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Old 04-24-2012, 10:54 AM   #1
Valafar123
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Default Math, Probability and Yu-Gi-Oh!

Warning: Math thread impending, some of it is advanced

Notes:
• I will be using "·" as a radix point instead of "." or ",". It's what I'm accustomed to. I also made this up on the spot, so some of the things posted may already have been calculated in the past.
• n is the number of cards drawn.
• d is the size of the deck.
• s is the group size.
• P(A) is the probability of drawing at least 1 copy of a card from group s.
• P(B) is the probability of drawing all cards from group s.

This thread is basically here to help people understand how probability works. I've seen similar things in the past, but they only give direct probabilities as opposed to the actual methods used.

I'd also greatly appreciate any help, errors or additions people have to offer.

TL;DR Useful things

The probability of opening with a card that you run one copy of: 15%

The probability of opening with a card that you run two copies of: 28·077%

The probability of opening with a card that you run three copies of: 39·433%

Formula for the probability of drawing a card from s:
Quote:
http://i.imgur.com/3Ge8e.jpg

P(A) = 1 - ∏((d-s-r+1)÷(d-r+1))

n over ∏, r = 1 under ∏ and r as a subscript for ((d-s-r+1)÷(d-r+1))

n is the number of cards drawn
r is the starting number and has to be 1
d is the number of cards being ran in the deck
s is the size of the group you are trying to find
Formula for the probability of opening with all cards from s: (IN TESTING: MAY NOT WORK)
Quote:
P(B) = (n-1)! ÷ dCs

or

P(B) = [s! × (n-1)! × (d-s)!] ÷ d!
(NOTE: this formula stops working once the samples are disrupted, e.g. something like Pot of Duality or RotA is used.)

Number of opening hand combinations (assuming one copy of every card) : 3,383,380

Permutations

This is the most common way of handling probabilities in this game. It would help if I had a tree diagram to help explain this, but I guess I'll improvise. I'll throw in some P() notation for those who are inquisitive, but you don't need to know that.

First an example with a fair two-sided coin toss. We're going to throw the coin twice.

I'll give you the results first and then show how it's calculated.

2 heads: 25%
1 head and 1 tail: 50%
2 tails: 25%

This is because:

The chance of getting a head is 0·5 and the chance of getting a tail is also 0·5.

⠀⠀P(H) = 0·5
⠀⠀P(T) = 0·5

⠀⠀P(H₁∩H₂) = 0·5 × 0·5 = 0·25

∩ means that the two events are intersected. You could read P(H₁∩H₂) as "The probability of getting a head on the first throw and then getting a head on the second throw".

This basically means that the chance of getting a head and then another head is 0·25 or 25%.

The same is also true for tails: P(T₁∩T₂) = 0·5 × 0·5 = 0·25

All the possible options must add up to 1, hence

⠀⠀1-(0·25 + 0·25) = 0·5

⠀⠀∴ P(H∩T) = 0·5

You can also calculate this by doing:

⠀⠀P(H₁∩T₂) = 0·5 × 0·5 = 0·25
⠀⠀P(T₁∩H₂) = 0·5 × 0·5 = 0·25

but this takes more time than the previous calculation.

Which give you the other two possibilities on the tree diagram. Added together, they make 0·5, which is what we got through our calculations.

Now we're going to apply this working to Yu-Gi-Oh!.

Say you have 1 copy of a card in your deck and you run 40 cards. Something like Monster Reborn; or whatever you like. Calculating the chance of drawing it in an opening hand confuses a lot of people.

The probability of the first card in your hand being Monster Reborn is 1/40, or 2·5%.

The first reaction you may have is to simply Multiply it by 1/40 5 more times, but this is incorrect. The sample sizes change because we're handling something that does not replace itself like a coin does. A coin will always have a 1/2 chance of landing on heads or tails, but a deck changes for every card removed.

You also have to rule out drawing Monster Reborn twice (which is impossible because you're using 1 copy).

You could calculate every single permutation on a tree diagram and add them together to get P(A), but that is extremely long winded and hideously inefficient. Like before, the fastest way to calculate this is to calculate P(A′) (the probability of drawing 0 cards of the group) and then doing 1 - P(A′) to get P(A).

