Conquer Club

My last lessons in chance possibilities where a long time ago, so I forgot how to calculate chances like this one:

I have territory A, B, C, D and E. In an unlimited fort. game.
On Territory A I have 50 armies. On B, C, D and E there are 1.

I know that in the next round my opponent will try to eliminate me, how do I fort my armies ?
My guess is to fort them all equel. And have 10 armies on A, B, C, D and E. But can anyone show me my advantage with a calculation ??

Thanks,
Frank.

2,2,2,2,46 but your only talking tenths of a percent better odds. if your opponent has 60 armies. 2,2,2,2,46 gives him 79.3% 10,10,10,10,14 gives him 79.9%

I think this is more of a Strategy question, than a Q&A type question.


--Andy

You reach maximum defence probability when you minimise the number of times someone attacks you while you are defending with 1 army. Dice calculation for the attacker is always based on throwing three dice. Hence, you need to have two armies in each territory with the rest in the last one. For example, if you have three in each territory and the rest on the last one, you are more likely to lose since at least once you need to defend with one dice for each territory. This will increase the number of times you defend with one dice and reduces your overall rate of success.

As a strategy, place your bulk at the end furthest away from your enemy. You eliminator may lose a lot of armies before getting to your main force and subsequently give up the attack.

http://gamesbyemail.com/Games/Gambit/BattleOdds Use this link to calculate odds. 1 thing to keep in mind is that when you have 4 army's on a territory you only enter 3 into this program because that is the number of armies that can attack.

Depends on how your territories are positioned. Generally, best odds is to keep your largest army as far away from the enemy's largest army as possible, with 2s in between

e_i_pi wrote:Depends on how your territories are positioned. Generally, best odds is to keep your largest army as far away from the enemy's largest army as possible, with 2s in between

Thanks, but can you explain why ?

Why do you have better ods with:
-A=20 B=2 C=2
-then with A=18 B=3 C=3
-or with A=14 B=5 C=5

Could someone explain it with a calculation ?

Your best odds are going to occur if they never roll against a single troop. The best way to achieve this is by putting 2's on all territories except one. The reason is because the only way to roll against a one on those territories is if they win 1 and lose 1. The odds of this happening at any time in the attack are around 48%. As you get more troops on the territory, the odds slowly approach 50%. That means that you will eliminate more of their men with 2 territories than with any other configuration.

In foggy games, though, this means little, as they will likely just blow through your 2's without a second thought, not knowing you have a huge stockpile at the back. In that case, you want to put a lot of troops at the front so they will think twice about trying to eliminate you.