Probabilities and Combinations

[Leodip]

What this topic is about: Probabilities of drawing something
Our most usual example is a deck of "d" cards, drawing "h" cards, with "m" copies of card M, "s" copies of card S and "r" copies of card R. What we study is what the probability of drawing M, S AND R, with and without mulligan (for mulligan, we'll consider "p" zero-cost cards).
Given that for any proposed problem there may be different solutions, I'll be listing the various ways proposed in as-ordered-as-possible paragraphs.

The Leodip Way! (mulligan-less)
The first spoiler is a quote, the first time I explained how 3 cards combine. The second one contains the final formula. I may as well make a spreadsheet, but I'll need a little time because I have yet to include mulligan. Next thing to be done.

Spoiler for Quote:

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 msr.
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.

Spoiler for Formula:

Given that it was way easier, I finally decided to use reverse, in order to make the formula more "elastic" and efficient. The ending formula goes like:
P=(((d-m-s-r)!/((d-m-s-r-h)!*h!))-((d-m)!/((d-m-h)!*h!))-((d-s)!/((d-s-h)!*h!))-((d-r)!/((d-r-h)!*h!))+(d!/((d-h)!*h!))+((d-m)!/((d-m-h)!*h!))-((d-s)!/((d-s-h)!*h!))-(((d-m-s)!/(d-m-s-h)!*h!))+((d!/((d-h)!*h!)))+((d-m)!/((d-m-h)!*h!))-((d-r)!/((d-r-h)!*h!))-((d-m-r)!/((d-m-r-h)!*h!))+(d!/((d-h)!*h!))+((d-s)!/((d-s-h)!*h!))-((d-r)!/((d-r-h)!*h!))-((d-s-r)!/((d-s-r-h)!*h!)))/(d!/((d-h)!*h!))

If you want to test some way you've thought of, you may try using this to check your result. It is a YGO software, so it doesn't account for mulligan, but for mulligan-less systems this is good enough.

Suggestions accepted.

Spoiler for Hidden:

For those who didn't follow the creation of the topic, it may be a little bit confusing. What actually happened is that this was initially a discussion in the topic Ask a simple question etc..., but, according to CG, it needed its own topic, so he transferred the posts and stuff from that topic into a new one. You'll, indeed, see lots of inexistent quotes and persons speaking about something that doesn't exist. That's perfectly normal. It isn't if someone talks about UFOs or such, but that's kinda OT.

[ColorlessGreen]

Ok, then, let's start asking:
Forgetting mulligan, which will be implemented later in the formula, I know how to calculate the probabilities of getting a specific card in at least a copy in your starting hand (first turn) of 7 cards, by using "d" as number of cards in the deck and "n" as number of cards in the starting hand and "k" as the number of copies you have of the card you need, "T" as the total number of combinations of possible hands, and "c" as the number of hands with that card we need and "P" as the probabilities of having a "good" hand we have a formula like this: T=d!/n!(d-n)!; c=1-((d-k)!/n!(d-k-n)!); P=c/T
After lots of self-studying sessions, I understood how to do this with two card: if you calculate the number of possible hands with neither card1 (M) nor card2 (S), you'll have T' (T'=(d-M-S)!/n!(d-M-S-n)!). Inverting this (1-T') you'll obtain MS, the number of hands with either M or S or even both. Now, if you calculate the number of hands without S (S=(d-S)!/n!(d-S-n)!) and subtract it to MS you'll obtain M', which is the number of hands without S but with M. Now you'll simply have to calculate MS-M' and you'll obtain P.

However, I couldn't get it to work with more than 2 cards, but I definitely know that's possible.

Sadly that's not one I can answer off the top of my head. I'll think about it and if the answer comes to me, I'll let you know, but in the mean time I hope someone who's much better at the more complex probabilities comes along with the answer.

edit: Honestly if I were trying to answer this question for myself, rather than sitting down and trying to work out the exact formulas for an arbitrary number of specific cards in a starting hand, I'd probably just write up some little simulator that shuffles a given deck a hundred thousand times or something like that and increments a counter if all X cards are in the top seven. But I like making computers do the statistics work via simulations a lot more than I like trying to puzzle my way through the really complex formulae.

