Matt Bayes
United States O'Fallon MO

I can't sleep. so I've done the math to figure out how many possible games you can play with all 27 factions currently available.
first we look at 1P who has the all 27 factions open them them. that's 351 possible mixtures open to him. we can find this by simply taking 27 and adding each number below it. 26+25+24+23+22+21.... And so on. another way to look at it would be like this
AB BC CD DE AC BD CE AD BE AE
notice how each row has one less pair then the one before it
Now things get a little more complex. 2P has less to choose from, we would take 25 and add the numbers below it 24+23+22+21+20... Or we could just take 351 and subtract (26+25) from it, giving us 325
Basically for the letter system above the first row is taken out and so is the second
If we keep this up we end up with 1P 351 2P 300 3P 253 4P 210
Then we just multiple each together to see how many possible games they can make.
So 2 players can have 105,300 different games 3 players 26,640,000 4 players 5,594,589,000
and all together one player can play 5,621,334,300 completely different games not counting bases or any other Expansions set to come out this year
thanks for reading. ps I'm half asleep so my math my be off. If so please tell me



I only have 25 factions so far, and I think I have everything.
There are 27 dividers in the Big Geeky Box, but bases and madness cards are no factions.

J
United States Alexandria Virginia

Lol someone always has to do this so lets have some fun. I'm assuming 27 factions no repeats though I think there are actually only 25 currently available.
2 player
The way I'd do this is to first figure out how many ways you can pick 4 different factions from the initial 27 which would be 27 Combination 4 Then of those 4 we need to know how many ways they can be divided among 2 players so 4 Combination 2 But half of those are repeat (Player 1 having AB Player 2 having CD same as Player 1 having CD Player 2 having AB) so we need to divide by 2 making the final equation: (27!*4!)/(4!*(274)!*2!*(42)!*2))= 52650
Hmmmmmmmm our answers are a little off. Why is that? Well your way doesn't consider Player 1 having AB Player 2 having CD the same as Player 1 having CD Player 2 having AB. Here I'll illustrate it your way pretending there is only 4 factions instead of 27.
(3+2+1)*(1)=6
Below shows the 6 but at the same time shows the 3 repeats
Player 1 AB AC AD ___BC BD ___CD Player 2 CD BD BC ___AD AC ___AB
3 player
27 Combination 6 6 Combination 2 4 Combination 2 6 Repeats
(27!*6!*4!)/(6!*(276)!*2!*(62)!*2!*(42)!*6)) = 4440150 again different by the "repeat" magnitude of 6.
4 player
27 Combination 8 8 Combination 2 6 Combination 2 4 Combination 2 24 Repeats
(27!*8!*6!*4!)/(8!*(278)!*2!*(82)!*2!*(62)!*2!*(42)!*24))=233107875
again different by the Repeat magnitude.

Mark Turner
United Kingdom Farnham Surrey

A slightly easier way to do the per player calculation is faction x factions1 / 2.
Therefore, for one player, we do 27*26/2=351.
Etc.



backspace8908 wrote: I can't sleep. so I've done the math to figure out how many possible games you can play with all 27 factions currently available.
Reminds me of playing Imaginary Smash Up, where you imagine the ideal possible scenario. I try to win within 1 turn.



To do the correct math you should calculate n!/2*(n2)! in order to determine the maximum number of possible unique combination you can play as. For 27 is 351 for example


