[Ogre] Optional Ogre tread damage rule

[David Morse]

On Wed, 22 Dec 2004 15:47:32 -0600 (CST), Henry J. Cobb <hcobb at io.com> wrote:
> "David Morse" <dcmorse at gmail.com> wrote:
> > Example: An Ogre Mark V has lost 6 treads, leaving 14 treads between it
> > and M2.  On a turn when it tries to move three hexes, roll a 20-sided
> > die,
> 
> Non-cubic dice in Ogre?  Die Die Die, it's always six sides to the die.
> 
> > I suppose this 'orribly screws Ogres.
> 
> Yeah, how about this instead?
> 
> Take the number of threads in a movement block, divide by three and round up.
> 
> So Mark V: seven dice.
> Mark III: five dice.
> Mark I: 2 dice.
> 
> Roll that many dice and add the remaining tread units and if you reach the
> full amount needed you get that final move.
> 
> So a Mark V with 42 tread units needs to roll 18 or higher on seven dice
> to move the third hex.

(Aside: If this were Ogre Miniatures, the 42-tread Ogre would already
be moving 5" - M2.5 in hex terms)

That Ogre, just 2 treads away from M2, will move M3 94% of the time! 
(or so says my possibly buggy program).  You could divide by 5 instead
of by 3, decreasing the chances to 14%.  (not unfair IMO since the
Ogre only has 10% of the treads between M3 and M2).

However, the chances of a "stall" are still non-linear with tread
damage, (except in the Mk I case), because whenever you roll a large
number of dice there's the bell-curve distribution that strongly
favors the median outcome over the outliers.

To see any payoff at all for targetting treads requires leveling even
MORE firepower than it would take to strip ALL secondary batteries off
the Ogre.  Thats why in most scenarios the treads are the last to go. 
Certainly once GEV overruns are in effect, you'd have to be insane to
fire at treads when there are any guns left, because once you've
stripped the guns it takes only a turn or two to immobilize the Ogre.