Fishing in the Desert, or Where Not To Build Some

[Avogadro]

I agree with whip, no doubt this game is streaky.
You guys are good at math can some ont tell me the odds of this "A warrior fortified in my land(flatland), it is regular. I hit it with 5 regular archers in a row...Warrior survives and is now a vet who lost but 1 pip."
Am I paranoid or am I smart never to play this ladder person again?

BTW another good article, I also am surprised at what I find in opponents town

[Earl Harewood]

the probability of one warrior defeating an archer that attacked it on flat land without losing a hp is, 0.037 or something to that effect, am i wrong?

[Avogadro]

so 5 regular archers?
btw this opponent is no longer on the ladder

[Don K Hotay]

yeah, probability is low for just 1 archer not being able to take of even 1 hp off a regular warrior (1/27 ~ 3.7%). For 5 regular archers in a row, man, that's damn unlucky (or lucky): (1/27)^5 or appoximately 7/1000000 % . But if you say your warrior lost a hitpoint, then it's a little more likely ;D

[Avogadro]

HIs warrior. He was in my land so I hit it with 5 archers. He became a vet lost 1 hit-point and took my capital beginning next turn.

[Don K Hotay]

ouch. I've had similar things happen with jags and impies, but never a regular old regular warrior.

[LordPhan]

First off a little known fact is Plains and Grassland do have a defensive modifier. 10%

Secondly about the streakyness that is due to the 'random' numbers. I put them in quotes because they arn't random at all. It's all done via a mathematical formula based on the current Milisecond. So if the game makes a large set of predetermined "random" numbers, since the processors nowadays can make multiple calculations per milisecond. What you get is, streaky numbers.

[LordPhan]

And to answer the question, the odds of a Reg archer attacking a Reg warrior on grassland or plains, is 75.7270% vs 24.2730% for the warrior.

If it was across a river it would change to 67.7389%
for the archer and 32.2611% for the warrior.

[Avogadro]

so the odds of 5 archers losing?

[LordPhan]

Sorry I don't know the formula for that.

[Don K Hotay]

A regular warrior can win a battle against an archer in three different ways. The warrior can lose 0 hp, 1 hp, or the warrior can lose 2 hp. Add those up and you'll get something around 25%. But the probability of a warrior losing 0 hp is about 4%. For a warrior not to lose a single hitpoint against 5 archers is about 7/1000000.

[Earl Harewood]

ive ALWAYS hated probability, now i know why! lol

[Don K Hotay]

So you are a fan of probality. Well lets say that the lucky regular warrior has succesfully defended against the unlucky regular archer's attack, while fortified on a plains square. What is the probability then that it has not lost any HPs knowing that it has won the battle. It is fortifed, so it has a defense of 1.1 as Phanny pointed out ;D

[cmdishr]

Since they are independent results the odds dont change for each occurance...in other words the odds of a 1hp 2hp warrior beating a archer are?? these odds would be exactly the same for each battle. of course unless there is a battle weariness factor that Im not sure about.

[Don K Hotay]

Well remember, each battle consists of the rng duking it for each hitpoint. In the case of the fortified regular warrior on plains square against an archer, each hp of the warrior has a 2/3.1 (~64.5%) chance of being taken out by an archer with an attack of 2 (numerator). The archer has to take out 3 hps to win the battle, so that's when you begin multiplying these independent events (losing an hp). If the archer lost no hp in the attack, then it would be (.645)^3; if the archer lost 1 hp, then it would be (.645)^2 * (.355), etc...