

With Secret Wars Vol. 1 here, Here is what you have here if you also have: The Core Set, F4, PtTR, GotG, Villains and Fear Itself. (No 3D Cards)
61 Playable Heroes/21 Villains 19 Masterminds/5 Commanders 9 Henchmen/4 Backup 25 Villains/9 Adversaries 36 Schemes/11 Plots 26 Special Bystanders
I think I got that right. Counting by hand.
I'm no mathematician but I'm pretty sure there are more combos than one person could ever play.

Tim McCormley
United States SD California
OUCH!

Euloque wrote: I'm no mathematician but I'm pretty sure there are more combos than one person could ever play.
Yeah. And don't forget the system also includes "Legendary Encounters," which also increases the number of combos.
TIm

James Drury
Canada Longueuil Quebec
The only stupid question... is the one you don't ask!

I am a math enthusiast so I will take a stab at the calculations
This is assuming you are playing a standard 3 player game, with no house rules. For the sake of simplicity, I'm combining Hero and Villain terms into 1 unit. This equation also does not include any exceptions added by schemes, such as additional Villain groups.
24 Masterminds * 47 Schemes * 1 Villain (this is 1 because the Mastermind always forces a specific group out) * 33 Villains remaining * 32 Villains remaining * 13 Henchmen * 82 Heroes * 81 Heroes remaining * 80 Heroes remaining * 79 Heroes remaining * 78 Heroes remaining
24*47*1*33*32*13*82*81*80*79*78 = 50,702,213,815,418,880 possible different game setups.

Nathan M
United States Knoxville TN

And that's only 3player games!
Bearing the following in mind: 1. Some masterminds always lead henchmen, 2. Some schemes choose villains for you, 3. Some schemes add extra villains or henchmen, 4. Some schemes add extra heroes.
You could probably play from now until the end of time and never play every possible combination. And there is another big box coming this fall/winter.



blipadouzi wrote: 24*47*1*33*32*13*82*81*80*79*78 = 50,702,213,815,418,880 possible different game setups.
Don't forget 16 more heroes from aliens as well as 64 possible 3 act movie setups.

James Drury
Canada Longueuil Quebec
The only stupid question... is the one you don't ask!

jodokast4 wrote: And that's only 3player games!
Exactly. I tried to be as minimalistic as possible to drive in how truly huge the number is lol
Quote: You could probably play from now until the end of time and never play every possible combination. And there is another big box coming this fall/winter.
Well, let's do another quick equation 50 quadrillion (rounded number of possibilities) divided by 7 billion (rounded population of Earth) = 7,142,857
It's mathematically impossible for every scenario to ever be played within a single lifetime. Every person on Earth would have to play the game 7 million times each.
That's an average of 245 games every day for 80 years per person.
Ok, enough geeking out

James Drury
Canada Longueuil Quebec
The only stupid question... is the one you don't ask!

therealtheshader wrote: blipadouzi wrote: 24*47*1*33*32*13*82*81*80*79*78 = 50,702,213,815,418,880 possible different game setups.
Don't forget 16 more heroes from aliens as well as 64 possible 3 act movie setups.
I left them out on purpose.
Though I could easily see Wolverine, Captain America, and so on beating up a Xenomorph... I do not see Ripley and her crew having a snowballs chance in hell in dealing any damage whatsoever to Apocalypse, Thanos, Red Skull or even Loki.
In the Marvel universe, Ripley's team(s) would just be bystanders. At best, Marine Sidekicks (like a henchmen group but on our side).



blipadouzi wrote: That's an average of 245 games every day for 80 years per person.
Of course, within those 80 years, it's reasonable to expect that a few more expansions will come out.



This topic should be highlighted on the game page under "Replayability".
Also, please factor in the order of the cards that show up in the city/HQ and obviously the shuffling of the player decks.
Everyday I'm shuffling...

James Drury
Canada Longueuil Quebec
The only stupid question... is the one you don't ask!

Shuffling, or deck randomization is easy to calculate. 5 Heroes * 14 cards per hero = 70 cards.
70! = 1.19785 × 10^100
or if you prefer the long version... 11,978,571,669,969,891,796,072,783,721,689,098,736,458,938,142,546,425,857,555,362,864,628,009,582,789,845,319,680,000,000,000,000,000
Note: for those that do not know, ! is used to calculate factorials. Eg. 5! = 5*4*3*2*1 = 120

Theorel Masheriel
United States Georgia

You're doublecounting quite a bit there.
For the hero deck randomization: there are 10 cards with 5 copies each, and 5 cards with 3 copies each. Any change of order within those cards is nonunique. So, I think the correct answer here is: 70!/(((5!)^10)((3!)^5)) ~ 2.4879x10^75
For the setup options, you're counting different orders of selections as different. So you're saying: Hydra, Brotherhood is different from Brotherhood, Hydra.
So, we need to divide by the number of orderings of villains/heroes: which is 2 for villains, and 5! for heroes. (so 240) That drops it to only 211 trillion options (which is still plenty substantial) But that does drop the number of games each person needs to play solo threehanded to play each setup to only ~29,000.