[Leodip]

Ok, then, let's start asking:
Forgetting mulligan, which will be implemented later in the formula, I know how to calculate the probabilities of getting a specific card in at least a copy in your starting hand (first turn) of 7 cards, by using "d" as number of cards in the deck and "n" as number of cards in the starting hand and "k" as the number of copies you have of the card you need, "T" as the total number of combinations of possible hands, and "c" as the number of hands with that card we need and "P" as the probabilities of having a "good" hand we have a formula like this: T=d!/n!(d-n)!; c=1-((d-k)!/n!(d-k-n)!); P=c/T
After lots of self-studying sessions, I understood how to do this with two card: if you calculate the number of possible hands with neither card1 (M) nor card2 (S), you'll have T' (T'=(d-M-S)!/n!(d-M-S-n)!). Inverting this (1-T') you'll obtain MS, the number of hands with either M or S or even both. Now, if you calculate the number of hands without S (S=(d-S)!/n!(d-S-n)!) and subtract it to MS you'll obtain M', which is the number of hands without S but with M. Now you'll simply have to calculate MS-M' and you'll obtain P.

However, I couldn't get it to work with more than 2 cards, but I definitely know that's possible.

Sadly that's not one I can answer off the top of my head. I'll think about it and if the answer comes to me, I'll let you know, but in the mean time I hope someone who's much better at the more complex probabilities comes along with the answer.

edit: Honestly if I were trying to answer this question for myself, rather than sitting down and trying to work out the exact formulas for an arbitrary number of specific cards in a starting hand, I'd probably just write up some little simulator that shuffles a given deck a hundred thousand times or something like that and increments a counter if all X cards are in the top seven. But I like making computers do the statistics work via simulations a lot more than I like trying to puzzle my way through the really complex formulae.

Maybe that's me, but I simply prefer mathematic formulae rather than empirical testing, even because I'd like to transpose this to other games and on my general studies of statistics.
Oh, well, I think I'll try again. I'll let you know.

[ColorlessGreen]

Ok, then, let's start asking:
Forgetting mulligan, which will be implemented later in the formula, I know how to calculate the probabilities of getting a specific card in at least a copy in your starting hand (first turn) of 7 cards, by using "d" as number of cards in the deck and "n" as number of cards in the starting hand and "k" as the number of copies you have of the card you need, "T" as the total number of combinations of possible hands, and "c" as the number of hands with that card we need and "P" as the probabilities of having a "good" hand we have a formula like this: T=d!/n!(d-n)!; c=1-((d-k)!/n!(d-k-n)!); P=c/T
After lots of self-studying sessions, I understood how to do this with two card: if you calculate the number of possible hands with neither card1 (M) nor card2 (S), you'll have T' (T'=(d-M-S)!/n!(d-M-S-n)!). Inverting this (1-T') you'll obtain MS, the number of hands with either M or S or even both. Now, if you calculate the number of hands without S (S=(d-S)!/n!(d-S-n)!) and subtract it to MS you'll obtain M', which is the number of hands without S but with M. Now you'll simply have to calculate MS-M' and you'll obtain P.

However, I couldn't get it to work with more than 2 cards, but I definitely know that's possible.

Sadly that's not one I can answer off the top of my head. I'll think about it and if the answer comes to me, I'll let you know, but in the mean time I hope someone who's much better at the more complex probabilities comes along with the answer.

edit: Honestly if I were trying to answer this question for myself, rather than sitting down and trying to work out the exact formulas for an arbitrary number of specific cards in a starting hand, I'd probably just write up some little simulator that shuffles a given deck a hundred thousand times or something like that and increments a counter if all X cards are in the top seven. But I like making computers do the statistics work via simulations a lot more than I like trying to puzzle my way through the really complex formulae.

Maybe that's me, but I simply prefer mathematic formulae rather than empirical testing, even because I'd like to transpose this to other games and on my general studies of statistics.
Oh, well, I think I'll try again. I'll let you know.

