# Recent content by Magpie

It really isn't

Sure but what that gives you is the probability that at least 1 and no more than 3 leaders are created. It doesn't tell you the most likely (expected) number of leaders that would be created per turn, which is 0

To work out what the expected number of leaders created you have to work out the probabilities for the various number of leaders created. Then from that you can work out what the most likely outcome is The table is : So the most likely outcome is no leader at all, second most likely is 1...

That's not how probability is calculated though. The formula of the situation is the binomial probability function that I posted earlier

I think you add +1 for every squad equivalent overstacked 16 HS = +5 ???

Yes so the first thing you need to calculate is the probability of rolling 1 or more 2's out of 16 attempts. That is the formula I posted earlier.

"What you have calculated is the p >=1 event occuring " Isn't that exactly what you are needing to know ? In 16 trials what is the probability of at least one event Event being an Orig DR of 2
8. ### Followup Commissar question

Not a lot of nuance in mediocrity I guess

The probability of at least 1 occurrence though is p = 1 - (1 - .027)^16 -> that's about 0.35 Where do you get 16/36 from ?
10. ### Followup Commissar question

Because 0 leaders are just doing the minimum, not being inspirational.
11. ### Commissar question

"For example, All level 0 DRM leaders are featureless vanilla. None of them ever lose concealment for being a leader in a hex because they don't have a drm so don't use it, although technically I guess they are mandated to use a drm of 0. Not very nuanced in either case. " I guess the 0 is an...
12. ### Commissar question

Which is exactly why I am suggesting a similar thing would apply to the Commissar.