Favorite SW and Ordnance in ASL,and why?

Philippe D.

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I'm rather of the opinion overall that he best method of attack for a FT toting infantry unit is to assault move out of LOS 1 location from being in LOS and <=2 hexes from a key enemy position, then never move him again. If the enemy takes the bait and holds his fire, no attack DR is ever made and the FT never depletes, while you run circles around his position with everything else. If he shoots at something else moving by, you move the FT up and then zap him with a FT attack DR.
It all depends on what your real goal is, and what threats you can use. Sometimes you'll have your FT three hexes from your target, and manage to move less threatening units adjacent safely because the threat of the FT moving just into long range is strong enough; and sometimes it will be a better move to have the FT move adjacent [often under some additional cover] so as to draw a lot of fire and then make it possible for other units to close in.

And then, some players are pessimists, and will always assume they'll X out the FT as soon as they fire it; those will tend to keep it as a threat for longer, because not firing it often means not malfunctioning it :)
 

von Marwitz

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Favorite SW: German MMG
  • 3PP (better than many MMGs)
  • Range 12 (better than most MMGs)
  • still able to fire dm as LMG and only 2PP then (better than any other MMG)
  • 5 FP (better than any other MMG - and, hey, IIFT rules!)
  • B12 (reliable hardware made in Germany)
  • FL capability (I use these extensively)
  • Spraying Fire Capability (though rarely used, can be good in combination with German squads & good leader in the AFPh vs. enemy units in low TEM)
Favorite Ordnance: German 8.8cm PaK 43
  • TK27 (enough for anything and yet some)
  • 16FP IFT / 18FP on IIFT for HE (which is good HE punch)
  • ROF 2 (fast for such a powerful gun)
  • Fast Traverse (so no lowering of ROF due to CA change and still decent to fire from Woods)
  • Normal Target Size while Unlimbered
  • Can be fired while Limbered
  • LL (excellent long range accuracy)
I am not completely sure if it may set up in Buildings (as it is normal Target Size while Limbered but large Target Size while Unlimbered).

von Marwitz
 
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Paul M. Weir

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I would say that the average number of attacks a FT can make is about three - that is, if the guy carrying survived incoming enemy fire.
The expected number of shots is 6.

The formula for expected number of events where an attempt depends upon the success of the previous attempt (like RoF or X#) is 1/(1-p) where p is the probability of a single successful event. 1/(1-p) is the sum of the infinite series 1 + p + p^2 + p^3 ...

With RoF you always have 1 guaranteed shot so is 1 + p + p^2 + ... = 1/(1-p)
RoF
[1]: 1/(1-1/6) = 6/5 = 1.2
[2]: 1/(1-1/3) = 6/4 = 1.5
[3]: 1/(1-1/2) = 6/3 = 2
[4]: 1/(1-2/3) = 6/2 = 3
[5]: 1/(1-5/6) = 6/1 = 6
[6]: 1/(1-1) = 1/0 = ±∞

OK, 1/0 is technically undefined but in practice is ±infinity depending whether you approach from the positive or negative side of 0. :D

x#: as per RoF.
x12: 1/(1-35/36) = 36/1 = 36
x11: 1/(1-33/36) = 36/3 = 12
x10: 1/(1-30/36) = 36/6 = 6
x9: 1/(1-26/36) = 36/10 = 3.6
x8: 1/(1-21/36) = 36/15 = 2.4
x7: 1/(1-15/36) = 36/21 = 1.714

That's regardless of opportunities for the use of a FT, sometimes the FT can't get there for various reasons but if you had unlimited opportunities ...

So 1/(1-p) is a very simple formula. Indeed RoF and x# are very similar except that RoF is 1D6 based and x# is 2D6 based and RoF is a success number while x# is fail based. If both were 1D6 or both were 2D6 based a RoF of [N] would be the equivalent of a x(N+1). An alternative way to look at it is 1/q where q is the probability of failure in a single event, as demonstrated by the 36/ (1, 3, 6, 10, 15, 21) progression in the x# table calculations.

EDIT: I messed up the difference of 1 between RoF (guaranteed 1st shot) and x# (with no guaranteed 1st shot). Fixed.
EDIT: I had it right 1st time, wrong 2nd time as Philippe pointed out.
 
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kcole4001

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I'm never afraid of Xing a FT or PSK, just afraid of never getting to range.
If I have a valid target, especially with an FT, unless it's a HS picking it's nose in an unimportant spot, I'm rolling the attack.
More often than not the toting inf has to endure all sorts of insult and injury just trying to get within 2 hexes of anything relevant.

I lost one to a HS double break on a 1 FP snap shot in a street, and everyone else who tried to recover broke and ran.
I gave up after 3 or 4 attempts, can't remember, but that extra -1 gets me every time.

As regards the PSK, I don't worry about it's X potential, if it's a decent shot, fire away.
Took out a tank very nicely with one a while back playing an SK scenario, I didn't remember them being that effective, but they do work well!
 

Robin Reeve

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Thanks Philippe - and Paul for the number crunching (in the old times I learned to do it when I followed a programmer-analyst formation, but that science is long forgotten for me) !
I have even better reasons to spray napalm on my cardboard opponents.
I remember a Russian 7-0 which did marvels in Stalingrad with its FT.
 

Paul M. Weir

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My bit of maths is a fraction misleading in that it assumes an infinite number of possible attempts if the weapon does not break first.

The 1/(1-p) being the sum of 1+ p + P^2 + p^3 + p^4 + ....

