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Cosmic Encounter» Forums » General

Subject: How many different alien battles... rss

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Shadow ruxer
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Well, I have been brushing up on my math after neglecting the trickier stuff for years and I started to get into combinations...
Needless to say I wanted to see what the limits of Cosmic Encounter currently are for a game that is advertised as having limitless possibilities laugh

If you have a math calculator you just use nCr. N=number of options and R=number of selections OR you would write (in the case of having all of the aliens [135] and playing a three player game).

You would divide the factorials 135!/(135-3)! 3!

Essentially N!/(N-R)! R!

Anyway, here is the interesting part... For a 3 player game of cosmic encounter there would be 400,995 different alien combinations. Adding just one additional player (a fourth alien) raises this amount to a staggering amount of over 13MILLION alien combinations, 13,232,835 to be exact.

I just thought you all would find this interesting goo

Does Cosmic Encounter stand up to this limitless argument? And this is just considering the aliens into account and not the cards, decisions, bluffing and other factors that make this game pure and utter chaos.
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Rus
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Cool, but technically speaking, not all of those combinations are allowed

So, what is the real answer? Is there a list of illegal combinations?
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Shadow ruxer
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Ah, that is true... I did it without considering the illegal combinations, and to be honest. I have no idea how to put that into consideration whistle...If say two aliens in a group of three make the group illegal, they prevent every other combination within that group of the two conflicting aliens. So, from the top of my head it would be 133 subtracted for each illegal pair up.
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Chris
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the_shadow447 wrote:
Ah, that is true... I did it without considering the illegal combinations, and to be honest. I have no idea how to put that into consideration whistle...If say two aliens in a group of three make the group illegal, they prevent every other combination within that group of the two conflicting aliens. So, from the top of my head it would be 133 subtracted for each illegal pair up.


In my group each person plays 2 alien powers.. So ... That seems like a lot more...
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Luca Giordano
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the_shadow447 wrote:
Ah, that is true... I did it without considering the illegal combinations, and to be honest. I have no idea how to put that into consideration whistle...If say two aliens in a group of three make the group illegal, they prevent every other combination within that group of the two conflicting aliens. So, from the top of my head it would be 133 subtracted for each illegal pair up.


What illegal combinations ?
 
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Homer Simpson
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Gambler vs. Sorcerer for example.
 
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Paul Strauss
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Also Sadist vs Zombie or Healer.
 
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James Williams
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The box advertises Cosmic Encounter as a game of infinite "possibilites." Since "possibilites" isn't a word, you can't really disprove the statement. ;D
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Just a Bill
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No, I said "oh, brother," not "go hover."
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homertve wrote:
Gambler vs. Sorcerer [*]
daspore wrote:
Sadist vs. Zombie or Healer

• Coordinator vs. Dictator
• Magician vs. Oracle [*]
• Masochist vs. Healer (not printed, but generally recognized as needed)
• Explorer, Pygmy, Roach, Symbiote... vs. no extra player color
• Grumpus vs. no inexperienced players at the table to pick it

* However, the Gambler/Sorcerer and Magician/Oracle restrictions are pretty easily eliminated with simple house rules.

Another consideration, of course, is design overlap. For example, is the Clone-Fungus-Sniveler-Zombie combination significantly different from the Clone-Fungus-Sniveler-Bulwark combo? Not really.

I never pay much attention when somebody states how many permutations there are. The math is never accurate, and it's moot anyway; for well over three decades, the answer has been simply "more than you could ever possibly exhaust."
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Roberta Yang
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the_shadow447 wrote:
Ah, that is true... I did it without considering the illegal combinations, and to be honest. I have no idea how to put that into consideration whistle

Use inclusion-exclusion!

Let's assume a four-player game with the first four expansions and one power per player. By default, there are 135!/(131! 4!) combinations, which is the 13,232,835 number you found. However, some are illegal, so we need to subtract them off.

Let's be conservative and say there are 6 illegal pairs: Gambler/Sorcerer, Sadist/Zombie, Sadist/Healer, Coordinator/Dictator, Magician/Oracle, Masochist/Healer. (With only four players and eight colors, we have enough to support all the powers that require extra colors.) Each of these pairs can be involved in (133!)/(131! 2!)=8778 potential "games" of four aliens, so we need to subtract off 6*8778 games that we previously counted but now know to be illegal.

But wait! We've now subtracted off games like the Coordinator/Dictator/Magician/Oracle game twice. That's no good at all! Even if it's illegal for two different reasons, we can't subtract it twice - it's not so illegal that it counts as a negative game and cancels out another legal game! From the 6 illegal pairs, we get 15 illegal "pairs of pairs". 13 of them (each pair except for the two below) occur only in a single possible set of four aliens, so we add 13 back in. The remaining two pairs of pairs are represented by trios of powers: Zombie/Sadist/Healer and Sadist/Healer/Masochist. Each of these trios can be paired with any fourth alien, so we add back in another 2*132 games.

Now there's only one last problem: look at the Zombie/Sadist/Healer/Masochist game. We initially counted it once, then subtracted it off three times for containing three illegal pairs, then added it back in three times for containing three pairs of illegal pairs (it was one of the 13 as Zombie/Sadist + Healer/Masochist, and it was also one of the 132 possibilities for each trio). That means in total we've counted this game once. But it's an illegal game and shouldn't be counted - so we'll subtract 1 to get rid of it once and for all.

This brings our total to (135!)/(131! 4!)-6(133!)/(131! 2!)+13+2*132-1, which comes out to 13,180,443. So getting rid of the illegal alien combinations has cost us about 50,000 of our games.

That's how inclusion-exclusion works. To find out how many elements (games) of a set have zero of a certain collection of undesirable properties (illegal alien pairs), you first count the total number of elements, then look at each property and subtract off the number of elements that have that property, then look at each pair of properties and add back in the number of elements that have that pair of properties, then look at each triple of properties and subtract back off again the number of elements that have those three properties, then look at each quadruple of properties and add back in... and so on.

Now do similar calculations for games of each other player size (taking care to avoid extra illegal cases for the extra-color aliens in larger games), multiply the total for whether you're using optional modules like tech, rewards, and hazards, and then update all your calculations for the Cosmic Dominion aliens...
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Just a Bill
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No, I said "oh, brother," not "go hover."
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salty53 wrote:
Now do similar calculations for games of each other player size (taking care to avoid extra illegal cases for the extra-color aliens in larger games), multiply the total for whether you're using optional modules like tech, rewards, and hazards, and then update all your calculations for the Cosmic Dominion aliens...

And then publish it all, so I can feel better about myself when I waste time on minutiae.
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Shadow ruxer
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thanks for the input everyone and thanks for all that great mathmatic info salty
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Don't forget to multiply by the number of combinations of Technology in the game, and the Hazards, and Space Stations, and x4 for the combinations of Reward Decks (once CD is out), and seating order, and x2 for whether you're using free-wheeling flares or not, and of course you have to factor in all the different extra flares that are in the deck besides the ones corresponding to the powers.
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