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 Magic: The Gathering Magic General Opening hand statistics--check my math, please!
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Switch to Forum Live View Opening hand statistics--check my math, please!
5 months ago  ::  Jan 10, 2013 - 12:56AM #1
Rush_Clasic
Date Joined: Mar 10, 2003
Posts: 12,344
I like to think I have a good grasp on a variety of mathematical skills, but in truth my understanding is somewhat rudimentary.

Suppose I have a deck of 4 s and 56 s. I am attempting to calculate what the odds are that I will draw at least 1 in the first 7 cards.

The number I calculated was roughly 65%. The process I used was to calculate the odds of drawing all 4 Bolts, just 3 Bolts, just 2 Bolts, just 1 Bolt, and finally 0 Bolts, adding all the results together. The specifics are below:

numbers Show
Total combinations overall: 60*59*58*57*56*55*54=1,946,482,876,800

Combinations drawing 4 Bolts: 4*3*2*1*56*55*54=3,991,680
Combinations drawing 3 Bolts: 4*3*2*56*55*54*53=211,559,040
Combinations drawing 2 Bolts: 4*3*56*55*54*53*52=5,500,535,040
Combinations drawing 1 Bolt: 4*56*55*54*53*52*51=93,509,095,680
Combinations drawing 0 Bolt: 56*55*54*53*52*51*50=1,168,863,696,000
Total of all these combinations added: 1,268,088,877,440

That total divided by the overall combinations:  0.651477

Going into this, I intuitively felt that the number would be a little lower than 50%. After working on it, I'm relatively certain that my calculations work better than my intuition. Are my processes correct? Is there a previous thread I can be pointed to where these calculations have been done before? I'm sure there are, but I was curious to know if anyone had a link handy.

Thanks for any help.

P.S. - As a bonus pratical exercise, I'm recording the stats in an actual field test. I'll post results later.
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5 months ago  ::  Jan 10, 2013 - 1:03AM #2
ilvmymustang
Date Joined: Jan 4, 2004
Posts: 1,042
Not going to be rude here....but honestly....common sense should tell you that it ain't 65%.

Instead of trying to explain it here,   www.wikihow.com/Calculate-Probability  is a very simplified site explaining how to do it properly.

Some people get so bogged down with the numbers in the math, that they just forget to set it up correctly.  Your math looks sound.  Your problem set-up however.....is incorrect.

Hope this helps.
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5 months ago  ::  Jan 10, 2013 - 1:05AM #3
bubba0077
• blame me for the weather
Date Joined: Feb 27, 2002
Posts: 9,841
The formula is readily available and is known as the hypergeometric distribution. It is implemented in the Magic Math spreadsheet available in my sig. Here is the probability of drawing four desired cards in sixty in the first seven cards:

 0 60.0500% 1 33.6280% 2 5.9344% 3 0.3804% 4 0.0072%
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5 months ago  ::  Jan 10, 2013 - 1:43AM #4
bajatmerc
Date Joined: Mar 3, 2010
Posts: 590
I think you should just explain it.

It isn't too complicated to explain. Though even wikipedia has it messed up for casual people because they use the binomial coefficient to descibe what is going on. Then they link to a separate page to talk about what that is.

In the below notation the binomial coefficient is incidated as (a,b). It looks like coordinates which could be even more confusing.

The general formula is as follows:
N = size of population
M = # of items in population with property "E"
N-M = # of items in population without property "E"
n = number of items sampled
k = number of items in sample with property "E"

P(k) = (M,k)*(N-M,n-k) / (N,n)

In general, the (a,b) means combinations of "a items, taking b of them:" (a,b) = a! / (b!(a-b)!)

a! is like 1*2*3...*a

Though I think it is commonly known. Most calculators have the ! key. It is intended to quickly calculate the permutations of combinations from a start point with many, down to the last.

P(k) =    M! / (k!(M-k)!) *  (N-M)! / ((n-k)!((N-M)-(n-k))!)   /  N! / (n!(N-n)!)

Though for practical use, that spread sheet is awesome.

I copy pasted most from www.talkstats.com/showthread.php/565-Hyp...
I was doing that on a laptop, and I had posted once before finished. Hope I didn't make an error.

That sig is awesome that guy above has.

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Concisely: I want a system where players don't have to pick between mechanics and roleplaying. I hope 5E fails asap so a better system can be made asap.
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5 months ago  ::  Jan 10, 2013 - 1:49AM #5
Rush_Clasic
Date Joined: Mar 10, 2003
Posts: 12,344
Where am I messing up in my calculations? If anyone could spot it, I'd be grateful.

