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I have 5 Hun armies of about 20-50 men that I can't get rid of because as soon as I try to fight them (even with just a single unit), they just withdraw off the battlefield. Even if they have already withdrawn once on the strategic map.
Assassins are no good either because even those that I have trained up to level 4 only have < 10% chance.
So how do I get rid of them, and how come the AI can withdraw in situations where I can't? Bug?
Try automated battles. It might cost you more in casualties but if the hun armies are so small then it shouldn't be a problem. Alternatively, you could completely surround them on the strat map so that they'll be forced to fight to the death.
I "fixed" that problem my gathering an all heavy cav army and rolled over Huns, Slavs, Lombardii you name it ~:)
Of course that might not be possible with factions that doesnt have as heavy heavy like Clibinarii but if it just horsearchers then you dont need that good cav to kill them in melee.
Edit: oh and concentrate on killing their leaders as they will run out of them at one point and then the faction is dead.
CBR
CBR, his problem is not beating them in battle, but the fact that they are somehow able to withdraw from it despite being out of movement points, which is likely a bug.
Yes I do find it odd how they would always withdraw. But maybe its because its only horsearchers. Are armies with generals also withdrawing like that?
CBR
That's odd, I've never seen anything like that before.
Especially thoose. I found out that auto-calc won´t work. They always escapes and since a general will regain casualities and grow until the next turn you´ll be faced with an never ending chase. Only way to get rid of them is to kill them off in a player controlled battle. I use a combo of 6 eastern archers (good hit points), one Bosphorian merc and 4 auxiliary cavalry for theese ops.Quote:
Originally Posted by CBR
Approach the general. Fire at him:
1.(90%) He attacks your infantry. Outflank with AC and smash into his rear.
2. He flees, pursuit with AC since they are faster you´ll likely to intercept him.
In my campaign I found that I could pretty much always engage the general. A few times I didnt get to kill him before the unit routed but I just surrounded the unit to prevent that from happening.
Edit: what battle difficulty are people using? I use normal (or medium whatever its called)
CBR
It's a horde. It can live off the land. Their ability to slink away is a major annoyance, but not a bug.
Bribe them, or offer them a ceasefire. If it costs to much to bribe a leader, assasinate the leader. So what if there's only a 10 percent chance? That means there's a 19 percent chance with two attempts, a 27 percent chance with three attempts, etc., etc.
The Goth remnants took my ceasefire offer, aren't doing me any harm and were were promptly attacked again by the Huns.
Actually your own units can also flee when they are out of MP.
I have had several units in transit get ambushed, they have more or less all managed to get away. But I always fight first, I find it a bit cheesy to run away at first sign of trouble. So I have lost not only good units but also a competent general to interdictions and ambushes.
Hmm, I guess you´ve to clarify this math for me. How can a 10% chance each time become a 27% one at the third attempt?~:eek:Quote:
Originally Posted by Doug-Thompson
When you fight a battle without MP, the "withdraw from battle"-button is greyed out. Units can only withdraw when they are beaten. The AI however can - and could in vanilla RTW - withdraw without fighting.
BTW, you can use a similar trick. If your army is in danger but has still MP, do not click the retreat option but let it come to battle. As soon as the battle starts, withdraw all your units. The consequence is that your army will retreat to save territory. The AI uses this exploit too.
He didn´t say 27% chance on the third attempt, but that chance with three attempts.Quote:
Originally Posted by PseRamesses
The chance of failure for one attempt is 0.9. For two attempts 0.9*0.9 and for three attempts 0.9*0.9*0.9=0.73.
Yes you can do that... But if you'll look in my early thread on BI I had a small battle with the Huns where they interdicted a unit of Hippo toxotai. I tried to withdraw them, but the Huns just came after me again (hey isn't that new?). So I fought and lost... Thought my men were doomed, but as it turned out I had a part of the unit escape and it has since been retrained.Quote:
Originally Posted by A.Saturnus
Try it out.
Maybee my understanding of English is really bad but how can you sum it up like that? First attempt 9%, 2nd 9%, 3rd 9% etc etc you can´t add each chance togheter to increase your chances. Then the math would look like this: 9/100, 18/200 and 27/300 = 9% each time, right?Quote:
Originally Posted by A.Saturnus
Or do he mean that using three different spies at the same time will actually increase the chances of killing a vic? So if I use 10 spies making 10 attempts at the same time I will actually have a 90% chance. This can´t be right.
A.Saturnus is right, PseRamesses. It's a little confusing, but it's the basis of probablity.
Suppose a coin is flipped once. The chance of it landing on heads is 50/50. Suppose you flip it again. What are its chances of landing on heads? They're 50/50.
So, what are the chances of tossing a coin twice and having it land on heads both times?