This is done by:

⠀⠀P(A) = 1 - ((39/40) × (38/39) × (37/38) × (36/37) × (35/36) × (34/35))

Which is exactly 15%.

The good thing about this method is that it shows the chance of drawing at least 1 Monster Reborn instead of simply the probability of drawing Monster Reborn. Assuming we run 2 Solemn Warning, this method shows the chance of drawing 1 SW or 2 SW.

This is done by:

⠀⠀P(A) = 1 - ((38/40) × (37/39) × (36/38) × (35/37) × (34/36) × (33/35))

Which is 28·077% to three decimal places.

And finally there's the chance of opening with a card that you're using 3 copies of.

This is done by:

⠀⠀P(A) = 1 - ((37/40) × (36/39) × (35/38) × (34/37) × (33/36) × (32/35))

Which is 39·433% to three decimal places

If you also run cards like Reinforcement of the Army, you can consider those cards to be additional copies of the card you're trying to find the probability of opening with. Also, you can adapt this to find groups of cards too.

The general form of the above calculation:

⠀⠀P(A) = 1 - (((d-s)/d) × (((d-1)-s)/(d-1)) × (((d-2)-s)/(d-2)) × (((d-3)-s)/(d-3)) × (((d-4)-s)/(d-4)) × (((d-5)-s)/(d-5)))

Which I conveniently converted to this formula:

⠀⠀P(A) = 1 - ∏((d-s-r+1)÷(d-r+1))

n over ∏, r = 1 under ∏ and r as a subscript for ((d-s-r+1)÷(d-r+1)).

n is the number of cards drawn
r is the starting number and has to be 1
d is the number of cards being ran in the deck
s is the size of the group you are trying to find

Here's a better laid out version of the working: http://i.imgur.com/hGAtz.jpg

(NOTE: this formula stops working once the samples are disrupted, e.g. something like Pot of Duality or RotA is used.)

Tutorial for my formula

Formula: http://i.imgur.com/3Ge8e.jpg

We have the formula P(A) = 1 - ∏((d-s-r+1)÷(d-r+1))

I'll explain this using a standard playing card deck. I'll go through all the variables.

We're going to find the chance of opening with at least 1 ace given that we open with 3 cards and are running a standard 52 card deck. We define this probability as P(A) as always. The A is technically irrelevant, any letter or symbol can be substituted. I chose A just because I lacked a better symbol and A is the first letter of the alphabet.

First we have n. n is the number of cards drawn, specifically n is the size of our opening hand. We're going to use a 3 card opening hand here. In Yu-Gi-Oh!, you would typically have n = 6. The n above the ∏ means that we will be repeating the bracket up to n times.

Secondly, r is the same as with ∑ notation. r is our starting number when working our way towards n. I have set the formula out so that r = 1.

d is the deck size. In a standard playing card deck, this is 52. Adjust this to Yu-Gi-Oh! by setting it to 40 (or however many cards you're using).

s is the group size, or the number of copies of a card you are using. In this case, we are using 4 aces. Adjust this to whatever number you require. For Yu-Gi-Oh!, this will be 1, 2 or 3.

The ∏ means we will be multiplying each quotient together starting at r and working our way to n.

Note: r=1, n∈ℕ, d∈ℕ, s∈ℕ, d>s, d≠0, d≥n, n≥1

Working:

: Worked Example 1

http://i.imgur.com/VYDCZ.jpg

①: Substitute the n in ∏ as well as d and s.

⠀⠀P(A) = 1 - ∏((52-4-r+1)÷(52-r+1))

3 over ∏, r = 1 under ∏ and r as a subscript for ((52-4-r+1)÷(52-r+1)).

②: Simplify.

⠀⠀P(A) = 1 - ∏((49-r)÷(53-r))

3 over ∏, r = 1 under ∏ and r as a subscript for ((49-r)÷(53-r)).

③: Convert ∏.

⠀⠀P(A) = 1 - [ ((49-1)÷(53-1))₁ × ((49-2)÷(53-2))₂ × ((49-3)÷(53-3))₃ ]

Note how r starts at 1 and increases until r=n.

④: Simplify and convert to fractions.

⠀⠀P(A) = 1 - [ 48/52 × 47/51 × 46/50 ]

Note how this is the same function as in the Yu-Gi-Oh! permutation example.

