Calculating Win% on multiple battles

NoAnswer

See You Space Cowboy...
Joined
Jun 28, 2003
Messages
123
Location
L.A.
Does anybody know how to calculate combined odds? I'm not sure what the correct term for this is, but I can explain what I mean.

Let us take an example of a fortified unit, already at Elite (can't get more HP back), with 1 HP in a city. Let's say it's an Infantry, and in this example I am attacking it with multiple available Veteran Longbowmen. The Infantry is fortified in a Walled City on a Hill. So the combat odds are

Chance of Longbowman winning: 54.296%.

My question is, how do you calculate the chances of my second unit winning if the first one fails? And further, if the 1st and 2nd both fail, what is the chance that at least the 3rd will win? etc

1st LB- this unit Victory: 54.296%. Total Chance of Victory this turn: 54.296%
2nd LB- this unit Victory: 54.296%. Total Chance of Victory this turn: ?
3rd LB- this unit Victory: 54.296%. Total Chance of Victory this turn: ??
4th LB- this unit Victory: 54.296%. Total Chance of Victory this turn: ???

Anybody know how to calculate that? To find out what is final percentage of my chances of defeating that unit this turn with 4 Longbowmen.
 
Let us take an example of a fortified unit, already at Elite (can't get more HP back)

Chance of Longbowman winning: 54.296%.

I assume that this percentage is true. I further assume that there no gain of HP and there are no other factors to consider like defensive bombardement. Chances of not winning the first battle are 1 - 54.296% = 0.45704.

The chances of not having won after 2 battles is 0.45704^2.

The chances of not having won after n battles is 0.45704^n

The chances of having won after n battles is 1 - (1 - 0.54296)^n

So changes of having won are 79.11%, 90.45%, 95.64% and 98.01% after 2, 3, 4 and 5 battles.
 
I assume that this percentage is true. I further assume that there no gain of HP and there are no other factors to consider like defensive bombardement. Chances of not winning the first battle are 1 - 54.296% = 0.45704.

The chances of not having won after 2 battles is 0.45704^2.

The chances of not having won after n battles is 0.45704^n

The chances of having won after n battles is 1 - (1 - 0.54296)^n

So changes of having won are 79.11%, 90.45%, 95.64% and 98.01% after 2, 3, 4 and 5 battles.

Thanks!
 
Top Bottom