First, you have to get heads on the first toss. So, 50 percent of the time, that will fail. You only have a 50 percent chance of getting the first result you want. Of those 50 percent, only half of those will also have a heads tossed on the second tossed. Therefore, 50 percent of 50 percent is 25 percent. You have a 25 percent chance of tossing two heads in a row.
Thx DT, it´s now clear to me what you meant by your original formula. As I said, english isn´t my native tounge, and I got the feeling you presented the probability for a kill with numerous assasins in that post and that I´d missed out on some stats regarding this. Sorry for the misunderstanding m8.Quote:
Originally Posted by Doug-Thompson
think about it this way. List out all the possible outcomes
On a 2 assassination attempt there are 4 results
SF
SS
FS
FF
So to have a successful assasination in 2 attempts, you would either have to have SF, FS, or SS. What are the chances of those? You alraedy know that the chance of succeeding on the first attempt is 10%. And on the case that the first one don't succeed (90%), you could still be successful by succeeding on the second try, and the possiblity of that is 0.9 x 0.1, which is 0.09. Add them up, you get 0.19 (19%)
You could do that with 3 attempts too, but if you notice, the chance of success is just 1 minus the probability of failing every attempt, since you fail the assassination if you don't succeed on every attempt. So on a 3 attempt assasination, it'll just be 1 - (0.9^3), which is 0.271
But, I don't think the assassination possibility is very accurate. I think it's more of a arbitrary value. I've had a success rate of 50%, but after 30 attempts with reloading, he still fails everytime. I think the computer determine whether the assassination would be successful before you could actually attempt it, so the number actually mean nothing because the computer already determined whether you would succeed or not.
% is not that simple.Quote:
Originally Posted by Taiwan Legion
The fact that you try more does not mean that you increase or decrease your %
The assasin will still only have 10% chance as he is independant of the 1st and the 1st assasin No longer enters the equasion,
At 10% chance that would mean 10 assasins have a 99.8% chance of sucsess (or similar)
Where in actual fact they have 10% chance,
And there all independant of the others
Don't leaders who survive assassination attempts get traits or ancillaries making them harder to kill ?
Yip common mistake with the % chance, the coin being the best example. Its all because the first toss which is of course 50/50 heads or tales is completly independent of from the 2nd , 3rd , nth toss, which all have the the same 50/50 chance heads /tales.
As for reloading and trying again does the game not have some kind off "seed" to produce the same results if you try this method?
Indeed... Throw a million dices and you never get a single 6, what is the chance of it actually comming up with the million and first throw? One in six of course.Quote:
Originally Posted by red comyn
And since leaders tend to get a whole lot of protection very fast it is not a good idea to try to assassinate him with low chances.
Shambles, everything Taiwan Legion said is indeed correct. If you use 10 assassins, each of the assassins has 10% chance to succeed. Each of them. That means, the expected value of successes is 1. The chance that none of them succeeds is 34,7%.
Yes. That's a very good point. I should have mentioned that.Quote:
Originally Posted by Magussen
It's actually a trait-chain that keeps improving the more unsuccessful attempts are made. So if you try to use bad assassins for the job, the victim will quickly develop from watchful to obsessed by security and from scant trust to paranoid, making him damn near impossible to assassinate even with highly skilled agents.
I don't have the cash to buy them, so I suppose I'll have to keep chasing them around with single units until I can get some assassins trained up.
It's a pain because I had a single army defeat almost the entire hoard in 1 round and now I just want to clean them up so I can get on with things. The AI kept throwing 1 stack at a time against my army which was in a good defensive location.
6 Goth raiders, 8 Goth spearmen, 1 Goth lancers and the general. They went from completely green to nearly all having a silver chevron and I only lost about 30-40% of the men in the process. :duel:
No I think I stand correct. If you think about it, my calculation does base on the fact that each event is independent. Lets assume that the general don't get new traits to keep it simple, so each assassin has a .1 chance of success. And how does 10 assassin with .1 probability have .998 expected value? E(.1) = 0.9^10 because basically you fail the whole assassination scheme if all of your 10 attempts fail. And you can't just use 0.1 to multiply, because technically, if your first attempt fail, you still succeed if ANY of your subsequent attempt succeeds. You would be ignoring them if you just use .1 to multiply.Quote:
Originally Posted by Shambles
Since each of the attempts is independent, and to find the expected value of independent events, you multiply the probability together. I mean, if your first attempt fail, then your chance of success is even lower. since now instead of it being .9^10 for failure, it's only .9^9 now. So the probability of failure and success changes as you carry out the assassination.
I certainly hope that I'm correct. I'm an actuarial minor, and if i'm wrong, then I have to seriously think about my future ~:eek: