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Date: Sat, 9 Jan 1999 09:44:00 -0700

Subject: RE: [FTFB] "Needle" fighters?

```
>>Jared E Noble wrote:
[snip]
>The squadron gives up it's normal attack and instead:
>- Declare the intended trench run (but not it's target)
>- Resolve PDS
>- Nominate the target system
>- Roll a die and compare it to squadron strength
> (Turkey groups count as 1 less, Groups with ace as 1 more)
> -- If above squadron strength, nothing happens.
> -- If equal to squadron strength target system damaged.
> -- If below to squadron strength target system destroyed.
So the probabilities for various squadron sizes would be:
Remaining
Fighters % no effect % damaged % destroyed
--------- --------- --------- -----------
6+ace 0% 0% 100%
6 0% 16.7% 83.3%
5 16.7% 16.7% 66.7%
4 33.3% 16.7% 50%
3 50% 16.7% 33.3%
2 66.7% 16.7% 16.7%
1 83.3% 16.7% 0%
So there's always a 1/6 chance of damaging a system, but the
chance of destroying one covers the full range.
If this turns out to be too powerful (i.e., too many "destroyed"
results),the damage rules could be changed to
-- If above squadron strength, nothing happens.
-- If equal to or below squadron strength, target system damaged.
-- If below half the squadron strength, target system destroyed.
This would give the following probabilities:
Remaining
Fighters % no effect % damaged % destroyed
--------- --------- --------- -----------
6+ace 0% 50% 50%
6 0% 50% 50%
5 16.7% 50% 33.3%
4 33.3% 33.3% 33.3%
3 50% 33.3% 16.7%
2 66.7% 16.7% 16.7%
1 83.3% 16.7% 0%
So the chance of destroying a system outright never exceeds 50%, but
the chance of damaging a system varies from 1/6 to 100%, instead
of being a constant 1/6.
There would still be an advantage for an ace squadron with
these probabilities, in that the probabilities of damaging or
destroying a system remain higher as the squadron strength
drops (through unfortunate encounters with PDS).
Note that I did not round the 1/2 values; half a squadron strength
of 3 is 1.5, so "below 1/2 squadron size" is 1, or 16.7%, but "equal
to or below squadron size" includes 3, 2, or 1; exclude 1, since that's
a "destroyed" result, and you get 2/6, or 33.3%.
This second distribution appeals to me more than the first, but I
have no idea how either one would work in play. Definitely needs
some playtest.
- Sam
________________________________________
Samuel Reynolds
http://www.primenet.com/~reynol
reynol@primenet.com
samuel_reynolds@csgsystems.com
```

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