⑤: Solve.

⠀⠀P(A) = 1 - [ 0·78262 ] to five decimal places

⠀⠀P(A) = 0·21738 to five decimal places

⑥: Convert to a percentage.

⠀⠀P(A) = 21·738%

Hence the chance of opening with at least one ace is 21·738%.

: Worked example 2

http://i.imgur.com/X5WIE.jpg

Binomial Coefficients

This is something I stumbled across that works nicely here.

nCr can be used to calculate combinations.

⁴⁰C₆ = 3,383,380 opening hand combinations (assuming 1 of each card or that each card is treated as being different).

⠀⠀⁴⁰C₆ = 40!/6!(40-6)! = 3,383,380


You can also use this to find out things such as the probability of all 5 of your immediate hand being Exodia pieces.

⠀⠀1/(⁴⁰C₅) = (5/40) × (4/39) × (3/38) × (2/37) × (1/36) = 1/658,008

Which is 0·000152% to three significant figures. Note that 0·000152% ≠ 0%, so don't call your opponent a cheater if they insta-win; it can happen.

Last edited by Valafar123 : 04-27-2012 at 11:45 AM.
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Old 04-24-2012, 11:15 AM   #2
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*Rates five stars*
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Old 04-24-2012, 11:46 AM   #3
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*Rates five stars*


Who would have thought that A-Level maths has a use.
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Old 04-24-2012, 11:47 AM   #4
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You can do that without A-Level maths.
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Old 04-24-2012, 11:57 AM   #5
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God damn, Tapatalk undid the sigmas. -.-
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Old 04-24-2012, 12:02 PM   #6
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I'm gonna have to start redirecting people here when they get their probabilities wrong. Nicely laid out and easy to understand (although I am doing A Level Maths so that could have something to do with it.) Definitely useful stuff if you want to go into that much detail.

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God damn, Tapatalk undid the sigmas. -.-
And the not equal to in the last line.
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Old 04-24-2012, 01:35 PM   #7
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Originally Posted by AcselOfDarkness View Post
I'm gonna have to start redirecting people here when they get their probabilities wrong. Nicely laid out and easy to understand (although I am doing A Level Maths so that could have something to do with it.) Definitely useful stuff if you want to go into that much detail.

Additional Comment:



And the not equal to in the last line.
Thanks.

Yeah, you'd need a good understanding of maths to get this.

And it undid the intersections. -.-
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Old 04-24-2012, 01:42 PM   #8
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Yay, Valafar said I have a good understanding of Maths. And damn Tapatalk, undoing all the special characters (just seen the &there4 halfway through), luckily it's still understandable.

I think from now on when I'm in Maths I'll just work on Yu-Gi-Oh related maths, rather than the actual stuff. As much as I enjoy Maths, this is a different level of interesting (not to mention relevant to my real world scenarios )
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Old 04-24-2012, 01:44 PM   #9
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Love this. Good job here sir

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Yay, Valafar said I have a good understanding of Maths. And damn Tapatalk, undoing all the special characters (just seen the &there4 halfway through), luckily it's still understandable.

I think from now on when I'm in Maths I'll just work on Yu-Gi-Oh related maths, rather than the actual stuff. As much as I enjoy Maths, this is a different level of interesting (not to mention relevant to my real world scenarios )
On topic with this... I am actually using some of the stuff I am learning in stats to calculate some Yugioh statistics. Its fun to apply your knowledge to something you love
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Old 04-24-2012, 01:51 PM   #10
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On topic with this... I am actually using some of the stuff I am learning in stats to calculate some Yugioh statistics. Its fun to apply your knowledge to something you love
Nice. I did some Statistics at GCSE level, the teaching was rubbish and I was ill at the time so I didn't really like it. I don't do any Statistics no until next academic year, at least this gives me something to look forward to.
A friend of mine (also a Yu-Gi-Oh player) is debating whether to take Maths next year, I might show him this and see if it helps him make a decision. I expect it's a bit off putting if you don't like Maths or just don't understand.
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Old 04-24-2012, 01:58 PM   #11
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Nice. I did some Statistics at GCSE level, the teaching was rubbish and I was ill at the time so I didn't really like it. I don't do any Statistics no until next academic year, at least this gives me something to look forward to.
A friend of mine (also a Yu-Gi-Oh player) is debating whether to take Maths next year, I might show him this and see if it helps him make a decision. I expect it's a bit off putting if you don't like Maths or just don't understand.
That's the big thing with, you need to have a good teacher, or else it doesn't matter. I am lucky to have had an amazing calc teacher and stats teacher.

