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Offline LeodipTopic starter

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Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092566#msg1092566
« Reply #12 on: August 14, 2013, 05:11:08 pm »
To be edited, I'll edit this post when I'll be back on my pc and have some time, in the while time, you can continue here the discussion from the Studies and Statistics Board's general discussion topic.

Offline rob77dp

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092572#msg1092572
« Reply #13 on: August 14, 2013, 05:16:11 pm »
Mis-understandings are happening on all fronts here, I think.

Allow me to re-request:  please give an example for solving and it will hopefully show our disconnects in communication.  Numbers and formulas are so much less abstract than typing the English language.
:)
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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092573#msg1092573
« Reply #14 on: August 14, 2013, 05:19:01 pm »
Mis-understandings are happening on all fronts here, I think.

Allow me to re-request:  please give an example for solving and it will hopefully show our disconnects in communication.  Numbers and formulas are so much less abstract than typing the English language.
:)
That was the trouble, I don't know the formula (or specifically, I know one, but it's too long to write via cellphone and not too customizable). The situation was that, but I don't know the formula itself.

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092577#msg1092577
« Reply #15 on: August 14, 2013, 05:22:56 pm »
Turns out it made the earliest post the OP instead of the new OP. Oh well, still Leodip's so he can still edit important information into it.

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092578#msg1092578
« Reply #16 on: August 14, 2013, 05:31:16 pm »
Turns out it made the earliest post the OP instead of the new OP. Oh well, still Leodip's so he can still edit important information into it.
Thanks.

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092587#msg1092587
« Reply #17 on: August 14, 2013, 06:32:29 pm »
Hmmmm, I'll do my own example:

Ready Pharaoh test-only
Hover over cards for details, click for permalink
Deck import code : [Select]
71d 71d 71d 71d 71d 71d 72i 72i 72i 72i 72i 72i 72i 72i 72i 72i 72i 72i 7q4 7q4 7q4 7q4 7q4 7q4 7qu 7qu 7qu 7qu 7qu 7qu 8ps

*One-game trial against AI3 to test functionality only generated 6-turn 70 HP victory playing first against the Steam Machine AI3!*

-snip- (Removed - left-over statement from a draft prior to final post.)

Order to accomplish this...
1. Mummy - 6 in deck needing to draw 1 after generating minimum 3 Death quanta.
2. Rewind - 6 in deck needing to draw 1 after playing minimum 1 Mummy.
3. SoR - 6 in deck needing to draw 1 after making first Pharaoh (note: this deck can pump out Scarabs even if no SoR is had but for sake of calc's let's require SoR'ing for the combo).

Variables:
Deck size:  30 cards
Mummy x6
Rewind x6
SoR x6

I want speed (and keeps the example short-ish) so I will presume going-first (seven card opening hand) wanting a 2nd turn Pharaoh and 3rd turn Scarab-making Readied-Pharaoh.

Turn 1:  Get 3 :death... Here, x2 minimum Death Pend's first turn (4 :death) which is {30|12|7|2;>=2} = 0.8749 no-mulligan which goes to 0.8749 * (1 + {30|12|7|1,<1}) --> 0.8749 * (1 + 0.0156) = 0.8885.

Turn 2:  Need at least one Mummy in-hand on 2nd turn which is {30|6|8|1;>=1} = 0.8743 (no mulligan effect included here... this is possibly discuss-able as to if and then how to do it).  Need at least one Rewind in-hand on 2nd turn (the 1 :time is 'guaranteed' by the Mark) which is the same as a Mummy = 0.8743.

Turn 3:  Need at least one SoR in-hand (Mark + minimum x2 Pend's from Turn 1 calc 'assure' the 3 :time required to summon) which is {30|6|9|1;>=1} = 0.9086.

Putting it all together is multiplication (as I see it) for 0.8885 * 0.8743 * 0.8743 * 0.9086 = 61.71%.  Three draws out of five, when playing first with the above deck, you will have a Readied-Pharaoh pumping out super-cool Scarabs by the end of your third turn.