Tim McCormley
United States SD California
OUCH!

blipadouzi wrote: I left them out on purpose.
Though I could easily see Wolverine, Captain America, and so on beating up a Xenomorph... I do not see Ripley and her crew having a snowballs chance in hell in dealing any damage whatsoever to Apocalypse, Thanos, Red Skull or even Loki.
OTOH, if they can do damage to an "Alien," they could certainly do damage to the Green Goblin. Your typical Alien is way tougher than any run of the mill human villain.
Tim

James Drury
Canada Longueuil Quebec
The only stupid question... is the one you don't ask!

Theorel wrote: You're doublecounting quite a bit there.
For the hero deck randomization: there are 10 cards with 5 copies each, and 5 cards with 3 copies each. Any change of order within those cards is nonunique. So, I think the correct answer here is: 70!/(((5!)^10)((3!)^5)) ~ 2.4879x10^75
For the setup options, you're counting different orders of selections as different. So you're saying: Hydra, Brotherhood is different from Brotherhood, Hydra.
So, we need to divide by the number of orderings of villains/heroes: which is 2 for villains, and 5! for heroes. (so 240) That drops it to only 211 trillion options (which is still plenty substantial) But that does drop the number of games each person needs to play solo threehanded to play each setup to only ~29,000.
I agree with you 100%, but I had to come up with a formula in the simplest possible form that everyone would understand and wouldn't just skip over



Theorel wrote: You're doublecounting quite a bit there.
For the hero deck randomization: there are 10 cards with 5 copies each, and 5 cards with 3 copies each. Any change of order within those cards is nonunique. So, I think the correct answer here is: 70!/(((5!)^10)((3!)^5)) ~ 2.4879x10^75
For the setup options, you're counting different orders of selections as different. So you're saying: Hydra, Brotherhood is different from Brotherhood, Hydra.
So, we need to divide by the number of orderings of villains/heroes: which is 2 for villains, and 5! for heroes. (so 240) That drops it to only 211 trillion options (which is still plenty substantial) But that does drop the number of games each person needs to play solo threehanded to play each setup to only ~29,000.
Oh, well that's much more manageable.
C'mon folks, roll up them sleeves and get to it!

Brett Chaney
United States Iowa

Hehe, I love combinatorics, and I'd like to give a crack at it...
Note, for my combinatorics notation, below, consider (n¦k)=(n!/((nk)!*k!).
For a 3player game, assuming one chooses 3 of 31 available enemy groups, the 1 henchmen group assigned to the mastermind, 1 of 47 schemes, 1 of 32 Masterminds, and 1 of 47 schemes, and add onto that 31, for the scenario where you get a second mastermind in the latest expansion (assuming that's the right way to adjust for that? You might doublecheck my math...)
But I get:
(31¦3)*1*(47¦1)*(32¦1)*(86¦5)+31=235,442,518,144,991≈0.2 quadrillion combinations... or if one were to dig through that problem space at the rate of 1 game per day...
~0.6 trillion days of onegameperday, or ~143.2 times the age of the Earth (assuming ~4.5B years), or 47 times the age of the Known Universe (assuming 13.7B years).
Well, that’s much more manageable, hehe…
Get to it, gang! ~Brett Chaney

Brett Chaney
United States Iowa

Actually, yeah, that's wrong, you have to adjust the above equation to say (46¦1) schemes for the value of the schemes parameter, and then add to that the full equation again, for the 2 masterminds scheme, albeit this time with the mastermind parameter to read (32¦2) and without the scheme parameter, as that would simply be (1¦1)=1...
Now, I'm getting:
[(31¦3)*1*(46¦1)*(32¦1)*(86¦5)]+[(31¦3)*1*(32¦2)*(86¦5)]= 308,079,039,700,320≈0.3 quadrillion combinations...
Which, to queue the dangitIneedaframeofreferencetoevenfathomsuchanumber... If one were to dig through that problem space at the rate of 1 game per day...
~0.8 trillion days of onegameperday (at the Gregorian Calendar rate of 365.2425 days/year, and this is also assuming no temporal, relativistic effects are experienced while digging said problem space), or ~187.4 times the age of the Earth (assuming ~4.5B years), or ~61.6 times the age of the Known Universe (assuming 13.7B years).
Well, so I suppose we should construct a astroengineeringscaled computer, e.g. the size of a Dyson Sphere, i.e. fueled by the solar energy of the contained star, in order to have something potentially capable of navigating the large, albeit finite, problem space of potential games, and every unique possible way each of those games could feasibly play out, in order to truly determine the optimal strategy, whether the game is truly balanced, how often it's winnable, or what have you, hehe...
However, perhaps with that kind of power, there's other mathematical problems/issues which would be more likely to receive the funding of the future's Galactic Research Consortium... However, perhaps they'd let some gamers run the simulation during down time, say over a weekend...
Cheers, ~Brett Chaney

Adelin Dumitru
Romania Bucharest

Euloque wrote: With Secret Wars Vol. 1 here, Here is what you have here if you also have: The Core Set, F4, PtTR, GotG, Villains and Fear Itself. (No 3D Cards)
77 Playable Heroes/21 Villains 23 Masterminds/5 Commanders 12 Henchmen/4 Backup 31 Villain groups/9 Adversaries 44 Schemes/11 Plots 37 Special Bystanders
I think I got that right. Counting by hand.
I'm no mathematician but I'm pretty sure there are more combos than one person could ever play.
I altered the numbers so that they reflect the content added in SW 2


