Hunger in the year 3058: Feeding Sarna from Kaifeng

Someone challenged me to justify low JumpShip counts in BattleTech. 
I believe I've done so. 
(These other threads are also relevant.)
Warning! This is like my other "bean counting" posts except EVEN MORESO.

Chargers and Banshees and Loss Rates

tl;dr?

• The Charger and Banshee show the same rate of loss per year, which lets us estimate that there are about 1155 Firestarters (or one per battalion), 100 Clints, 150-450 Assassins and 350 Victors walking around in 3025.  
• Stingers use a different and simpler ratio, which lets us estimate that there are 2500 Archers, 1100 Jenners, and probably 100 (but maybe 700) Atlases in 3025. 
• The Wasp's figures are only possible if it were abandoned early on and then revived after the Star League fell.

DropShip Engines

In FASA's 1980s DropShips and JumpShips sourcebook, the engines on the Triumph, Condor, Fortress, Overlord and Excalibur all equal [ship mass]*[thrust]/15 - 50, and the engines on the Seeker, Monarch and Behemoth are close to that.

The Avenger, Buccaneer and Union are at [mass]*[thrust]/15 - 75, while the smallest ships, the Leopard, Fury and Gazelle are right around [mass]*[thrust]/15 - 100. The curve

[mass*thrust]/15 - [(14494*mass*thrust - 11150000) / (311*mass*thrust - 1279000)]

fits most of those engines. I don't expect to fit them all perfectly--when I worked through the JumpShip construction data I found typos, copy errors, and weirder inconsistencies which make that impossible. I'm pretty confident that the Mule and Mammoth engines have a digit wrong, and that the Vengeance's engine was calculated from a thrust of 3, instead of the thrust 4 its listed at. (I'm not the first to notice this about the Vengeance, but danged if I can remember who I've heard it from before.)

I'd feel pretty good about my equation, except that the ships which don't fit are all fast- the Achilles at thrust 8, the Seeker at thrust 5, and the Intruder at thrust 4 (in a band dominated by thrust 3 ships). So I don't think I'm accounting for thrust correctly.

I'd hoped to solve this by Thanksgiving, but que sera sera.

Draconis Combine: Probable World Populations

[Edit, 2019 Feb 28: I have refined my opinions on this topic. I no longer believe the Combine has room for more worlds of 500 million or more people; I now believe Ningxia's 23 million is average for the Combine's 379 "typical" stars, and the lowest 312 of those average 1 million people apiece while the highest 67 average 113 (or maybe 77.8) million apiece. /Edit]

Total Population: 90 billion (75% concentrated on administrative worlds)
Average Population per World: 260 million (30-40 million for non-capitals)
Minimum Population per "Inhabited" World: 1 million

Supposing that a single Behemoth DropShip can satisfy one day's food and water needs for 12.5 million people, and that a round trip takes 1 month, then just 700 Behemoths could supply all the food and water needs for 1/3rd of the Combine's 350 inhabited worlds. (Essentially every world with 5 million or fewer people.)

Ningxia, with 23 million inhabitants, would be something like the 120th most populous world in the Combine. There's about 25 worlds ranging from 100 million to 600 million inhabitants, an industrial world somewhere around 750 million, and another around 1.1 billion.


How Do I Figure?

MW1e: Personnel

MechWarriors see an average of 250 combat encounters (almost half are artillery bombardments, aerospace strafing runs, infantry or minefields)* annually. A career lasts 26 years on average, and about 1600 end (retirement, dispossession or death) each year.

There's 5/6ths as many technicians as active MechWarriors, and their military careers last about 11 years on average.

There's 1/6th as many scouts as active MechWarriors, and they last about 14 months.

Sphere-wide...
About 80 regiments are deployed as full regiments.
About 350 battalions are deployed as separate battalions.
About 600 companies (350 full strength, the rest 9-11 Mechs) deploy independently.
More than 1800 other groups of 8 or fewer 'Mechs are deployed independently.

That's almost 2900 deployments, about 2000 of which are defensive (25/36, using my trick with the events table). The Inner Sphere only has 1700 settled star systems! Too few for a single deployment per system, and too many for a single deployment per world (assuming better than 1.2 inhabited worlds per system). Best explanation, I think, is that relatively few worlds have 'Mechs in permanent garrison, and that each garrison consists of one large group plus multiple small groups, scattered across the world.

MW1e: Mech Production Rates

The old House Marik sourcebook says the Free Worlds League manufactures "an estimated 500 'Mechs per year," and the old House Liao sourcebook lists eight manufacturers totaling 400 'Mechs per year. The other three House books don't say how many 'Mechs they build.