Notes:
1. One could add in steps to figure the odds to pump out THREE Scarabs instead of two on the third turn by adding in chances for a 3rd Pend on-or-before the 2nd turn (8 cards in for this example) to get the extra time needed.

2. This deck is not "ideal" in design towards the goal in my calculation.  It is quite suited to a solid example using a _functioning_ deck.
« Last Edit: August 14, 2013, 06:40:52 pm by rob77dp »
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Offline LeodipTopic starter

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092874#msg1092874
« Reply #18 on: August 16, 2013, 09:37:31 am »
@Rob, that wasn't what I was trying to accomplish, but still, it's a pretty good study.

Wall of text ahead:
Spoiler for Hidden:
Ok, here's my version of the whole thing:
Considering your deck up here, let's say I want to calculate the probabilities of drawing AT LEAST 1 of each Mummy, SoR and Rewind, calling them, respectively M, S and R (yeah, they follow my standard names, too, lucky ^^).
How do we do this? First of all, the probabilities of something happening are FavorableSituations/TotalSituations (P=C'/C). Let's start calculating the Total Situations. To do so the formula is pretty easy and is like d!/((d-n)!*n!) where "d" is the number of cards in your deck and "n" the number of cards you draw (7, we're assuming we're always going first). Calculating this you obtain 2035800. For who may be asking, d! calculates the various combinations your deck may have. Dividing it by (d-n)! will let you calculate the various combinations your hand could have, where the order of the card matters. Dividing it back for n! makes it so that the order doesn't matter.
Now that we have C, let's go to C'. In case that's just one card, there are two ways which I like to call Reverse and Specific:
-Reverse is the clearest one, simply, you calculate the probabilities of that NOT happening, and then you invert them. Basically, you do 1-Z/C, where Z is the number of combinations in which you don't get the card you need. For one card (M), Z is simply (d-m)!/((d-n-m)!*n!), where "m" is the number of copies you have of M. This happens because you are calculating the combinations the deck may have without M. Doing so you'll obtain 346104. Calculating the whole 1-Z/C you obtain 0,8299911583, so about 83%, pretty giant, uh?
For one card, Reverse is probably the best you can use, but for more than one card, it isn't optimal. However, after researches, I found a way to calculate the probabilities of two cards keeping Reverse: to do so, you'll have to get C', the number of situations without neither M nor S. As you've seen, it is simply (d-m-s)!/((d-m-s-n)!*n!) with 31824 as result. With T-T' you obtain MS, the number of combinations with either M or S or even both. In this case the number is 2003976. Now simply calculate the number of hands without S like (d-s)!/((d-n-s)!*n!)), which is 346104. Now MS-S' will get you M', the number of hands with M but without S. M-M' will get you ms, which is the number of combinations in which you have both M and S.
However, with more than two card, this is not  possible unless another way is found.
-Specific is the longest one which requires a basic understanding of the formula to modify and adapt it to other situations, but can let you calculate pretty much all you may need. As always, the result P is given by C'/C. While C is calculated as before, what changes is C'. Using Specific, you'll have to manually calculate the situations in which you have exactly one M, two M, three M, 4 M, 5 M and 6 M and add them up. You do this via m(d-m)(d-m-1)(d-m-2)(d-m-3)(d-m-4)(d-m-5)+m(m-1)(d-m)(d-m-1)(d-m-2)(d-m-3)(d-m-4)+m(m-1)(m-2)(d-m)(d-m-1)(d-m-2)(d-m-3)+m(m-1)(m-2)(m-3)(d-m)(d-m-1)(d-m-2)+m(m-1)(m-2)(m-3)(m-4)(d-m)(d-m-1)+m(m-1)(m-2)(m-3)(m-4)(m-5)(d-m).
That's pretty long, isn't it? And consider that if it were 8 cards, you'd have to change alot once again, and what if you wanted to calculate SoBra? Deadly, you'd have to change pretty much everything. The good thing is that it probably isn't too hard to calculate more than one card, but I still am not sure on how to do that. For two cards, you write the formula above and subtract to it the same formula but substituting all of the "d" with "d-s".
If you try to do so with 3 cards, what you'll get is the number hands that have intersections between the cards you're studing. I have figured out another way, but that's even harder. You calculate MSR like we did before (calculate Z and do C-Z). Now you'll have to calculate MSR', which is the number of hands that have M, S and/or R, but never together, and then subtract it to MSR. To do so, you'll have to calculate all of the intersections like we did before with Reverse (you can do that with Specific, too, if you want to do so). You'll basically obtain the intersections between M and R, M and S and R and S. However, you've calculate MSR among those, and you did so 3 times. What we'll do then is calculate M, S and R singularly, sum them up and then subtract the intersections to it. Explaining what happened is pretty difficult, basically consider 3 circles, M, S and R. When they overlap, they create 3 spaces with just one layer (M', S' and R'), 3 spaces with two layers (M'S, M'R, R'S) and 1 space with three layers (C'). If you subtract the three intersections, you'll remove one layer from each of the spaces with two layers, and will remove all of the three layers from C'. You'll obtain MSR', the number of hands with M, R and/or S, but never with them all together. Now MSR-MSR' and you'll obtain.
I don't have much time to check this last one, so if someone were so kind to do so for me, I'd be more than happy. Compare your result with this, you should obtain 53.65%, not considering Mulligan, which will be implemented when we find the final efficient formula.

