my simpler math for the lunar has it having 2 WB dice and 2 lance dice against a sword moving away. this on average is .66 hits from batteries and 1 hit from lance for a total of 1.66 hits. on average. given that it takes 2 hits to kill this means 0.83 chance of a sword kill from an lunar.
against an eldar escort it is 1 WB die and 2 lances for an average of 0.5 hit from the battery and 1 hit from lance, reduced to 0.166 hits after holofield saves for a total of .666 hits on average. given that it only take 1 hit to kill this remain 0.66 eldar kill from a lunar.
versus closing escorts it comes to 0.99 hits from batteries and 1 from lance for a total of 1.99 hits vs the sword (0.99 chance of a kill from a lunar) versus 1 hit from the batteries and 0.166 damage point from the lance on the eldar escort (1.16 chance of a kill). so here the lunar is more effective against the eldar, but this situation should rarely if ever happens with MSM.
This is absolute nonsense, as Sigoroth points out. Did you even think about your maths? Take the first part: A Lunar does an average of 1.66 hits, this I'm not disputing, but you then say it has an 83% chance of a kill as a result. So in spite of being not-enough-damage-for-a-kill on average, it will be killed more than 8 times in 10? Did this not ring alarm bells?
And a 1.16 chance of a kill on an eldar escort? There's no such thing as a chance with greater certainty than 1!
This is how you do it properly:
There are 16 unique combinations of four weapons either hitting or missing. 11 of these result in the destruction of at least 1 escort:
0B, 1WB 1WB, 2WB, 1L, 1WB&1L, 1WB&1L, 2WB&1L, 1L, 1WB&1L, 1WB&1L, 2WB&1L, 2L, 1WB&2L, 1WB&2L, 2WB&2L
Many of these results are indistinguishable though: 1WB twice, 1L twice, 1WB&2L twice, 2WB&1L twice, 1WB&1L four times, giving 9 unique outcomes, with the probabilities:
P(0) = P(WBMiss)*P(WBMiss)*P(LMiss)*P(LMiss) = 2/3 * 2/3 * 1/2 * 1/2 = 0.1111
P1WB) = 2*P(WBHit)*P(WBMiss)*P(LMiss)*P(LMiss) = 2 * 1/3 * 2/3 * 1/2 * 1/2 = 0.1111
P(2WB) = P(WBHit)*P(WBHit)*P(LMiss)*P(LMiss) = 1/3 * 1/3 * 1/2 * 1/2 = 0.0277
P(1L) = 2* P(WBMiss)*P(WBMiss)*P(LHit)*P(LMiss) = 2 * 2/3 * 2/3 * 1/2 * 1/2 = 0.2222
P(1WB&1L) = 4* P(WBHit)*P(WBMiss)*P(LHit)*P(LMiss) = 4 * 1/3 * 2/3 * 1/2 * 1/2 = 0.2222
P(2WB&1L) = 2* P(WBHit)*P(WBHit)*P(LHit)*P(LMiss) = 2 * 1/3 * 1/3 * 1/2 * 1/2 = 0.0555
P(2L) = P(WBMiss)*P(WBMiss)*P(LHit)*P(LHit) = 2 * 2/3 * 2/3 * 1/2 * 1/2 = 0.1111
P(1WB&2L) = 2* P(WBHit)*P(WBMiss)*P(LHit)*P(LHit) = 2 * 2/3 * 2/3 * 1/2 * 1/2 = 0.1111
P(2WB&2L) = P(WBHit)*P(WBHit)*P(LHit)*P(LHit) = 1/3 * 2/3 * 1/2 * 1/2 = 0.0277
Now you can either read the chance of destroying an escort straight off of there by adding up the 11 results that represent at least 2 hits, or you can collate and tidy the results further first:
P(0Hits) = P(0) = 0.1111
P(1 Hit) = P(1WB) + P(1L) = 0.3333
P(2Hits) = P(2WB) + P(2L) + P(1WB&1L) = 0.3611
P(3Hits) = P(2WB&1L) + P(1WB&2L) = 0.1666
P(4Hits) = P(2WB&2L) = 0.0277
And:
P(0 Escorts Killed) = P(0hits)+P(1Hit) = 0.4444
P(1 Escort Killed) = P(2Hits) + P(3Hits) = 0.5277
P(2 Escorts Killed) = P(4Hits) = 0.0277
P(At least 1 Escorts Killed) = 1-P(0 Escorts Killed) = 0.5554
These are nowhere near your figures. You can tell these are correct, because if you weight each probability by the number of htis it represents and add them together, you get an average value of 1.667 hits as expected.
If you only want to know the chances of "At least one" for a given number of trials, I use the following procedure:
P(Failure) = 1-P(Success)
P(0 Successes) = P(Failure)^Trials
P(At least 1 Success) = 1-P(0 Successes) = 1-P(Failure)^Trials = 1-[1-P(Success)]^Trials