You need 600 ADM and 400 DIP for your first colonist. Starting with your 3/2 ruler with NF on ADM and +2 advisor both in ADM and DIP (you need to disband everything to have +3, wouldn't be really worth it and you wouldn't have the required amount of DIP) you'd get +10 ADM and +6 DIP a month. So you'd see at average one 6/6/6 ruler with 6/6/6/6 leader stats in somewhere around 1 500 000 complete playthroughs as a dutch republic. Furthermore, if you say, you get one 6/6/6/00 rolls and you have a Dutch republic and only electing Statists for the entire time, you'd get 94 elections in a complete playthrough, meaning you'd get 1 6/6/6 ruler in roughly 3.64 playthroughs. If you're rolling generals you would get 1 6/6/6/6 general per 400000 rolls (rough estimation). That would cost 50 billion military points, which equal to assuming you have a 3 military ruler with +1 military advisor all the time - you'd then generate in a whole game 31598 Military points - 1 582 379 complete playthroughs to get only SEVEN 6/6/6 ruler with 6/6/6/6 leader stats.

Considering he didn't even had 100 AT and no siege bonus, I'd guess the real probability of rolling 6 siege is maybe at 1 in 400 tries, perhaps even less.Īssuming ruler stats and leader stats don't correlate, we have a probability of somewhere around 7*10 -9, meaning, you would get 7 of them in a billion rolls. What about siege though? I rolled in my Nepal game with +2 siege in 40 to 50 tries 2 6 siege generals at 100 AT.

How likely is it to roll a general at 6/6/6/6? According to this and assumimg fire, shock and maneuver pips are identically and indepently distributed, the probability of having 6/6/6/x leader would be somewhere between (6/100) 3 = 216 per million and (1/10) 3 = 1000 per million. There might be another distribitution, which could be weighted more towards the mean value 3, in that case it would be even more unlikely. Even if you ignore the other stuff and just focus on the stats of leader and general: How likely is it to roll a 6/6/6 ruler? IF those stats are independently uniformly distributed, that would be a probability of (1/7) 3 = 1/343, that is somewhere around 29 per thousand.

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