However, from the Attack by Natives table I suspect that at least some tables in the first edition MechWarrior RPG (hereafter MW1e) are carefully non-arbitrary, and from my survey of Notable Pilots I know that numbers in MW1e can correlate systematically with numbers in other books. (Humans, writers especially, like to be consistent and will reuse numbers when they can.) In particular I want to look at the chart where players roll a d36 to choose their starting faction. Chances/36:
7 - Davion
6 - Kurita
5 - Steiner
5 - Marik
4 - Liao
2 - Bandit King
7 - Unaffiliated (i.e. e.g.,  periphery)

I doubt it's a coincidence that Marik gives you 5 chances and produces 5x100 'Mechs/year, or that Liao gives you 4 and produces 4x100 'Mechs/year. I think it's reasonable to take Davion production as 700 'Mechs/year, Kurita 600/year, and Steiner 500/year for a total of 2700 'Mechs produced annually across the whole Inner Sphere.

I also want to look at the "Standard Enemy Forces Table." It's easy to find the chance of any 'Mech turning up in a given lance type, and I've roughly calculated the chance of each lance type turning up during a "Battlefield Encounter."

If I take the ratio of Valkyries to other light 'Mechs encountered as also being the ratio of Valkyries to other light 'Mechs manufactured annually, and assume there are [30% of 2700] light 'Mechs produced annually, MW1e yields an annual production of 128.4 Valkyries. This is close to the 130 cited in TR:3025.

This vindicates my assumption about annual Davion, Kurita and Steiner production rates, and shows that the lance table is at least a reasonable guide for individual 'Mechs. It doesn't have every 'Mech on it, of course, so I'm hoping (for instance) that more precise calculations will push the Valkyrie entry up to 146/year to cover both Valkyrie and Spider production.

BattleForce says that 30% of all 'Mechs are light 'Mechs, but I'm reluctant to assume that same 30% also applies to annual production. Hopefully more precise calculations will let me discard that 30% weighting factor.

MW1e: Attack By Natives

The first edition MechWarrior RPG has a general weekly encounter table. You have a chance to be attacked by things, to receive reinforcements, and a number of other events, all being more or less likely depending on your current assignment.

There's no table to determine the players' current assignment, but you have to figure that most forces are in garrison, so the chances of getting some other kind of duty should correspond with the chance of an equivalent event happening to a garrison. If the Garrison Events column gives you a 3/36 chance of being attacked by natives, then I figure that's equivalent to a 3/36 chance of being assigned to Pacification Duty.

Now, whatever assignment you're on, if you roll the "attack by natives" event and get attacked while in your 'Mechs, you roll 2d6 (then modify the result depending on your assignment) to see what support they get. I wanted to know just how often the natives had 'Mech support.

I noticed something peculiar.

Instead of Vanilla 2d6

My work back here rolling "lower of two dice" made me wonder if I could emulate a 2d6 curve that way. " 7 + (lower of 2d8) -- (lower of 2d8) " turns out to be almost indistinguishable from a 2d6 curve! Were I actually using this die mechanic for something, I'd be tempted to have results of 0 and 14 "explode" with further "lower of 2d8" rolls.

Tabletop play seems to have a tension between providing interesting results (usually means increasing numbers of dice) vs. keeping record keeping reasonable (means keeping pip totals small). World of Darkness and Shadowrun solve this by rolling lots of dice and counting each die as a pip; Betrayal at House on the Hill is similar, in that its dice are numbered from 0 to 2. ...I guess my personal holy grail here would be a simple, manual die mechanic that'll produce a smooth 1/x curve.

2d6 is really easy to use on the tabletop. It does have some problems--cannot roll higher than 12 or lower than 2, "lucky" streaks are pretty common, and "unlikely" results may be more or less common than players think appropriate.

My first thought is that I can break "lucky" streaks by rolling lots of times and discounting all results until a given value has been discounted a certain number of times. (The quota for each value would be proportionate to the chance of rolling that value.) This would be terrible to execute manually but could work with a computerized die roller.

My second  thought is to replace 2d6 with 4d6 / 2. This makes middling results more likely and extreme results much less likely.

I'm not sure I'd want to use any of these as a straight replacement for 2d6 rolls in (say) BattleTech, but I think it might be interesting to designate individual pilots as cautious (4d6/2) or swingy (quotas).
(I'd like to eventually do my own computerization of BattleTech, and I have some thoughts on how I'd modify specific BattleTech mechanics, but that's not a project I can start soon.)

How valuable is range

After an afternoon of trying to calculate exactly how often two randomly placed 'Mechs would be any given distance from each other, I noticed that the distribution was going to look something like sqrt(x^2 + y^2 - r^2), where x and y are the length and width of a rectangular board, and r is the distance between any two randomly selected points on that board. (The actual equation may feature arcsins, and would have some sort of scaling factor, but I figure this is close enough to fudge the rest.)

Pretty sure it only works this way because the maps have square borders... there are just a ton of ways that you can place two 'Mechs at range 1, and only a little fewer at range 2, but as your range gets longer you start getting trapped going corner-to-corner. There's not many ways to do that, so the curve drops off sharply. It zeroes out when you finally go off the map.

A circular border would have a different distribution (sqrt(xy)-r, maybe?) and the curve for an infinite map might instead march continuously upward.

Of course, this is just the distance between two randomly selected points. In a boardgame, those points won't be random, and the terrain is rarely so perfectly flat and open. For BattleTech, when the scenario prevents kiting, I'd guess that range decreases in usefulness more or less linearly from range 1 and zeroes around range 30. (You could calculate these values exactly, given a known distribution of terrain and combatants and victory conditions, but it'd be hassle because those things change so often.)

