Cowman, why the divide by 2? The rest all makes sense to me.

.95136=1-(5*.6*.4^4/2+.4^5)

with 75%:

- if there is a spot i consider very important, i attack with 7
- if there is a spot i consider vital, i attack with more than 7

i've seen a 9 or 10 attack fail to take a 2 once. 7 very rarely doesn't take the 2.

it would require 14 armies attacking, at 75% luck, to have 100% chance to win killing 2

13*.6=7.8
7.8-(7.8*0.75)= 1.95

minimum armies killed would be 1.95, thus s light chance to not kill 2, using 13

i think fizzer wrote somewhere about rounding up when doing these calculations. maybe that means 13 kills 100% too

I'm fairly certain that all subsets are not rounded.. ie .95 armies killed = 95% chance of success... but we haven't had a debate involving something like this for months, so it is probably buried somewhere I won't bother to find, lol

Something is off here.

In the first, he has a -7 cumulative offensive luck, in the second one he has -13. These values are too big (I could calculate how improbable it is).

I tried it myself in single player and I got -6.4

Divided by two because if x=1 there is a 50% chance that y=2 and a 50% chance that y=1.

Your minimum kills would be the minimum of possible events in Y when x=0. This would imply(from the formula in a previous post) minKills= floor(np(1-L)). Fixing p=.6 and L=.75 this means minKills = floor(.15*n).

Thus from this I would agree with Perrin's analysis.

The wiki suggests that in the event of a fractional kill, that kill takes place with the probability of the fraction. I.E. 1.95 kills results in 1 kill 5% of the time and 2 kills 95% of the time.

Mathwolf... the huge negative makes sense given the context. He is only ever making the same attack with a high chance of success. As such there is never an opportunity to have positive luck (safe players don't get positive luck by the very fact they are safe).

He would have to repeatedly attack 3 v 2 or something similar in order to see positive luck impact the game.

Makes sense cowman... had completely forgotten about that aspect. Was trying to figure out where it came from forgetting about the luck aspect and instead simply doing the binomial distribution consideration.

Thanks for the answer.

Question..,
What about the good-ole **You've Been Screwed** factor? It sure seems to play a big part in real life..,