For just moment, think of a Battlemech in the game as a single group of points, rather than as 8 separate locations that each absorb a smaller amount of damage. Here the Birnbaum-Saunders approximation works very well, and if we were playing different game where the goal is only to reach that overall total, like in the card game cribbage, then we would be done. A cribbage board has only one “track” for scoring points.
Battletech though, has a much more complicated sort of board (the Mechsheet); Each location has it’s own track, and most tracks spill over in into the next track when it is full (i.e.: damage transfer). That too can be approximated, but writing the Mechsheet as a mathematical formula is MUCH more complex. I’ll get to that too, but today I want to do something more simple and intuitive. We can learn something just by considering the median survival time of each mech location. This is the point at which there is a 50% chance that the
But first a bit of notation. The unit of time here is actually the number of damage groups, as assign by rolling on the Hit Location Table. We usually think of time as turns or phases during the turn, but the damage groups are a better measure of a mechs progress towards destruction. The word “hit” has a lot of different associations in Battletech, so I will try to use “Damage group” (or maybe “shock” - the engineering term) to avoid confusion with “to-hit” rolls and “hit location”. Further, everything that follows assumes that the to-hit roll succeeded, so I am counting time as damage groups that actually hit the mech, not the number of turns passed or to-hit rolls attempted. Also, if I fail to mention it, I am only considering Front Hit Locations at this time, a mech with a standard fusion engine, and ignoring damage transfer.
The formula for the Birnbaum-Saunders probability distribution gives the probability that a certain event will happen by some time t. While this formula might be bit intimidating for some people, the formula for the median survival time is simple. If a given location, let’s say the right arm (RA), can absorb 25 points of damage before it is destroyed, and the mech is taking damage that averages 5 points per group, then we can expect the RA will survive until it is hit by 5 damage groups. From the Front Hit Locations table, the probability of hitting the RA is 5/36 or 13.9%, and the median time is equal to 25/(5*0.139) = 36 damage groups or shocks. In general, this formula is:
Median Time = D / (d x p)Where D is the total amount of damage (internal structure and armor) a location can absorb, d is the average damage per group/shock, and p is the probability of hitting that location.
Now consider a mech that has armor allocated so that locations can absorb damage as follows: Arms 25 each, Legs 20 each, Right/Left Torsos 25 each, Center Torso 35 and Head 5. What I have done is to set up each location so that it can absorb damage proportional to the probability of that location being hit. Further, the median survival of each section will be exactly 36 shocks. Recall this is ignoring all damage transfer, and 36 shocks is also the maximum amount of damage this mech could possibly absorb with every location completely destroyed. In actual play, this mech would very likely be destroyed before 36 shocks.
Well that was a lot of work to set up, but if you are still with me, we can now consider what this tells us. Here each location has the same life expectancy, but this is far from ideal. What we really want is for the head and Center Torso Locations to survive. We also want two functional legs so it can move and fire for as long as possible before it is destroyed. To accomplish this, we need each leg to have a survival time at least as long as the center torso. The center torso is more critical than the side torsos, is it should have a longer survival than the side torsos. If either side torso is destroyed we also lose the same-side arm, so it doesn’t make much sense to have the arms survive longer than the torso sections. Finally, we don’t want the head to get blown off, but we are strongly limited in the maximum armor that can be place there, so it may be beyond our control. This suggests the each location be armored so that the median survival times for are ordered as follows (in descending order):
PS: Under the assumption I have made here, the Ostsol and Ostroc battlemechs may have near-optimal armor distributions (an educated guess). There may also be more than one way to be "optimal" depending on what goals you might use.
[Battletech images from the excellent ClassicBattletech.com gallery.]