If you arrive at an answer, please do post it in here. If you acquire a formula that can be scaled to an arbitrary number of cards or otherwise improves on the 1-2 card variant that had been posted in the past, I would say it deserves its own thread on this board.

I'm sure there is an answer, and if we're lucky there's someone here who's more up to date on complex probabilities than I am who can help you put it together.

[Leodip]

Yeah, thanks. BTW, I think I got to the point, but I'm from my phone. As soon as I have access to my pc (the day after tomorrow), I'll turn it into a formula and see if it works. Hopefully, it should work for all the 3+ combination of cards.

[rob77dp]

Ok, then, let's start asking:
Forgetting mulligan, which will be implemented later in the formula, I know how to calculate the probabilities of getting a specific card in at least a copy in your starting hand (first turn) of 7 cards, by using "d" as number of cards in the deck and "n" as number of cards in the starting hand and "k" as the number of copies you have of the card you need, "T" as the total number of combinations of possible hands, and "c" as the number of hands with that card we need and "P" as the probabilities of having a "good" hand we have a formula like this: T=d!/n!(d-n)!; c=1-((d-k)!/n!(d-k-n)!); P=c/T
After lots of self-studying sessions, I understood how to do this with two card: if you calculate the number of possible hands with neither card1 (M) nor card2 (S), you'll have T' (T'=(d-M-S)!/n!(d-M-S-n)!). Inverting this (1-T') you'll obtain MS, the number of hands with either M or S or even both. Now, if you calculate the number of hands without S (S=(d-S)!/n!(d-S-n)!) and subtract it to MS you'll obtain M', which is the number of hands without S but with M. Now you'll simply have to calculate MS-M' and you'll obtain P.

However, I couldn't get it to work with more than 2 cards, but I definitely know that's possible.

I use a Hypergeometric Distributino calculation to determine draw probabilities when I'm building decks or wanting to figure the "odds of THAT happening" in a game with a lucky/un-lucky draw.

I use this online calc tool: Hypergeometric Calculator
- Population size = Deck size you are drawing from
- Number of successes in population = quantity in draw-stack of the card you are finding the draw chance for
- Sample size = how many cards to be drawn for this calculation
- Number of successes in sample (x) = quantity you want to draw or calc the odds of drawing
+ P(X = x) --> probability of drawing exactly 'x' copies
+ P(X < x) --> probability of drawing fewer than 'x' copies
+ P(X <= x) --> probability of drawing no more than 'x' copies
+ P(X > x) --> probability of drawing more than 'x' copies
+ P(X >= x) --> probability of drawing at least 'x' copies

For mulligan usage, here is an example of how I do it:

30 card deck
6 Nova with 24 non-zero summon cost cards

30 | 6 | 8 | 1 for the text-box values on the calculator...  '>= x' indicates 87.43% chance I draw at least one Nova in going-second scenario.  Another way to state this value is 100 - 87.43 = chance to have all non-zero cost cards resulting in ONE mulligan (re-draw) on which the chance is 100 - 87.43 I get all non-zero again.  Hence, the chance to have an eight-card (going second) hand with zero Nova is (1 - 0.8743)^2 = 0.0158 (or 1.58%) including mulligan.

I have been using this calculator and "statistic strategy" in applying it to Elements draw chances for almost a year now.  Hopefully it is accurate and my method correct.  (I could give more examples, but it is probably best for each to try it out and post questions I can field and answer on a case-by-case basis.)

...
OK, one more basic example: 30 card deck with 12 pill/pend and going-first presumption wanting 3 or more pillars.
30 | 12 | 7 | 3
which yields 59.7% chance of occurence.  To include mulligan (usually I don't when a deck is 12/30 zero-cost cards) simply change the '3' to '1' to determine how often mulligan gets triggered.  This is 1.56% of the first-draws will give a mulligan.  From here, 59.7% of those 1.56% re-draws will give at least 3 pill/pend's.  Result is 0.597 * (1 + 0.0156) = 60.6%.  If your deck design and play-style will be fine with having at least 3 pill/pend draws ~60% of the time opening first, then your deck is good on that front as 12 out of 30.  (This is not to preclude QI balancing and considering play styles and other factors.)

