Recommend
 
 Thumb up
 Hide
6 Posts

Smash Up» Forums » General

Subject: late night math rss

Your Tags: Add tags
Popular Tags: [View All]
Matt Bayes
United States
O'Fallon
MO
flag msg tools
mbmbmb
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
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Wim D
Belgium
flag msg tools
Avatar
mbmbmbmbmb
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.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
J
United States
Alexandria
Virginia
flag msg tools
Avatar
mb
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!*(27-4)!*2!*(4-2)!*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!*(27-6)!*2!*(6-2)!*2!*(4-2)!*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!*(27-8)!*2!*(8-2)!*2!*(6-2)!*2!*(4-2)!*24))=233107875

again different by the Repeat magnitude.
3 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Mark Turner
United Kingdom
Farnham
Surrey
flag msg tools
badge
Avatar
mbmbmbmbmb
A slightly easier way to do the per player calculation is faction x factions-1 / 2.

Therefore, for one player, we do 27*26/2=351.

Etc.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Smashup Fanatic
msg tools
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.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Johny D
msg tools
mbmbmb
To do the correct math you should calculate n!/2*(n-2)! in order to determine the maximum number of possible unique combination you can play as. For 27 is 351 for example
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Front Page | Welcome | Contact | Privacy Policy | Terms of Service | Advertise | Support BGG | Feeds RSS
Geekdo, BoardGameGeek, the Geekdo logo, and the BoardGameGeek logo are trademarks of BoardGameGeek, LLC.