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### V4

#### wy_mentat

*Thursday 13th September 2007, 8:27 GMT*

still no v4 :-(

WY still retired

#### krang

*Wednesday 19th September 2007, 20:50 GMT*

I'm sorry MY... been very busy for the past few months.

I would like to say that I will get around to it soon, but I have no idea what I'm doing this weekend, let alone for the next month.

Although tiki is doing a good job at giving me the odd reminder.

#### comrade tiki

*Sunday 23rd September 2007, 23:07 GMT*

...You know what, Mentat? I now think that some of the equations should be simplified and re-balanced. Especially your quote:

"...in real world attacker unit must be 3 (least 2) times bigger."

Victory probability of two equal armies is 1/3 chance of attack success with a new equation heavily inspired by your past calculations (not exact, but a close mirror).

(a^2) / (a^2 + d^2 + d^2) = attacker's chance out of 1

#### wy_mentat

*Tuesday 25th September 2007, 15:57 GMT*

I like it, that would make game more balanced and realistic. lets try that!

other thoughts: I considered a bit lower winning chance equation:

(a^2) / (a^2 + 4*d^2) = attacker's chance out of 1

examples:

(1) if a=d, then winnining chance would be 20% - I wouldn't consider to attack, unless situation is so urgent, that I would consider attack as my last chance.

(2) if a=2*d, then winning chance would be 50% - This is a fair 50-50 chance to win. But I wouldn't risk with that, because I, as attacker can choose an attacking place, time and better winning rates, when attacking with more units.

(3) if a=3*d, then winning chance would be 69,23%

- so in this equation I would choose a simple rule form myself: "... attacker unit must be 3 (least 2) times bigger."

So this equation would be realistic, but I think that for dominationgame, your equation is better. Because my last equation suggestion might change game too slow; and that wouldn't be fun at all.

#### j lorenz

*Tuesday 25th September 2007, 19:59 GMT*

...Heh, you're right, with that ratio I'd imagine nobody would even go to war! :-P

With the equation as-is (2*d^4) the defenders are already twice as likely to win if evenly matched. Making it a landslide may indeed be more realistic but take away from the benefits of violence.

I, too, am interested in testing the new combat eq. out.

If you have any time to tinker, what do you think about this for conquering?:

cos( arctan( 4 / a ) )^4 = conquered probability out of 1

#### wy_mentat

*Wednesday 26th September 2007, 7:29 GMT*

btw, Tiki's offered equation

[(a^2) / (a^2 + d^2 + d^2) = attacker's chance out of 1],

winning chances are these:

if 3*a=d, then winning chance would be 5,26%

if 2*a=d, then winning chance would be 11,11%

if a=d, then winning chance would be 33,33%

if a=2*d, then winning chance would be 66,67%

if a=3*d, then winning chance would be 81,82%

#### wy_mentat

*Wednesday 26th September 2007, 7:42 GMT*

and if U want 50-50 chance with Tiki's equation, then attacking units must be about 1,4142135624 times bigger:

if a=1,4142135624*d, then winning chance would be 50,00%

;-)

#### comrade tiki

*Wednesday 26th September 2007, 19:22 GMT*

I HAVE tested the chances of success... but you've passed me up when it came to computing a perfect 50-50 chance! Why would you want to know how to have a half-chance of defeat, anyway? Send more battalions, crush the enemy! :-)

(I'm sure that sending 1.4142135624 times the battalions for only a half chance of victory is known as a "bad long-term strategy")

#### wy_mentat

*Thursday 27th September 2007, 5:49 GMT*

"Why would you want to know how to have a half-chance of defeat, anyway? Send more battalions, crush the enemy!"

- more battalions from what? yes, send more battalions than 1.4142135624 times bigger unit, then U get better than 50-50 chance to win.