[Leodip]

I already knew that, only that it doesn't work for more than 1 card. For more than one card I use this, and, while it works awesomely, I wanted to learn the process behind it. If I were to post the topic CG mentioned (and I will, as soon as possible),  I would use this and then add the mathematical process once I figure that out.

[rob77dp]

I already knew that, only that it doesn't work for more than 1 card. For more than one card I use this, and, while it works awesomely, I wanted to learn the process behind it. If I were to post the topic CG mentioned (and I will, as soon as possible),  I would use this and then add the mathematical process once I figure that out.

For drawing combo's, just do the hypergeo-calc for each card in the hand (doesn't HAVE to be 7 or 8 either - if by "2nd turn" I go with 9 drawn, for example) and then combine 'em up!

Combo needs card A + B + C.

Odds:  A - 45%, B - 35%, C - 35%.  Combo all in-hand is .45 * .35 * .35 = 5.5%.

I used this one hypergeometric calculator to do all the numbers on my (pre-1.32 SoR nerf) Ready FoPhar deck for that combo-draw.  Your YuGiOh calc link seems a bit time consuming (perhaps no more than several calc's then combined together like I suggest above) if you have to type it all out.  Then again, maybe this is simply a result abiding by more-than-one-path-to-success.

[ColorlessGreen]

I already knew that, only that it doesn't work for more than 1 card. For more than one card I use this, and, while it works awesomely, I wanted to learn the process behind it. If I were to post the topic CG mentioned (and I will, as soon as possible),  I would use this and then add the mathematical process once I figure that out.

For drawing combo's, just do the hypergeo-calc for each card in the hand (doesn't HAVE to be 7 or 8 either - if by "2nd turn" I go with 9 drawn, for example) and then combine 'em up!

Combo needs card A + B + C.

Odds:  A - 45%, B - 35%, C - 35%.  Combo all in-hand is .45 * .35 * .35 = 5.5%.

I used this one hypergeometric calculator to do all the numbers on my (pre-1.32 SoR nerf) Ready FoPhar deck for that combo-draw.  Your YuGiOh calc link seems a bit time consuming (perhaps no more than several calc's then combined together like I suggest above) if you have to type it all out.  Then again, maybe this is simply a result abiding by more-than-one-path-to-success.

The reason I didn't answer A*B*C to the original question (even though I'd typed out a response saying to do that) is that it isn't exactly A*B*C, since each of A, B, and C are calculated without considering that by the time you have drawn A, you have less cards left in the deck from which to draw B and C.

The result from A*B*C is in all likelihood extremely close to the actual answer, to the point of being functionally identical for real-world purposes, but I do not believe it is actually 100% correct.

Again, though, I'm not super up to speed on complex probabilities, so everything I just said could be wrong.

edit: To clarify, each of A, B, and C is the probability that the card will appear in the first seven cards drawn from a (let's say) 30 card deck, when what we want is the probability that A will appear in the first seven cards, multiplied by the probability that B will appear in the remaining six, multipllied by the probability that C will appear in the remaining five.

Coming up with that clarification may have pointed me at an answer, though. If we just used the calc to calc A(7,30), then B(6,29), then C(5,28), then multiply the results together, we might have the correct answer. Or I may have failed to account for something.

At any rate, this has moved beyond the "quick question" realm. Leodip, please make a thread for this topic at your earliest convenience and I'll move this discussion into that thread, even if we haven't yet arrived at an answer (you can edit the answer into the OP of the thread when one is found).

[rob77dp]

Spoiler for quotes:

I already knew that, only that it doesn't work for more than 1 card. For more than one card I use this, and, while it works awesomely, I wanted to learn the process behind it. If I were to post the topic CG mentioned (and I will, as soon as possible),  I would use this and then add the mathematical process once I figure that out.

For drawing combo's, just do the hypergeo-calc for each card in the hand (doesn't HAVE to be 7 or 8 either - if by "2nd turn" I go with 9 drawn, for example) and then combine 'em up!

