Odds Of Winning an Attack?

[Rosso87]

Who has the better chance of winning an attack, the attacker or the defender?

Say you have 10 troops attacking a country with 10 troops on it defending. What are the odds you'll win?

How many troops should you lose if it followed the expected outcomes, for example if you flip a coin 10x then 5-5 heads-tails if it followed as expected - obviously it doesn't but you can work it out theoretically. I just wondered what the odds of winning or losing on the dice as a defender or attacker are? Is it say stacked 3:2 advantage to attacker, or 3:2 advantage to defender etc.

thanks

[MartijnF]

I was wondering about this too, so I ran some simulations and put the results online. I made a topic about it here: viewtopic.php?f=6&t=102421&start=0

To answer your question, if an attacker would use 10 troops to engage an area defended by 10 troops, the chances of success would be about 48%. The larger the attacking force is (with equally larger defending troops), the more favorable it is for the attacker. So while a 10 vs 10 battle holds a 48% chance of succes, a 15 vs 15 attack will be succesful in almost 54% of the times, a 20 vs 20 attack holds a success rate of about 58% etc. In a 50 vs 50 battle, the attacker will defeat the defender in more than 70% of the cases!

[AndyDufresne]

Ahoy Rosso87.

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Assault Odds.


--Andy

[ljex]

I was wondering about this too, so I ran some simulations and put the results online. I made a topic about it here: viewtopic.php?f=6&t=102421&start=0

To answer your question, if an attacker would use 10 troops to engage an area defended by 10 troops, the chances of success would be about 48%. The larger the attacking force is (with equally larger defending troops), the more favorable it is for the attacker. So while a 10 vs 10 battle holds a 48% chance of succes, a 15 vs 15 attack will be succesful in almost 54% of the times, a 20 vs 20 attack holds a success rate of about 58% etc. In a 50 vs 50 battle, the attacker will defeat the defender in more than 70% of the cases!

yes this is because with lower numbers, eg 10 vs 10, you will more often come to attacks like 3 vs 2 which is lower odds than 4 vs 2.

[40kguy]

the attacker gets the advantage. but a 10 on 10 will not be a big difference just a little. all tho 100v100 the attacker has a much bigger advantage.

[ljex]

the attacker gets the advantage. but a 10 on 10 will not be a big difference just a little. all tho 100v100 the attacker has a much bigger advantage.

its not that the advantage changes, its that the advantage has more time to take effect probability wise

[rdsrds2120]

the attacker gets the advantage. but a 10 on 10 will not be a big difference just a little. all tho 100v100 the attacker has a much bigger advantage.

its not that the advantage changes, its that the advantage has more time to take effect probability wise

This is a true statement. The Law of Large Numbers says that as we, in context, roll more and more times, we will get really close to the expected win:loss ratio, which is calculated to be better for the attacker.

-rd