If I'm not wrong, what I did with the Specific, can be done with Reverse, too.

Let me know.

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092922#msg1092922
« Reply #19 on: August 16, 2013, 04:31:19 pm »
:D
Spoiler for Rob's wall of text reply:
@Rob, that wasn't what I was trying to accomplish, but still, it's a pretty good study.

Wall of text ahead:
Spoiler for Hidden:
Ok, here's my version of the whole thing:
Considering your deck up here, let's say I want to calculate the probabilities of drawing AT LEAST 1 of each Mummy, SoR and Rewind, calling them, respectively M, S and R (yeah, they follow my standard names, too, lucky ^^).
How do we do this? First of all, the probabilities of something happening are FavorableSituations/TotalSituations (P=C'/C). Let's start calculating the Total Situations. To do so the formula is pretty easy and is like d!/((d-n)!*n!) where "d" is the number of cards in your deck and "n" the number of cards you draw (7, we're assuming we're always going first). Calculating this you obtain 2035800. For who may be asking, d! calculates the various combinations your deck may have. Dividing it by (d-n)! will let you calculate the various combinations your hand could have, where the order of the card matters. Dividing it back for n! makes it so that the order doesn't matter.
Now that we have C, let's go to C'. In case that's just one card, there are two ways which I like to call Reverse and Specific:
-Reverse is the clearest one, simply, you calculate the probabilities of that NOT happening, and then you invert them. Basically, you do 1-Z/C, where Z is the number of combinations in which you don't get the card you need. For one card (M), Z is simply (d-m)!/((d-n-m)!*n!), where "m" is the number of copies you have of M. This happens because you are calculating the combinations the deck may have without M. Doing so you'll obtain 346104. Calculating the whole 1-Z/C you obtain 0,8299911583, so about 83%, pretty giant, uh?
For one card, Reverse is probably the best you can use, but for more than one card, it isn't optimal. However, after researches, I found a way to calculate the probabilities of two cards keeping Reverse: to do so, you'll have to get C', the number of situations without neither M nor S. As you've seen, it is simply (d-m-s)!/((d-m-s-n)!*n!) with 31824 as result. With T-T' you obtain MS, the number of combinations with either M or S or even both. In this case the number is 2003976. Now simply calculate the number of hands without S like (d-s)!/((d-n-s)!*n!)), which is 346104. Now MS-S' will get you M', the number of hands with M but without S. M-M' will get you ms, which is the number of combinations in which you have both M and S.
However, with more than two card, this is not  possible unless another way is found.
-Specific is the longest one which requires a basic understanding of the formula to modify and adapt it to other situations, but can let you calculate pretty much all you may need. As always, the result P is given by C'/C. While C is calculated as before, what changes is C'. Using Specific, you'll have to manually calculate the situations in which you have exactly one M, two M, three M, 4 M, 5 M and 6 M and add them up. You do this via m(d-m)(d-m-1)(d-m-2)(d-m-3)(d-m-4)(d-m-5)+m(m-1)(d-m)(d-m-1)(d-m-2)(d-m-3)(d-m-4)+m(m-1)(m-2)(d-m)(d-m-1)(d-m-2)(d-m-3)+m(m-1)(m-2)(m-3)(d-m)(d-m-1)(d-m-2)+m(m-1)(m-2)(m-3)(m-4)(d-m)(d-m-1)+m(m-1)(m-2)(m-3)(m-4)(m-5)(d-m).
That's pretty long, isn't it? And consider that if it were 8 cards, you'd have to change alot once again, and what if you wanted to calculate SoBra? Deadly, you'd have to change pretty much everything. The good thing is that it probably isn't too hard to calculate more than one card, but I still am not sure on how to do that. For two cards, you write the formula above and subtract to it the same formula but substituting all of the "d" with "d-s".
If you try to do so with 3 cards, what you'll get is the number hands that have intersections between the cards you're studing. I have figured out another way, but that's even harder. You calculate MSR like we did before (calculate Z and do C-Z). Now you'll have to calculate MSR', which is the number of hands that have M, S and/or R, but never together, and then subtract it to MSR. To do so, you'll have to calculate all of the intersections like we did before with Reverse (you can do that with Specific, too, if you want to do so). You'll basically obtain the intersections between M and R, M and S and R and S. However, you've calculate MSR among those, and you did so 3 times. What we'll do then is calculate M, S and R singularly, sum them up and then subtract the intersections to it. Explaining what happened is pretty difficult, basically consider 3 circles, M, S and R. When they overlap, they create 3 spaces with just one layer (M', S' and R'), 3 spaces with two layers (M'S, M'R, R'S) and 1 space with three layers (C'). If you subtract the three intersections, you'll remove one layer from each of the spaces with two layers, and will remove all of the three layers from C'. You'll obtain MSR', the number of hands with M, R and/or S, but never with them all together. Now MSR-MSR' and you'll obtain.
I don't have much time to check this last one, so if someone were so kind to do so for me, I'd be more than happy. Compare your result with this, you should obtain 53.65%, not considering Mulligan, which will be implemented when we find the final efficient formula.

If I'm not wrong, what I did with the Specific, can be done with Reverse, too.

Let me know.

The formula you are using seems quite similar to the one used on the Hypergeometric calc.  This calc also accounts for draw-chance changes "real time" (you get one you need, there is one fewer to "get" next draw simulation etc.).

Hypergeometric Distrubution
Hypergeometric Distrubution]
Hypergeometric Formula. Suppose a population consists of N items, k of which are successes. And a random sample drawn from that population consists of n items, x of which are successes. Then the hypergeometric probability is:
h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ]

Combination
Computing the number of combinations. [nCr]The number of Combinations of n objects taken r at a time is
nCr = n! / r!(n - r)!

Note:  I have been using {x|N|n|k} notation with x = deck size, N = card count in deck you are wanting to 'draw', n = cards drawn, and k = number of copies drawn of the card you are wanting to draw.

Final Forumla:  h{x|N|n|k} = [k! / (x! * [k - x]!)] * (N - [k! / (x! * [k - x]!)] - x) / [N! / (n! * [N - n]!)]

