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.
