Originally by: bldyannoyed

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So what you are essentially saying is that the chance of getting permajammed by a Falcon (or indeed any ECM) that is using MORE THAN ONE jammer is significantly higher than the numbers would at first suggest?

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Originally by: lecrotta

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Originally by: bldyannoyed

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So what you are essentially saying is that the chance of getting permajammed by a Falcon (or indeed any ECM) that is using MORE THAN ONE jammer is significantly higher than the numbers would at first suggest?

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The chance of jamming a ship by using multiple jammers of the race that ship just happens to be has always been high or what would be the point of fitting a rack of a single racial instead of a mix?. Although the trade off in doing so limits your effectiveness against other races significantly.

I am not sure what this has proven tbh as the figures are simple and obvious, a 6 rack of a single races jammers can jam a low number (2) of that particular races ships very effectively is hardly a epiphany imho.

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Quote:

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This results in:
Classical caluculation:
both targets permajammed during the first 40 seconds:
0.58618164
During the first 60 seconds:
0.4487
During the first 120 seconds:
0.2013

Using the real probability estimation (estimated by a few million runs..)
During the first 40 seconds:
0.7744
during the first 60 seconds:
0.681472
During the first 120 seconds:
0.4644

So the real probability of beeing permajammed by a falcon for 2 minutes in the situation the model is seeted in is more than twice as high as the wrong and simple formula returns.

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Quote:

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And this is not even factoring in overlapping jams and relock time

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Quote:

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So happy flaming

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Originally by: chrisss0r

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As I stated already numbers alone will never prove if something is overpowered or not.
Though my personal opinion is that falcons are overpowered it has nothing to do with this calcs.

i just wanted to proof that you need the bayesian calcs in small gang situations and get the numbers the whole discussion is partly based on straight. As i already said everyone who is able to put more variables into the code is very welcome. Due to my incompetence when it comes to code i can't do it

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Quote:

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This means, that, IF BOTH opposing battleships are using neither ECCM, nor sensor boosters, in the end your opponent will have 70% of a battleship damage against your full Battleship damage(100%). You have a clear advantage, as you should, or else what would be the point in having a balanced gang. This advantage is not huge, though, and you have nothing near a permajam in either opponent.

Now, if both battleships were using ECCMs, your gang (Falcon+Battleship) would be in trouble. You can repeat the calculations for a 0.25 base chance and you will see that in this case, the two battleships will have a clear advantage. Showing that the correct counter DOES WORK.

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Originally by: Cohkka

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Nice post, keep up the good work.

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Originally by: Etho Demerzel

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Lets suppose that you have two gangs. One is composed by 2 battleships, the second is composed by a falcon and one battleship. By your own results in 120 seconds you will be able to jam both ships 46.44% of the time. If you facto lock time from battleship to battleships without any help from modules, we can overestimate this jam time to be, lets say 65%.

This means, that, IF BOTH opposing battleships are using neither ECCM, nor sensor boosters, in the end your opponent will have 70% of a battleship damage against your full Battleship damage(100%). You have a clear advantage, as you should, or else what would be the point in having a balanced gang. This advantage is not huge, though, and you have nothing near a permajam in either opponent.

Now, if both battleships were using ECCMs, your gang (Falcon+Battleship) would be in trouble. You can repeat the calculations for a 0.25 base chance and you will see that in this case, the two battleships will have a clear advantage. Showing that the correct counter DOES WORK.

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Originally by: Etho Demerzel

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Now, if both battleships were using ECCMs, your gang (Falcon+Battleship) would be in trouble. You can repeat the calculations for a 0.25 base chance and you will see that in this case, the two battleships will have a clear advantage. Showing that the correct counter DOES WORK.

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Originally by: Arthur Frayn

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How much to ruin all your holes, luv?

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Originally by: Camilo Cienfuegos

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I would be interested to see if this is the case. Looking at the above figures, I have a feeling that if the 2 battleship gang has any advantage, it will be marginal (~5%) at best. I suspect that the Falcon + Battleship combination might well have the upper hand by a percentile or two.

Key word there is suspect of course.

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Originally by: Terianna Eri

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Originally by: Etho Demerzel

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Now, if both battleships were using ECCMs, your gang (Falcon+Battleship) would be in trouble. You can repeat the calculations for a 0.25 base chance and you will see that in this case, the two battleships will have a clear advantage. Showing that the correct counter DOES WORK.

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Can we see the calculations for 0.25 jamming chance, 6 available jammers, against 2 targets?

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Originally by: chrisss0r

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#!/usr/bin/python
# -*- coding: utf-8 -*-
import random

class moo:
def main(self):
# initializing of success counter
amount_successes = 0

# outer loop for repetition
for i in range(1000):

# first flip. 0 is heads, 1 means tails
c1 = self.flip()
c2 = self.flip()

# loop for remaining throws
amount_throws = 0
while amount_throws < 4:
if c1 == 0:
c1 = self.flip()
amount_throws += 1
if c2 == 0:
c2 = self.flip()
amount_throws += 1

if (c1 == 1) and (c2 == 1):
amount_successes += 1
break

print("Amount of successful tries: "+str(amount_successes))

return

def flip(self):
return random.randint(0, 1)


if __name__ == "__main__":
moo().main()

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Even if that was the case and they barely break even in targetable damage. You will end with 100% damage + the damage of non target weapons against 100% damage, and 200% EHP against 100% EHP. An lets not forget that the falcon will be at 120 km tops because it will be using multispecs, and thus reasonably vulnerable to cruise missiles, for example.

