Intensity level and dice

[kaobang]

hello,

I'm new here and after reading the instructions and in particulary the chapter II: "Assaulting your ennemies", I have some questions.

It said there are 6 levels of intensity, does it mean that there is an algorithm that modify the probabilty of the dice in function of the intensity (I guess the number of attackers/defenders) ?

basically:
Do I have the same chance with my dice with:
case 1: I attack with 4 troups, he defends with 2 troups
case2: I attack with 40 troups, he defends with 2 troups
?

thank you for your help

[The Neon Peon]

Okay, lets go back before the law suit:

Assault = Attack
6 Levels of Intensity used = 6 Sided Dice rolled

So no, you have higher odds attacking with more armies to less armies. (unless it is a 2 v 1) If you have 40 v 2... you will rarely lose (odds are about 40 in 10 million)*, but the odds of losing a 4 v 2 are a lot larger.

* A very rough estimation that e_i_pi can correct, I am sure. (if he happens to go in this forum)

[Geger]

Hm.. you loose 19-times until you have only 2 troops. The probability is (0.293)^19 = 7.21 x 10^(-11)

[The Neon Peon]

Hm.. you loose 19-times until you have only 2 troops. The probability is (0.293)^19 = 7.21 x 10^(-11)

Losing one 3 dice v 2 dice roll has odds of .46

Where did the .293 come from?

[kaobang]

Assault = Attack
6 Levels of Intensity used = 6 Sided Dice rolled

that's the key eheh, So the result is only based on dice

thank you for the answer !

[Geger]

Hm.. you loose 19-times until you have only 2 troops. The probability is (0.293)^19 = 7.21 x 10^(-11)

Losing one 3 dice v 2 dice roll has odds of .46

Where did the .293 come from?

from viewtopic.php?p=1719891#p1719891

[The Neon Peon]

Ah... so sully was wrong? (He was the one who told me the statistic I was using)