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Phil Miller
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Posted - 2007.12.12 20:07:00 -
[1]
Does anyone know the exact % that an ECM would jam a ship?
Lets say I have 25points and the enemy ship is at 28, what are my %s?
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Hannobaal
Gallente
Igneus Auctorita
GoonSwarm
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Posted - 2007.12.12 20:16:00 -
[2]
Roughly 89% in that example.
Each module works separately though. So, if you have two ecm with a strength of 10 each and you target ship has a sensor strength of 20, you get two separate 50% chances of jamming when you activate each module.
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James Lyrus
Lyrus Associates
Enuma Elish.
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Posted - 2007.12.12 20:19:00 -
[3]
ECMs are chance based per jammer.
The chance is very simply, jammer str / ship sensor str.
Using, of course, the appropriate jammer type for the sensor in question.
... and that's it.
Str 2 jammer, vs str 20 sensor = 2/20 chance of success (10%).
Jammers do not stack, in any way shape or form, so you need to treat multiple as a compound probability.
To take the example above, and lets say we have 4 jammers, the chance of a jam is 1 - the chance that all the jammers fail to jam.
Chance of failing on each jammer, is 0.9 (1 - 2/20).
Chance of failing on _all_ of them, is:
0.9 x 0.9 x 0.9 x 0.9 = 0.65 = 65%.
Therefore your chance to jam your opponent, is 35%, if you fire all 4 jammers at it.
In practice it's slightly more complicated, as their locking time is also a factor, as is whether you're staggering your jammers or not.
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Crane needs more grid
249km locking?
GMP and TNP |
Niques Leutre
Deep Core Mining Inc.
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Posted - 2007.12.12 20:20:00 -
[4]
Edited by: Niques Leutre on 12/12/2007 20:20:36
Each ECM module counts as it's own dice roll. So for example, say you had three [4 pts. Jamming Strength] Multispectral jammers. Your target is a cruiser with [15 pts. Sensor Strength].
Each of your modules has 4:15 odds of succeeding per cycle (chance-based, like a dice roll), and NOT a combined single-roll of 12:15 odds. You will always have at least a slim chance of failing each cycle due to the fact that ECM is still based on the luck of the roll.
Edit: James beat me to it, and with a longer post! 
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When Newbies Attack! -- A True Story |
Phil Miller
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Posted - 2007.12.12 20:36:00 -
[5]
Thanks, exactly what I was looking for.
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Dexton
Caldari
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Posted - 2007.12.12 22:44:00 -
[6]
 Originally by: James Lyrus
Chance of failing on _all_ of them, is:
0.9 x 0.9 x 0.9 x 0.9 = 0.65 = 65%.
Therefore your chance to jam your opponent, is 35%, if you fire all 4 jammers at it.
You sir did not pay attention in Math when they covered probability.
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Grimpak
Gallente
Trinity Nova
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Posted - 2007.12.12 22:48:00 -
[7]
Originally by: Niques Leutre
Edited by: Niques Leutre on 12/12/2007 20:20:36
Each ECM module counts as it's own dice roll. So for example, say you had three [4 pts. Jamming Strength] Multispectral jammers. Your target is a cruiser with [15 pts. Sensor Strength].
Each of your modules has 4:15 odds of succeeding per cycle (chance-based, like a dice roll), and NOT a combined single-roll of 12:15 odds. You will always have at least a slim chance of failing each cycle due to the fact that ECM is still based on the luck of the roll.
Edit: James beat me to it, and with a longer post!
this.
each jammer has it's own dice. so treat each jammer independently.
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planetary interaction idea! |
James Lyrus
Lyrus Associates
Enuma Elish.
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Posted - 2007.12.12 22:52:00 -
[8]
Originally by: Dexton
Originally by: James Lyrus
Chance of failing on _all_ of them, is:
0.9 x 0.9 x 0.9 x 0.9 = 0.65 = 65%.
Therefore your chance to jam your opponent, is 35%, if you fire all 4 jammers at it.
You sir did not pay attention in Math when they covered probability.
What did I miss?
2/20 jam str, with 4 jammers ...
1 - ( 1 - 2/20 ) ^ 4 = 0.3439
Kindly enlighten me?
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Crane needs more grid
249km locking?
GMP and TNP |
Sirius Problem
Darkness Inc.
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Posted - 2007.12.13 00:55:00 -
[9]
Edited by: Sirius Problem on 13/12/2007 00:55:28
Originally by: Dexton
Originally by: James Lyrus
Chance of failing on _all_ of them, is:
0.9 x 0.9 x 0.9 x 0.9 = 0.65 = 65%.
Therefore your chance to jam your opponent, is 35%, if you fire all 4 jammers at it.
You sir did not pay attention in Math when they covered probability.
He's using joint probablity because each jammer is independent, and its success or failure does not influence the success or failure of any other jammer. The formula is correct.
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I am Super Cool
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