So whats mathematically the best defense:

A) 4-2

B) 5-1

C) 3-3

Thanks!

When you say "4-2"'  Do you mean attacker has 4 sides and defender has 2?

I believe he means spreading out units behind a defense... should you have 4 on the outside with a 2 behind it, a 5 protecting 1's, or double thick even 3's...  

It kind of depends on the game you're playing...  Statistically, I imagine that 4-2 will end up being your best bet.  In heavy fog, I would do 5-1 and in a vicious battle of attrition, I would do 3-3.

Hey Yuri,

That’s a great question, those types of problems always
intrigue me - especially since that situation occurs often in the beginning of
Wargear Warfare games.  There are lots of ways to consider how to answer that.  The
first approach I took is to determine the probability of an opponent taking
both territories in each of the three scenarios.  It would be necessary to calculate the
probability for different starting strengths of your opponent, but I worked out
some probabilities assuming your opponent attacks from a territory having 6
armies with the intention of taking both the territories in question,

1^st territory         2^ndterritory        Odds to take 1^st             Oddsto take 2^nd       Odds to takeboth         

5                              1                        41%                              88%             35.6% (.41 x .88)

4                              2                        50%                              66%             32.6%

3                              3                        68%                             49%              33.5%

The odds are based on the attacker stopping the attack when odds no longer favor another attack.  For example they would not attack 3 v 2 (rolling 2 v 2) or 2 v 1 (rolling 1 v 1).  The odds to take the 2^nd
territory are based on the distribution of the successful attacks on the first
country.  For example, a certain percentage of the time the attacker will have 6 armies left after the first
attack, 5 armies some of the time, etc. Having said that, I just realized that I forget to deduct one army from each scenario after the first attack to account for leaving one army behind as
you move into the conquered territory.  I will recalc on Monday and post the new results. 
But I think the results above still give some insight.  4,2 and 3,3 are about the same with 5,1
having a slightly less effective defense. This I think is because the attacker has better odds against the 1 army territory on the second attack.

Ah thanks, that was exactly what i was looking for!

SquintGnome wrote:

Hey Yuri,

That’s a great question, those types of problems always
intrigue me - especially since that situation occurs often in the beginning of
Wargear Warfare games.  There are lots of ways to consider how to answer that.  The
first approach I took is to determine the probability of an opponent taking
both territories in each of the three scenarios.  It would be necessary to calculate the
probability for different starting strengths of your opponent, but I worked out
some probabilities assuming your opponent attacks from a territory having 6
armies with the intention of taking both the territories in question,

 

1^st territory         2^ndterritory        Odds to take 1^st             Oddsto take 2^nd       Odds to takeboth         

5                              1                        41%                              88%             35.6% (.41 x .88)

4                              2                        50%                              66%             32.6%

3                              3                        68%                             49%              33.5%

The odds are based on the attacker stopping the attack when odds no longer favor another attack.  For example they would not attack 3 v 2 (rolling 2 v 2) or 2 v 1 (rolling 1 v 1).  The odds to take the 2^nd
territory are based on the distribution of the successful attacks on the first
country.  For example, a certain percentage of the time the attacker will have 6 armies left after the first
attack, 5 armies some of the time, etc. Having said that, I just realized that I forget to deduct one army from each scenario after the first attack to account for leaving one army behind as
you move into the conquered territory.  I will recalc on Monday and post the new results. 
But I think the results above still give some insight.  4,2 and 3,3 are about the same with 5,1
having a slightly less effective defense. This I think is because the attacker has better odds against the 1 army territory on the second attack.

Good analysis Squint.

thanks squint! now i know what to do :P

In the post above are updated calcs from last week, and I added rows assuming the attacker starts with seven armies also. 

 The leftmost column shows how many units the attacker starts with and the next two shows how you place the defenders.  The columns after the Win % first column show the distribution of units when the attacker wins the first battle.  So, for example, in the first row if the attacker wins the battle, 22.5% of the time he will have six units left (this is only .405 x .225 = 9.1% of all attacks he makes).  Based on this distribution the attacker will have a certain probability to take the second territory, 68.4% in this case.  Multiplying both gives the probability to win both on a turn. 

So, the conclusion is that leaving only 1 in the second territory you wish to keep is not a favorable strategy.  Leaving 2 or 3 is about the same.

As an interesting side note Yuri you posted about a common starting position on Wargear Warfare where you place units against two territories holding 3 each.  This happens often in Australia or SA when you hold a territory, a neutral holds one, and your opponent holds two.  If you place all 4 there you have about a 1 in 3 chance to take both your opponents territories.