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Maximum armies per territory

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Maximum armies per territory

Postby BaldAdonis on Sun Aug 26, 2007 5:02 pm

I know this has been brought up before, and the problems people had with it were:
1) It doesn't work with escalating cards.
2) It will take too long to play.
3) LOL, that's dumb, I'd never play it!

Basically, the idea is to limit the number of armies to 12 per territory.
It's not meant to be played with escalating cards.
The game is more aggressive than others that get bogged down by over-building, and as far as I can tell, will be faster than standard fixed/no card games.
The attacks can't take more than 12 rolls to resolve, and almost always fewer. Although this point doesn't make a difference if you auto-attack, I like to see how my attacks go so I don't end up dropping to three and leaving my opponent with more.
Most importantly, I've been playing this way for ages, with modified fixed sets, and it works really well. The games are balanced, rely more on strategy and don't take all day to play.
If you don't want to play it, I wouldn't be bothered. There are plenty of escalating set games where you can put 200 armies on the Ukraine and stream roll everyone left. This option would be for players who want more strategy in the game.

If anyone's wondering, and maybe you'll like this idea better than 12/place, the modified fixed set gives 4 for mixed and 6, 8, 10 for the others. Since the mixed set is most common, giving it the least amount makes the game less luck-dependent.
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Postby Rocketry on Sun Aug 26, 2007 5:33 pm

I cant see this being brought in man - its too limiting. In my games it is usual to have twenty or thirty armies per state.

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Postby sfhbballnut on Sun Aug 26, 2007 6:21 pm

yeah, 12 armies is nothing, you'd neve be able to limit to that a legitimatly play, it'd be gimmicky for a while and then die out
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Re: Maximum armies per territory

Postby Kaplowitz on Sun Aug 26, 2007 7:54 pm

BaldAdonis wrote:1) It doesn't work with escalating cards.
2) It will take too long to play.
3) LOL, that's dumb, I'd never play it!


QFT

(yes, i had a good chance to use that)
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Postby misterman10 on Sun Aug 26, 2007 8:32 pm

Terrible idea, I absolutely hate it, it limits the gameplay way too much.
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Postby BeakerWMA on Sun Aug 26, 2007 9:01 pm

misterman10 wrote:Terrible idea, I absolutely hate it, it limits the gameplay way too much.


I agree.
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Postby BaldAdonis on Sun Aug 26, 2007 11:43 pm

What it limits is the ability to take one continent, turtle and build up your forces into an unstoppable wrecking ball. If you like escalating cards and making loads of armies, then this option isn't for you. It's not about having the biggest guns and the most patience, but the best laid plans. You might want to try it out, in a private game, before deciding it doesn't work.
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Postby XenHu on Sun Aug 26, 2007 11:49 pm

I think we can all agree it wouldn't work well on:

http://www.conquerclub.com/game.php?game=739898


:wink:

Can you imagine how often continents would trade hands.

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Re: Maximum armies per territory

Postby Robinette on Sun Aug 26, 2007 11:55 pm

BaldAdonis wrote:...maybe you'll like this idea better than 12/place, the modified fixed set gives 4 for mixed and 6, 8, 10 for the others. Since the mixed set is most common, giving it the least amount makes the game less luck-dependent.


um, if you do the math you'll see that the odds are the same and that this would not change the game to be "less luck-dependent"...
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Postby BeakerWMA on Mon Aug 27, 2007 7:31 am

The only way someone could turtle and buildup their forces is if no one attacked them. Attack, attack, attack.
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Re: Maximum armies per territory

Postby BaldAdonis on Mon Aug 27, 2007 10:13 am

Robinette wrote:um, if you do the math you'll see that the odds are the same and that this would not change the game to be "less luck-dependent"...


I did do the math, that's how I know they're not the same.

[the math]

Consider every arrangement of 5 cards. There are 243=3^5. This outline works the same for fewer cards, but 5 gives all the in-game possibilities.
The 5 cards resolve to one of 7 set choices: Red, Green, Blue, Mixed, Red or Mixed (ie RRRGB), Green or Mixed and Blue or Mixed. I'll write these as numbers like this [R G B M R/M G/M B/M]. For example [7 0 0 0 2 0 0] means that there are seven possible sets that end with only a Red set, and 2 that end with either a Red or a Mixed, which is what happens when you start with 3 Reds and draw two more cards (try it!).
Since there's no innate difference between Red and Blue, the number of sets is symmetric with regards to colors. So you can count all the sets that start with red, then flip the red and blue numbers to find the number that start with blue. Using the example above, this means that the sets starting with BBB are [0 0 7 0 0 0 2].

Now we just get counting. There's not that much to do, considering there are 243 sets to look at, because symmetry does almost all of the work.
Since we already know about sets starting RRR, go on to RRG- This can end as a Red set (RRGRR, RRGRG, RRGGR), a Green set (RRGGG), a mixed set (RRGGB, RRGBG, RRGBB) or a choice of Red or Mixed (RRGRB, RRGBR). And that's all, because there are only 9 choices. Since starting RRG is the same as starting RGR, we have
RRG = RGR = [3 1 0 3 2 0 0]
By symmetry around Blue, we get RRB = RBR = [3 0 1 0 2 0 0]
Next, we have RGG =[1 3 0 3 0 2 0] (check it if you want), so RBB = [1 0 3 0 3 0 0 2]
Finally, we have RBG = RGB = [0 0 0 6 1 1 1]. Including the RRR from above, that's all that start with Red, because there are only 9 choices for the second and third cards.
So the sum of all sets beginning with Red is [21 5 5 30 12 4 4]. By symmetry we get the sets starting with Blue and Green, and the total of all three, which is [31 31 31 90 20 20 20].

[/the math]

So there are 90 sets (~37%) which force you to use Mixed, and 60 (~25%) with Mixed or another colour. Since it's worth the most, people will choose it every time, effectively making the set resolution 31 Red, 31 Blue, 31 Green, 150 Mixed. The mixed set comes up ALOT more often than any other.

Making Mixed worth the least makes the sets resolve to 51, 51, 51, 90, still more Mixed than any single other choice, but no longer the outright majority, and now less desirable.
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