I believe this is correct based on what Mehul had posted about this before he sold the game. Feel free to update us on the formula used ;)
Printable View
Testing equal off to def sends would be of more interest to me.
It's not clear if 2 or 4 random numbers are used (I believe Mehul once said it was 4). It'd be 50-50 at 97% either way assuming this part is still correct. But at an even send, 2 variables would give you 84% success while 4 variables would give you 93% success. This should be enough difference to be able to tell which it is with enough careful attacks.
Yes, 4
To answer some of the questions in the previous Attack Gains Clarification post...
* There is no penalty for sending overkill
* The "random factor" simply adds and subtracts a small percentage from both the attacker and defender points. By doing it to both, it means that most likely random factor is no-change, and larger changes occur less often. For example, if the random factor was 100%, then we'd basically modify the formulas as such:
Offense = Offense * (1 + random-number-between-0-and-0.5 - random-number-between-0-and-0.5)
Same for defense.
Worst case, Offense would get a -50% disadvantage and a Defense would get a +50% advantage, meaning the random factor is 100%. However, the likelihood of this extraordinarily small as 4 random numbers are involved.
NOTE: THE RANDOM FACTOR IS NOT 100%
Mehul
dig dig dig
FWIW, the probability distribution of the random factor isn't flat, or
even a normal/bell curve distribution. It's triangular, resulting from
statement like "points = points - random(0.035) + random(0.035)".
Love you Bishop.
This. If anyone's interested in collecting the data (maybe EoA if you're in the ghetto?) I'll be happy to do the maths and work out what the underlying distribution is for the random factors. Now we know there are 4 of them and how they add it should be possible to work out with just the 1 extra data point (success at equal off/def). I am assuming the same distribution for each factor - if each had a different distribution it would be messy lol
Not to necro (well the thread wasn't really dead, just a little cold), but if anyone is interested I finally got round to working this all out (assuming the four random numbers are distributed uniformly which makes sense - else why have the two extra ones?)
It's late now but I'll post some results (and triple check my working) tomorrow probably. (It does pass the test of giving 0 at 90.44%, 50% at 97% and 100% at 104.04% so I'm fairly confident I've nailed it - else I wouldn't be posting!)
Some quick results:
sending 103% gives you 99.8% probability of success
sending 102% gives you 98.5% probability of success
sending 100% gives you 87.9% probability of success
Can't get my img to work, but this should link you to an image of the graph of RatioSent vs. Probability of Success: http://pasteboard.co/22MHaOrs.jpg
(line isn't smooth as I've just joined the points linearly)
Hope it's of interest to someone.