Also, thanks bubba.
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5 months ago  ::  Jan 10, 2013 - 2:37AM #6
ilvmymustang
Date Joined: Jan 4, 2004
Posts: 1,042
Although others have given you the answer already....if you want to understand WHY your math is wrong, consider the logic behind what your math is doing.

As the site I linked you to explains....you take the ratio of an "event" ie drawing a bolt happening and multiply them altogether.

So it would be 4/60 * 3/59 * 2/58 * 1/57

The fact that you would normally draw 7 cards together is irrelavant, as you draw 1 card at a time, and each time you draw a card, you are removing 1 from the deck of 60.  Whether or not you actually draw a bolt is irrelavant based on your question of probability of getting on in your opening hand.  If you had asked what is the probability of drawing a bolt on any given draw, that would be different.

To determine if you would have x bolts in hand, you would simply remove "events" or draws.

To that end....if you wanted to know the odds of drawing 2 bolts..... 4/60 * 3/59 or (.066667 * .05) also note that I rounded.... = approx 6% which matches what bubba posted.

Hope this helps even more!  lol
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5 months ago  ::  Jan 10, 2013 - 2:40AM #7
bajatmerc
Date Joined: Mar 3, 2010
Posts: 590
You said:
"
Total combinations overall: 60*59*58*57*56*55*54=1,946,482,876,800
Combinations drawing 4 Bolts: 4*3*2*1*56*55*54=3,991,680
Combinations drawing 3 Bolts: 4*3*2*56*55*54*53=211,559,040
Combinations drawing 2 Bolts: 4*3*56*55*54*53*52=5,500,535,040
Combinations drawing 1 Bolt: 4*56*55*54*53*52*51=93,509,095,680
Combinations drawing 0 Bolt: 56*55*54*53*52*51*50=1,168,863,696,000
Total of all these combinations added: 1,268,088,877,440
That total divided by the overall combinations:  0.651477

"

Your method for calculating the total combinations looks right to me. I didn't check the value.

I think you also got the combinations of drawing 0 bolts right in method. You can get the odds of not drawing a bolt.
1 minus those odds should give you the same result as the spread sheet for 1 bolt.

However the rest looks wrong.

Spoiler: Show
Concisely: I want a system where players don't have to pick between mechanics and roleplaying. I hope 5E fails asap so a better system can be made asap.
Spoiler: Show

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5 months ago  ::  Jan 10, 2013 - 2:53AM #8
Nylon
Date Joined: Sep 19, 2008
Posts: 202
Your calculation of 60*59*58*57*56*55*54=1,946,482,876,800 gives the total number of combinations assuming their order matters, so that, for example, c1+c2+c3+c4+c5+c6+c7 is considered to be a different combination than c7+c6+c5+c4+c3+c2+c1.

If you want the total number of unique combinations ignoring the order, the calculation is the one given by bajatmerc: (a,b) = a! / (b!(a-b)!). In this situation, a=60 and b=7, so the calculation is (60, 7)= (60*59*58*57*56*55*54)/(1*2*3*4*5*6*7)=386206920.

The number of combinations that don't include a lightning bolt is (56, 7)=(56*55*54*53*52*51*50)/(1*2*3*4*5*6*7)=231917400.

And so, the number of combinations that include at least one lightning bolt is 386206920-231917400=154289520.

And the probability of finding at least one lightning bolt in your opening hand is 154289520/386206920=0.399499625 ~ 40%.

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5 months ago  ::  Jan 10, 2013 - 7:31AM #9
Dragon_Nut
Date Joined: Feb 22, 2005
Posts: 2,136
Alternately, if you only care about getting greater than 0 lightning bolts, the math is incredibly easy if you remember a simple piece of probability: The chance of X happening is the same as one minus the chance of X not happening.

In this case, that means the chance of one or more lightning bolts is the same as 1 - P where P is your chance of drawing zero lightning bolts. And that's a method you may be more familiar with:
(56/60)*(55/59)*(54/58)*(53/57)*(52/56)*(51/55)*(50/54) = .601

So we get 1 - .601 or .399

So you'll draw a lightning bolt in your opening hand about 40% of the time.
Immature College Student
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5 months ago  ::  Jan 10, 2013 - 8:45AM #10
bay_falconer
Date Joined: Oct 12, 2010
Posts: 9,710
After that, things get tough, just because card draw spells, tutors, cycling, and cantrips.

Jun 27, 2012 -- 12:04AM, GM_Champion wrote:

Clever deduction Watson! Maybe you can explain why Supergirl is trying to kill me.

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