Gravities * Days^2 = Light Minutes

Hey all, just wanted to say this quick before I forget again:

Suppose you are floating at rest relative to the local star, then accelerate at a constant rate towards a planet, and when you're halfway to that planet you flip end over end and decelerate at that same constant rate, eventually coming to rest above the planet.

If you measure your acceleration in gravities (~9.81 m/s/s) and your time in days (~86400s), then Gravities * Days^2 = the distance you travel in units of 61 light-seconds. That's close enough to light-minutes to make a super easy rule of thumb.

Of course, in a "real" situation you'd want to account for the planet's orbital position and orbital motion, gradually change your acceleration to help passengers transition from the old planetary gravity to the coming planetary gravity, account for relativistic dilation on month+ trips, track fuel use and how the rate of use changes as fuel is expended... probably other things.

But hey! Super easy rule of thumb.

Math Puzzle

I spent way too much time today trying to figure out this math puzzle. The results he wants are equivalent to rolling 1d4 twice and keeping the lowest result:

He wants to use d2s exclusively, and you can reproduce a d4 with two 1d2s easily enough. If you're actually sticking to coins, for instance, you could say "flip a shiny penny, a shiny nickel, a dull penny and a dull nickel. Face-up coins get their full value, tails-up coins get zero value, take the lower value of shiny+shiny and dull+dull." That would give you 6 cents 1/16th of the time, 5 cents 3/16ths of the time, 1 cent 5/16ths of the time, and 0 cents 7/16ths of the time.

The trick is that he wants all d2s to have uniform face values, and because d2s only have two faces, all of those other tricks he mentions are going to come out the same as just adding the dice together. At least, everything else I've tried so far has come out that way, and I'm pretty sure you can't get the 1/3/5/7 ratio just by straight summing the roll.

Okay, [(3d2 - 2)*1d2]/2, round up, seems to work. It's not the friendliest looking thing, but I don't think the solutions can get much better.

Anyways, I was looking at that chart I drew of the 1/3/5/7 ratio, and I thought, hey, that kind of looks like my bullseyes. But, instead of summing two dice like the 2d6 rolls I graphed before, it chooses between two independent rolls! That's more flexible, mechanically; and it's nice to see that the stripes don't thin towards the rim. I guess I'm now hoping to find a 1dX roll that thins evenly and infinitely towards the center. So I can roll two of them for every roll, and allocate them between disconnected aspects of the action.

Weird thing about the chance one roll has to beat another:

A while ago, EastwoodDC posted a spreadsheet for calculating the chance of rolling any given result on a number of whatever sized dice. I started looking at the chances one roll had to beat another and I noticed something weird.

When rolling 1d6 against 2d6, the chance that they'll tie is equal to the chance of rolling exactly 7 on 3d6, and the chance that 1d6 will beat 2d6 is equal to the chance of rolling 6 or less on 3d6.

When rolling 2d6 against 3d6, the chance that they'll tie is equal to the chance of rolling exactly 14 on 5d6, and the chance that 2d6 will beat 3d6 is equal to the chance of rolling 13 or less on 5d6.

When rolling 3d6 against 4d6, the chance that they'll tie is equal to the chance of rolling exactly 21 on 7d6, and the chance that 3d6 will beat 4d6 is equal to the chance of rolling 20 or less on 7d6.

So I'm pretty sure that, anytime you roll Nd6 against (N+1)d6, the chance they'll tie is equal to the chance of rolling N*7 on (N+N+1)d6 and the chance that Nd6 will win is equal to the chance of rolling (N*7)-1 or less on (N+N+1)d6.

Messed around a little more, and it looks like 1d6 ties 3d6 as often as 4d6 will roll exactly 7; 1d6 beats 3d6 as often as 4d6 rolls 6 or less; 2d6 ties 4d6 as often as 6d6 will roll exactly 14; and 2d6 beats 4d6 as often as 6d6 rolls 13 or less. So I expect that, when rolling Nd6 against (N+2)d6, they'll tie as often as (N+N+2)d6 rolls exactly N*7, and that Nd6 will beat (N+2) as often as (N+N+2) rolls (N*7)-1 or less.

Wait. Those look the same. Can I just follow this pattern for any number of d6?

Nd6 ties (N+M)d6 as often as (N+N+M)d6 rolls exactly (7*N).
Nd6 beats (N+M)d6 as often as (N+N+M)d6 rolls (7*N)-1 or less.

And then, since the "7" there is a result of the average outcome of a single d6, can I extend it to other sizes of dice?

NdD ties (N+M)dD as often as (N+N+M)dD rolls exactly ((D+1)*N).
NdD beats (N+M)dD as often as (N+N+M)dD rolls ((D+1)*N)-1 or less.

Seriously just guessing at this point (could probably figure it out by hand with more spreadsheets, or by pure math from Pascal's Triangle) but I think this is going to work.


EDIT: VanVelding has tackled the math for this here and here.
EDIT ²: and here.