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Originally by: Camilo Cienfuegos

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If we assume (a common scenario) two close range battleships fitted with ECCM versus one close range battleship with a long range support Falcon, the battleships aren't going to be able to touch the Falcon (a mechanic which to my mind is working perfectly). If we're seeing roughly equal amounts of damage going out against both side, provided the single battleship doesn't foolishly decide to attack both his opponents simultaneously, he has every chance of finishing off one of those battleships before he himself goes down. That then leaves Battleship (admittedly almost dead) + Falcon (with three jammers now free). The chances of the now lone battleship winning despite it's full HP are far less as there are now six and not three jammers on him; the almost dead battleship could probably warp out at this point if the need is there, or provided there's good communication might be able to use the time purchased by the Falcon to repair whilst keeping his target pinned.

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Originally by: Camilo Cienfuegos

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It's crucial that we can demonstrate the exact figures so that we can see which side this falls on. If it falls in favour of the Falcon + Battleship, it's indisputable that there's a big problem there.

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#!/usr/bin/python
# -*- coding: utf-8 -*-
import random

class moo:
def main(self):
# initializing of success counter
amount_successes = 0

# outer loop for repetition
for i in range(1000):

# first flip. 0 is success, 1-3 failure
c1 = self.flip()
c2 = self.flip()

# loop for remaining throws
amount_throws = 0
while amount_throws < 4:
if c1 != 0:
c1 = self.flip()
amount_throws += 1
if c2 != 0:
c2 = self.flip()
amount_throws += 1

if (c1 == 0) and (c2 == 0):
amount_successes += 1
break

print("Amount of successful tries: "+str(amount_successes))

return

def flip(self):
return random.randint(0, 3)


if __name__ == "__main__":
moo().main()

Quote:

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i feel a BS + ewar ship VS 2 close range BS should be favorable towards the ewar/BS team as its a much more tactical setup than just tank/gank.

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Originally by: Camilo Cienfuegos

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If we assume (a common scenario) two close range battleships fitted with ECCM versus one close range battleship with a long range support Falcon, the battleships aren't going to be able to touch the Falcon (a mechanic which to my mind is working perfectly). If we're seeing roughly equal amounts of damage going out against both side, provided the single battleship doesn't foolishly decide to attack both his opponents simultaneously, he has every chance of finishing off one of those battleships before he himself goes down. That then leaves Battleship (admittedly almost dead) + Falcon (with three jammers now free). The chances of the now lone battleship winning despite it's full HP are far less as there are now six and not three jammers on him; the almost dead battleship could probably warp out at this point if the need is there, or provided there's good communication might be able to use the time purchased by the Falcon to repair whilst keeping his target pinned.

It's crucial that we can demonstrate the exact figures so that we can see which side this falls on. If it falls in favour of the Falcon + Battleship, it's indisputable that there's a big problem there.

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Originally by: Camilo Cienfuegos

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Quote:

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i feel a BS + ewar ship VS 2 close range BS should be favorable towards the ewar/BS team as its a much more tactical setup than just tank/gank.

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It certainly is, which is why if it's mathematically stacked in favour of them as well as tactically, I believe we have a problem.

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Originally by: Etho Demerzel

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Originally by: Camilo Cienfuegos

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If we assume (a common scenario) two close range battleships fitted with ECCM versus one close range battleship with a long range support Falcon, the battleships aren't going to be able to touch the Falcon (a mechanic which to my mind is working perfectly). If we're seeing roughly equal amounts of damage going out against both side, provided the single battleship doesn't foolishly decide to attack both his opponents simultaneously, he has every chance of finishing off one of those battleships before he himself goes down. That then leaves Battleship (admittedly almost dead) + Falcon (with three jammers now free). The chances of the now lone battleship winning despite it's full HP are far less as there are now six and not three jammers on him; the almost dead battleship could probably warp out at this point if the need is there, or provided there's good communication might be able to use the time purchased by the Falcon to repair whilst keeping his target pinned.

It's crucial that we can demonstrate the exact figures so that we can see which side this falls on. If it falls in favour of the Falcon + Battleship, it's indisputable that there's a big problem there.

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I agree that the numbers are important. And fortunately chrisss0r just provided them in hsi post above.

Now we have a base chance of 46% against 2 battleships with 1 ECCM each.

Before we start let me list my assumptions:

1) In average the drones account for 25% of a battleship damage
2) The battleships have no other form of untargeted damage as FoF, Smartbombs, etc
3) The relock time increases the effect fromt he ECM from 46% to 60% (look the charts you referred to for validation)

Now using a single battleship damage as 100% the total damage of the two jammed batleships will be 50% (from the drones) + 150% * 0.4 = 60%, or a total of 110% of a battleship damage. As you see the advantage is clearly on the 2 battleship groups, as it should be considering they have ECCM equiped. Note that a single module in each battleship turns the fight in their favor.

Now if you consider what effect 2 tracking disruptors (one in each battleships) used by a curse would have in the same fight you will clearly see why the Falcon is not the best buddy to the single battleship.

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