For figuring card-combo draw chances from that one simply needs to apply the formula with logic in a series of calculations to arrive at a final "big picture" result.  The YuGiOh calculator agrees with the hypergeometric calculator results on all simple calc's - such as drawing 1 Rewind in first 9 cards or drawing 2 Pend's in first 7 cards from the sample deck we're using here.

Spoiler for Screenshots showing similarity of results between YuGiOh and Hypergeometric calculators:


YIELDS



which matches a similar input to the Hypergeo Calc like this:



This leads me to conclude that the differences in our results lie in some or all of the following:

1. YuGiOh calculator does not do mulligans the same as one can "manually" with the hypergeo calc to simulate Elements' drawing.
2. YGO calc has some other "behind the scenes" methods based on game-mechanics/rules I am unaware of and that do not match Elements'.  (I have not played and do no know how to play YuGiOh... ...)
3. Differences (not sure at this stage which side, if not both) in calculations that are wrong or mis-applied formulas or bad math.

Further, I get your 53.65% chance on the YGO calc with input indicating drawing >= 1 SoR, >= 1 Mummy, and >= 1 Rewing from our sample deck in the opening hand.  This slightly disagrees with the hypergeo calc of h{30|6|7|1} ^ 3 = 57.18% (which I had hoped would match the YGO result).  They are close so I am guessing the YGO has some behind the scenes extras accounting for the few cases of drawing 6 SoR and 1 Rewind (or some similar outlier type draw) which I think the simplified hypergeo shown does not do correctly.

TL;DR - We are not quite yet matching up but I have an idea how to proceed.  In the near future I hope to post again on this with more helpful discussion and information.
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Offline LeodipTopic starter

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1092931#msg1092931
« Reply #20 on: August 16, 2013, 05:08:25 pm »
Let's start saying I find it pretty hard to use that notation, so, sorry if I don't use it.
Going to the point, no, I don't think the two games have differences on that, only the Mulligan, but I believe none of the programs take that into account, and the only differences between the two programs themselves are:
1.DBToolkit always rounds off the number to the second decimal cipher (wrote that right? lol)
2.While the Hypergeometric calculator gives off the result of the Specific formula I wrote ignoring the numbers lower than the number you've specified, the Toolkit is more user-friendly, not making you work with numbers you don't know the meaning.

Probably there's simply something wrong in the inputs, and not the outputs, because I believe both are pretty professionals, maybe you simply got confused with all those flying numbers, I do that a lot.

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1093517#msg1093517
« Reply #21 on: August 20, 2013, 03:56:19 pm »
I think Xeno beat us to creating the opening hands and draws calculator, Leo!

Read here and try it for yourself.
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Offline LeodipTopic starter

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1093543#msg1093543
« Reply #22 on: August 20, 2013, 05:42:07 pm »
Thanks Rob, but, yeah, I knew that already.
What I wanted was finding a way to mathematically calculate them. Not to say that Xeno's software is bad (instead, it's is surely the best for what it does, even though it calculates only single cards and not combinations), but I wanted to calculate it mathematically instead of empirically.
What's more, I wrote the way to calculate it, no matter the number of cards you want to imply into the formula, but I just can't find the way to simplify it to a formula that only requires the number of copies of the cards you want and the cards themselves and the number of cards in the deck alongside the number of draws. I'm just one step behind it, but that's one big step.

Spoiler for Hidden:
BTW, did the name of the subforum change? I like this one better, because it is more generic, way to go.

Offline ColorlessGreen

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Re: Probabilities and Combinations https://elementscommunity.org/forum/index.php?topic=50769.msg1093557#msg1093557
« Reply #23 on: August 20, 2013, 06:24:05 pm »
BTW, did the name of the subforum change? I like this one better, because it is more generic, way to go.

Yup. Higurashi and I have been working to clarify the exact purpose of this board, starting with that modification of the title. More information will come.

 

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