Rikki Tikki Traveller said:
Yes, I would include low population "garden" worlds as a "weird" category.
Say POP 3- and ATM 4-9 to give you search criteria.
Damn, not at at work. Luckily that's not too hard to run by hand to get probabilities, if not actual counts.
Pop 3- = unmodified 5 or less on 2d6: 1+2+3+4 or 10/36
ATM 4-9 is a bit trickier due to the dependency on SIZ, but I'm going to ballpark it at 60% since it straddles the mean and a bit more on one side - it's max value will be (5+6+5+4+3+2) =25/36...and since I think more SIZ/ATM rolls than not can result in in a 4-9, I say that will knock it down by no more than 50%.
So.....25/36 *1/2 = 25/72, call it 1/3 .
10/36 * 1/3 ~~ 9% conditional prob.
Now if the main anomaly is simply the profile, done. (which is to say, the anomalous question is "In a civilized and developed area, with people living on crap planets, why the hell don't more people live there")
call it 4/subsector on the average, and one certainly will roll up at least one, likely at least two in an given subsector.
If the issue is of being right close to (say) a hi pop hellhole world.....a bit trickier. assume a max of J-2 separation for the sake of ease. Start with the 4 garden worlds, we'll get 1/10 hexes., plenty of room to avoid overlap and complication. So we can worry about the single case -start with one vacant garden.
It has 6+12 hexes around it which qualify.
a simple hi pop hellhole of any type is a good match to define them as an anomalous pair. (ie Why the hell don't more people live there instead of there ?) Luckily, I have the stats for that above: hellholes, all atm types, hi pop = 4%.
So...if the pair is what you're interested in, the basic probability of at least on hellhole and one garden within 2 hexes it is ROUGHLY the probability of having at least one hellhole within the 18 surrounding J2 hexes of any given empty garden * the liklihood of at least one empty garden in that area. . ( This is an lower estimate, as for sanities sake, it ignores the possibility of of multiple examples occurring of either in the defined space).
The ugly math is thus:
(((1 - ((1- .04) ^18 )) * (1- ((1-.09) ^17)) ) / 2
The 17 is because you can't have two stars in the same space, and once you have the low value member, you go for the high value to maximize the liklihood estimate; the * 1/2 is because you can arrange two items in two ways, but in this case, order doesn't matter -so they both are equiv
I leave the actual calculation as an exercise, unless I brought my real calculator home.
Not much help, I admit........
[edit: Okay, the answer is .208]
Please remember THE NUMBERS FOR THE GARDEN WORLD ARE ESTIMATED BY REAL PROBABILITY TIMES BALLPARK PROBABILITY. So caveat emptor...and if someone runs it, and gets different, Bully ! Use that.