Was the city too large for walls, ainwood? I remember something obscure about not being able to put a wall around a city once it reaches a certain size.
AlanH said:
Do the multiplication of 5 by 0.69 repeatedly, and you'll see that it will take 11 attacker hits to get above 90% chance of killing the defender.
I see what you're trying to do, Alan. You want to multiply 0.69 five times, to simulate the chances that the Defender will win 5 battles:
0.69 x 0.69 x 0.69 x 0.69 x 0.69 = 15.6%
That's the odds of the Defender winning 5 attacks in a row, though. The actual formula you need for the odds has to take into account that the Defense can fail 4 times (it takes 5 failures to kill him). I think you need to use that old factorial formula for computing this one.
I wish I knew for sure.
The logic denyd followed is concrete, and seems reasonable to me. Here's another way to look at the problem:
Start with a 31% chance of losing a hitpoint (69% of not losing one).
1 attack = .31 hitpoints lost (on the average).
2 attacks = .62 hitpoints lost (on the average).
3 attacks = .93 hitpoints lost (on the average).
This passes a "reality check": if you've got a 31% chance of losing a single round of combat, you'd expect it to take an average of a little more than 3 attacks to lose 1 hit. Continue this line ...
17 attacks = 5.27 hitpoints lost (on the average)
So, in an average situation (50% of the time), it would take 17 attacks to kill a 5 hitpoint Defender. That's 17 attacks, 5 of which the attacker wins, leaving 12 attacks he loses. 50% of the time, you would kill 3 Vet attackers, and the fourth one would win.
If you kill 3 Vets 50% of the time, it just doesn't make sense that there would be a 0.02% chance of killing 6 Vets (particularly since we have a report of it happening). It does make sense that there would be a 25% chance of killing 6 Vets. I'm not saying 25% is the answer to the question; I'm just saying it has a good feel to it.