What math is he planning on taking?
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Old 04-24-2012, 02:04 PM   #12
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That's the big thing with, you need to have a good teacher, or else it doesn't matter. I am lucky to have had an amazing calc teacher and stats teacher.

What math is he planning on taking?
Yeah, back then the teacher was [censored] at teaching. Currently though, it isn't nearly as bad, I pick things up and understand it easily, partly because I have good teachers. Hell, they asked me to start mentoring other students in the class so I and they must be doing something right.

As far as I can tell, he was planning on taking AS Maths for a year (Core 1, Core 2 and Decision 1 Modules). I don't think he was going to go into year 2 though since it would be his third year at college.
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Old 04-24-2012, 02:11 PM   #13
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Yeah, back then the teacher was [censored] at teaching. Currently though, it isn't nearly as bad, I pick things up and understand it easily, partly because I have good teachers. Hell, they asked me to start mentoring other students in the class so I and they must be doing something right.

As far as I can tell, he was planning on taking AS Maths for a year (Core 1, Core 2 and Decision 1 Modules). I don't think he was going to go into year 2 though since it would be his third year at college.
I'm not that advanced at math :P. Though I would love to be, just a simple freshman at college.

HE should do it though if he loves math. And if it fulfills his passion for card games, then it should be obvious to do it.
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Old 04-24-2012, 02:19 PM   #14
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I'm not that advanced at math :P. Though I would love to be, just a simple freshman at college.

HE should do it though if he loves math. And if it fulfills his passion for card games, then it should be obvious to do it.
I remember back in September when I was knew at college, everything was so different from school and it was comprehensively the best time of my learning career so far. To think that was only 8 months ago and I'm nearly halfway through my entire college life.

The thing is, I don't think he loves Maths, not in the same way I do. He's more taking it as an extra subject, at the moment we are doing a huge IT course that counts as 3 A levels but I'm doing Maths as well. It's a useful subject to have, even if you don't like it and can be applied to so many basic scenarios in life (as this thread proves). If he loved it, he'd probably have taken Maths and/or Further Maths at the beginning of term. As a far as taking an extra subject goes though, Maths seems like the logical choice. The card games is just a bonus.
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Old 04-24-2012, 03:18 PM   #15
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Tell him not to do maths. ._.

I used to think I was good at maths, but A-Level is intense.
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Old 04-24-2012, 03:23 PM   #16
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Told you

I did not like the change one bit, so happy I dropped it last year.
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Old 04-24-2012, 03:34 PM   #17
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Just how your mind should work on every duel :P
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Old 04-24-2012, 09:48 PM   #18
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this is why i think yugioh should start being marketed to teenagers instead of little kids.

...

and my brain hurts now... i can understand some of it, but my forte was english, not math
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Old 04-24-2012, 10:00 PM   #19
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I... can... understand it... quite well...
Thanks Val.
*overloads and faints*
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Old 04-25-2012, 02:15 AM   #20
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Tell him not to do maths. ._.

I used to think I was good at maths, but A-Level is intense.
Fair enough, I don't find it too bad. Each to his own I guess. I'll pass that on anyway though.
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Old 04-25-2012, 03:42 AM   #21
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Found a problem with my formula. Will update OP with the working version.
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Old 04-25-2012, 03:59 AM   #22
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You better. Now I'm on my main PC I can read it all.
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Old 04-25-2012, 07:45 AM   #23
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The math itself doesn't look to intense. It's just written in text form instead of symbols so it looks more complicated.

Oh the Internet... Why do you hate my math?
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Old 04-25-2012, 09:45 AM   #24
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All fixed. I'll add a tutorial for the formula.

Additional Comment:

Some colour coding for good measure.

Last edited by Valafar123 : 04-25-2012 at 09:45 AM. Reason: Automerged Doublepost
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Old 04-25-2012, 11:57 AM   #25
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http://i.imgur.com/RFFic.jpg?1

That is supposed to be a summation symbol, right?
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