Combo needs card A + B + C.

Odds:  A - 45%, B - 35%, C - 35%.  Combo all in-hand is .45 * .35 * .35 = 5.5%.

I used this one hypergeometric calculator to do all the numbers on my (pre-1.32 SoR nerf) Ready FoPhar deck for that combo-draw.  Your YuGiOh calc link seems a bit time consuming (perhaps no more than several calc's then combined together like I suggest above) if you have to type it all out.  Then again, maybe this is simply a result abiding by more-than-one-path-to-success.

The reason I didn't answer A*B*C to the original question (even though I'd typed out a response saying to do that) is that it isn't exactly A*B*C, since each of A, B, and C are calculated without considering that by the time you have drawn A, you have less cards left in the deck from which to draw B and C.

The result from A*B*C is in all likelihood extremely close to the actual answer, to the point of being functionally identical for real-world purposes, but I do not believe it is actually 100% correct.

Again, though, I'm not super up to speed on complex probabilities, so everything I just said could be wrong.

Leodip, perhaps you could give an example for which you are attempting to make a calculation of draw chances.  I'll take a crack at solving it and post here and you and CG can critique.  Together we can achieve more! 

[Leodip]

I already knew that, only that it doesn't work for more than 1 card. For more than one card I use this, and, while it works awesomely, I wanted to learn the process behind it. If I were to post the topic CG mentioned (and I will, as soon as possible),  I would use this and then add the mathematical process once I figure that out.

For drawing combo's, just do the hypergeo-calc for each card in the hand (doesn't HAVE to be 7 or 8 either - if by "2nd turn" I go with 9 drawn, for example) and then combine 'em up!

Combo needs card A + B + C.

Odds:  A - 45%, B - 35%, C - 35%.  Combo all in-hand is .45 * .35 * .35 = 5.5%.

I used this one hypergeometric calculator to do all the numbers on my (pre-1.32 SoR nerf) Ready FoPhar deck for that combo-draw.  Your YuGiOh calc link seems a bit time consuming (perhaps no more than several calc's then combined together like I suggest above) if you have to type it all out.  Then again, maybe this is simply a result abiding by more-than-one-path-to-success.

Wheep, that's wrong. What you calculated by combining them is basically the chances of drawing in one hand card A, reshuffling and then probabilites of b, reshuffle and then C. Want a proof? Try to calculate the probabilities with the hypergeometric calculator of drawing 8 cards with exactly 6 copies of a given card and other 3 of another card, then combine them. It'll give you a number that's over the 0%, but that's wrong, because you can't possibly draw 9 cards over 8 draws. The YGO Calc actually works while being even pretty easy to use and efficient, but I wanted to calculate the formula by myself. As of now, I found a way to do so, but it is pretty hard (not easy) to customize. I'll eventually write this all.

@rob, ok, let's just say that I have 6 copies of M, 6 copies of S, 6 copies of R (those are standard names I use, sorry for making you use my standards) in a 30-cards deck. Forgetting mulligan, how do you calculate the probabilties of drawing at least one copy of each in 8 cards? I wouldn't want to use any kind of software, I'm trying to find out the solution. Have fun, I'll post my version the day after tomorrow in a dedicated topic, after I work on it for a little bit.

[ColorlessGreen]

edit: To clarify, each of A, B, and C is the probability that the card will appear in the first seven cards drawn from a (let's say) 30 card deck, when what we want is the probability that A will appear in the first seven cards, multiplied by the probability that B will appear in the remaining six, multipllied by the probability that C will appear in the remaining five.

Coming up with that clarification may have pointed me at an answer, though. If we just used the calc to calc A(7,30), then B(6,29), then C(5,28), then multiply the results together, we might have the correct answer. Or I may have failed to account for something.

At any rate, this has moved beyond the "quick question" realm. Leodip, please make a thread for this topic at your earliest convenience and I'll move this discussion into that thread, even if we haven't yet arrived at an answer (you can edit the answer into the OP of the thread when one is found).

Quoting myself since the conversation moved on quite a bit while I was thinking/typing.