Now what happens when you only have n possible shots? The expected value is then 1 + p + p^2 + p^3 + ... +p^(n-2) + p^(n-1). That is finite calculable but a pain for all except when N is small. Let us look at the leftovers from the infinite series and that will be p^:thumbsdown: + p^(n+1) + p^(n+2) + ... and that = p^n * (1+p+p^2+p^3+p^4...) = p^n * 1/(1-p). So the sum of n elements = 1/(1-p) - p^n*1/(1-p) = (1-p^n)/(1-p), n being the maximum opportunities for a shot.

So for a X10 FT
N
1 : 6/6 =1
2 : 11/6 = 1.833
3 : 91/36 = 2.528
4 : 671/216 = 3.106
5 : 4651/1296 = 3.589
6 : 31031/7776 = 3.991
7 : 201811/46656 = 4.326
8 : 1288991/279936 = 4.605

So you can see it slowly creeps up to the limit of 6 fairly slowly as the opportunities increases.

I made an error in my earlier formula posting, that has been corrected in place.
EDIT: my original was right, re-corrected. This one has been fixed as well
 
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Philippe D.

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My bit of maths is a fraction misleading in that it assumes an infinite number of possible attempts if the weapon does not break first.

The p/(1-p) being the sum of p + P^2 + p^3 + p^4 + ....

Now what happens when you only have n possible shots? The expected value is then p + p^2 + p^3 + ... +p^(n-1) + p^n. That is finite calculable but a pain for all except when N is small. Let us look at the leftovers from the infinite series and that will be p^(n+1) + p^(n+2) + p^(n+3) + ... and that = p^n * (p+p^2+p^3+p^4...) = p^n * p/(1-p). So the sum of n elements = p/(1-p) - p^n*p/(1-p) = p(1-p^n)/(1-p), n being the maximum opportunities for a shot.
There's obviously a problem with your final formula: for n=1 (a single firing opportunity), you are certain to get exactly one shot - so your formula should yield 1 in this case.

If you don't include a limit, the number of shots follows the geometric distribution with parameter 1-p, where p is the probability of not malfunctioning. The average of this is indeed 1/(1-p), which is then 6 for the FT (p=5/6).

Now if you cap the number at n, then this becomes the minimum of your geometric and n. Unless I am wrong, the expectation of this is 1/(1-p) - p^n/(1-p), that is, (1-p^n)/(1-p). Your numbers should just be divided by p, that is, multiplied by 6/5=1.2.
 

Paul M. Weir

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There's obviously a problem with your final formula: for n=1 (a single firing opportunity), you are certain to get exactly one shot - so your formula should yield 1 in this case.

If you don't include a limit, the number of shots follows the geometric distribution with parameter 1-p, where p is the probability of not malfunctioning. The average of this is indeed 1/(1-p), which is then 6 for the FT (p=5/6).

Now if you cap the number at n, then this becomes the minimum of your geometric and n. Unless I am wrong, the expectation of this is 1/(1-p) - p^n/(1-p), that is, (1-p^n)/(1-p). Your numbers should just be divided by p, that is, multiplied by 6/5=1.2.
You are right in that the FT is only removed AFTER resolving the attack in the case of a B10, so does count. I'll fix those post.
 
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witchbottles

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Moving FT into range and enemy LOS requires some skill indeed.
I know that there is a tradition here to moan about X# weapons being disabled on first shot. But that experience (EXC dice which roll 4, 5 or 6 more frequently than 1, 2 or 3) is typically the bad experience which marks memory.
If you play against me, be sure that I won't just keep my FT as a threat.
I will take the occasion to blast it against that pillbox or Fortified building Location - of course, not recklessly.
Because I know that it will run out of fuel only 17% of the time.
Because 83% possibility of at least a NMC (9 on the 12 IFT column if firing at double range) or a 2MC (on the 24 column) influences my choice much more than the fear of busting my weapon - not speaking of 1MC-NMC even if it Xes on the 24 FP column.
Of course, sometimes I will be proven wrong.
But in the long run, using my FT will make sense much more often than not.

As a side note, I don't see many players say that they refrain from using a PSK which has the same risk of Xing.
A PSK - while still a rather fearsome anti-armor weapon, is not nearly as dangerous to an infantry target in +3 TEM or better as a FT is. Probably why there is an inherently less hesitation to pull the trigger on the tank terror than there is to pull it on the Ronson. The FT while intact, presents a major threat to any enemy infantry on the map within 2 hexes of it, as well as any AFV that ventures within 2 hexes of it. Once it is dysfunctional, it loses all those capabilities. The PSK offers little fear to an enemy firmly entrenched or even advancing stealthily, through +3 or better TEM.

KRL, jon H
 

WuWei

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I had lots of fun with a lousy ATR in a scenario where all my squads were 527s. My opponent somehow overlooked that I could interdict from way back there. :cool:
 

Robin Reeve

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I recently had a German ATR kill a Russian BT tank.
A nice feat.
 

witchbottles

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To kill ANYTHING with a German ATR is...well, at least memorable!:D
the actual TK DR required is typically 1 more pip than an LMG shot - making it a bit more likely to get an effect with the ATR. That said, the LMG has the possibility of an ROF 2nd shot that the ATR never gets.
 

witchbottles

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I had lots of fun with a lousy ATR in a scenario where all my squads were 527s. My opponent somehow overlooked that I could interdict from way back there. :cool:
for a bolt action overgrown rifle, the 12 range makes them rather nice at times. :)
 
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