number of that card wanted in a hand...

Selenbrant

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Re: Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg94691#msg94691
« Reply #3 on: June 17, 2010, 11:44:38 pm »
I guess your equation is wrong, with c=1 the fraction is always reduced to "0", although it should be d/30 for d cards drawed.
I hope I am not wrong with this statement, maybe I have something on my eyes and its already late so be dont be angry with me if I am wrong.

My equation:
6 copies , 7 draws and 30 total to get AT LEAST 1:

p(X>0)=1-p(X=0)
=1 - 24/30 * 23/29* 22/28 * 21/27 * 20/26 * 19/25 * 18/24

the upper sequence: 24, 23, 22, 21, 20, 19, 18 is done with: (t-c)! / (t-c-d)! --> (30-6)! / (30-6-7)! --> (24)! / (17)! --> 24*23*22*21*20*19*18
the lower sequence: 30, 29, 28, 27, 26 ,25, 24 is done with: (t)! / (t-d)! --> (30)! / (30-7)! --> (30)! / (23)! --> 30*29*28*27*26*25*24

Your lower sequence will be 29*28*27*26*25*24*23, because the "-1".
I cant imagine that the "-1" will realise "exactly 1", but I will think about that tomorrow or the day after. Maybe you find my or your fault. ;)

Selenbrant

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Re: Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg94727#msg94727
« Reply #4 on: June 18, 2010, 12:20:37 am »
Ok, I cant help but calculating it...
For p(X>0) I didnt need hypergeometric, but for p(X=1).

My tables says for c=6, d=7 and t=30: 83,00(82,999)
I used hypergeometric..

p(X=1)39,669
p(X=2)+31,317
p(X=3)+10,439
p(X=4)+1,491
p(X=5)+0,081
p(X=6)+0,001
p(X>0)=82,998
My table isnt wrong, but your calculating seems to be, I get 39,669 for exactly 1.

PuppyChow

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Re: Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg94764#msg94764
« Reply #5 on: June 18, 2010, 01:09:53 am »
Actually, I'm calculating for "At least 1," (I should edit that in), and you're right that the "-1" isn't needed.

Selenbrant

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Re: Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg94993#msg94993
« Reply #6 on: June 18, 2010, 11:23:50 am »
http://elementscommunity.org/forum/index.php/topic,7893.0.html

I actually think we're probably all using the same formula to get the same answers, but that table puts it nicely too.  I was almost envisioning a 3D graph with the x axis being the number of cards wanted included in the deck, the y axis being the number of cards in the deck, and the z axis being the number of that card wanted in a hand...
Are you able to do this? I cant... ;). My knowledge in informatics is too limited. I elected mechanical engeneering and maths as advanced courses. But at least I can take the work typing all this numbers on. Too I guess for pillars it can be more intersting to know the chance getting at least 3 or 4.(I know about the QI and the QI is working pretty well, but it could be a nice addition)
Should I add some further tables? Tables are so cute.  :))

Actually, I'm calculating for "At least 1," (I should edit that in), and you're right that the "-1" isn't needed.
Ohh, sorry, I thought I read "exactly 1" or rather I misinterpreted your sentence:
Quote
So if I had six precognitions in a 30 card deck, and I wanted one in my hand of seven, then the probability would be:

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Re: Determine the right amount of copies; Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg1117417#msg1117417
« Reply #7 on: December 26, 2013, 07:00:26 pm »
This tables do NOT count the auto mulligan, therefore they are all wrong and with a considerably big error.

I couldn't make this tables counting the auto mulligan when I tried some months ago.

Admins: please remove this thread from the tutorials section.
« Last Edit: December 26, 2013, 07:04:36 pm by Chapuz »
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Re: Determine the right amount of copies; Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg1117418#msg1117418
« Reply #8 on: December 26, 2013, 07:14:22 pm »
Counting the automulligan shouldn't be hard. I think I can give it a shot.

Just to be sure, that's how the automulligan works (right?):
-Draw 7 cards
-If at least one of them is a 0-cost card, keep the hand
-else, draw 7 cards again. Always keep the new hand.
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Re: Determine the right amount of copies; Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg1117428#msg1117428
« Reply #9 on: December 26, 2013, 08:55:11 pm »
Well I don't really see the use of this thread anymore since we got this.
Does math meath for you.
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« Last Edit: December 26, 2013, 09:00:26 pm by Blacksmith »
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Re: Determine the right amount of copies; Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg1117561#msg1117561
« Reply #10 on: December 27, 2013, 12:48:18 pm »
A simulator doesn't do math. It collects empirical data. This is a math table used for a quick look; a simulator takes time.
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Re: Determine the right amount of copies; Draw Chance https://elementscommunity.org/forum/index.php?topic=8295.msg1118014#msg1118014
« Reply #11 on: December 29, 2013, 11:05:35 pm »
I managed to consider the effect of the mulligan (need to polish the end result, since it is not in terms of the right variables yet) and I can tell that the draw chances of all cards get affected, not only pillars.

In a nutshell, if you have a chance P of having a mulligan with a given deck, then using the mulligan multiplies the probability of a draw that has no 0-cost cards in the first 7 by P and the probability of all other draws gets multiplied by (1+P).

So those tables are all indeed flawed, since the draws for N>7 depend on starting hands of both types, them there is no quick and dirty fix. When I have the correct results I'll post them here.
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