Has every team possible in MvC2 and CvS2 been used at some point?

I’d say probably not.

Here’s the math:

MvC2 has 56 characters. Each of those characters has 3 different assist types. Now we’re looking at 168 distinct playable characters (this does not include Abyss).

CvS2 has 48 characters, but there are 6 different grooves to deal with (this does not include custom grooves). However, you must choose one groove for your whole team.

This is all under the assumption that teams each have a different character (ie. no Megaman/Megaman/Megaman with the same assist). However, since I’m treating each assist type as a different character, you could have all three Megamen(?) with different assists.

So, the math for Marvel goes like this: 168x167x166

Which is a whopping 4,657,296 different possibilities.

The math for CvS2 is a little different: 48x47x46x6

Which is 622,656

So, have more than 4million different teams been chosen in MvC2 or more than 600,000 in CvS2?

Are there any other games that can beat Marvel? Maybe KOF98UM or KOF2002UM? I haven’t checked them out yet.

you have to much time on your hands brah…

My sentiments exactly.

I have never seen a Storm, Magneto, Sentinel team in MvC2.

Well ain’t that Abyss. :coffee:

Even with different assists you can’t have the same person twice or three times on your team (at least at Evo where it counts).


I’m pretty sure they have. Not competitively, but otherwise, it’s kind of probably considering a) How long the game’s been out and b) how many people (worldwide) that have played it.

I think that math is wrong.

since once you pick the first character you automatically lose 2 more choices in MvC2 (can’t pick the same character nor the same assist type).

I also thought there was some dividing that was supposed to go on there too, like by the number of remaining options… But its been8 years since discrete math so I’m rusty at this.

I’m not talking about competition, where a named character can be on a team only once. I’m simply talking about possibilities. How many unique teams can be made if you only remove one choice for each member of the team?

The math is simple. Your first character slot can be one of 168 different unique characters (a unique character, imo, is a character that is somehow different than another - anti air Cable is different than projectile Cable who is different than projectile Roll, so that could be a possible team).

Your second slot has 167 options. You only subtract 1 from the original total because he/she is in the first slot. Again, you take away 1 more for the third, so you have 166.

Your first slot has 168 possibilities
Your second has 167
Your third has 166

You simply multiply them to get your team of three.

Now, if you want to use EVO rules, it does change a bit. For every slot, you need to remove 3 instead of 1 because once a version of a character has been picked, all three of his versions are removed from the possibility of being in the second or third slot. So it would go like this: 168x165x162, which is still very high at 4,490,640.

Now, for those saying this has happened, consider this.
Use this MvC2 randomizer: http://www.richmondtrials.com/rand.html
It needs to be clicked about twice (probably more than that) as many times as there are possibilities for each team to show up once, so you’re looking at clicks. For each click, a match has to be played, at an approximate average time of 45 seconds per match.

That means that the game would have had to have been played a total of 12 years 10 months 3 days and 8 hours (or 112,500 hours) straight to get every team to show up, if it was being played on random the whole time. You can halve the time to about 56,250 hours because the team has a chance of showing up on 1P or 2P.

Finally, if you want to consider each character to be unique, then it’s simply 56x55x54, which is 166,320 - considerably lower and more likely to have happened.

Hmm, nice, but what about using the combination factorial math?

I used that for my calculation in another thread, for MVC2 at least:

It depends on whether you count Mags/Storm/Sent, Storm/Mags/Sent, Sent/Storm/Mags etc. as the same team or different teams.

Just looking at the characters:
If Mags/Storm/Sent et. al. count as different teams, you have 56!/53! = 166,320 teams.
If Mags/Storm/Sent et. al. count as the same team, you have 56!/[3!(53!)] = 27,720 teams.

Since there are 27 combinations of assist type for any given team, you have:
166,320 * 27 = 4,490,640 or
27,720 * 27 = 748,440 teams.

If you’re allowed to pick the same character all three times, but with different assists, you end up with:
168!/165! = 4,657,296 or
168!/[3!(165!)] =776,216 teams.

This is Mahvel Math Babee!!

btw my guess is no. :coffee:

Here’s hoping Mag/Storm/Sent is beat out by some random ass team. Serv/Jill/Hood

Yeah that’s the more professional solution :sweat:

Hooray for still being in High School math classes :rofl:

i bet there are people out there who have heard “gonna take u for a ride” about 5 million times

I use Ryu (b), Guile (a), and Hayato (y). You can scratch that one off.

It’s a start.

It’s a simple math problem that takes about 30 seconds to think about.

the answer is Yes. I did it last night. It was pretty cool.