The ULTIMATE Death Star Challenge

Philippe D.

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Do you have to be able to demonstrate the require skill by actually stacking the counters, with instant "falling rubble" rules?
 

MAS01

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Correct about the German halftrack. I can fit a combo of 14 effective Heroes/LMG in there.

Well, here comes my final say:

For a Level 2 Building with Ground, 1st, and 2nd Level:

34 Heroes with 17 HMG for 119FP @ -34 DRM
34 Heroes with 17 HMG for 119FP @ -34 DRM
34 Heroes with 17 HMG for 119FP @ -34 DRM

In Bypass a SPW 251/1 halftrtrack with a 3FP AAMG plus loaded with 7x Hero, 7x LMG, using up 14 out of 15 available PP for another 21FP totalling 24FP.

This would yield a grand total of 381FP @ -102.


With Captured Use weaponry:

34 Heroes with 17 captured .50 Cals for 136FP @ -34 DRM
34 Heroes with 17 captured .50 Cals for 136FP @ -34 DRM
34 Heroes with 17 captured .50 Cals for 136FP @ -34 DRM

In Bypass a captured US M3 (HMG) halftrtrack firing through its rear target facing for 20FP plus loaded with 5x Hero, 5x LMG, using up 10 out of 10 available PP for another 15FP totalling 35FP.

This would yield a grand total of 443FP @ -102.

My idea to yet 'up' it was to add a SMC Rider to the halftracks per

"D6.2 RIDERS: Riders are Personnel being transported on the outside of an AFV (or Cavalry or motorcyclists). AFV Riders cannot be used until 1942—and even then only by the Russians. In 1943 and thereafter, all nationalities may use Riders. Riders are limited to tanks (not tankettes), TD, SPA, SPAA, Carriers (limited capacity; 6.81), and Assault Guns. Rider capacity is limited to a maximum of 14 PP (as per 6.1); within those timeframes a vehicle not otherwise granted Rider capability can carry one SMC as a Rider and the two PP he possesses. See A5.5 for equivalents. Rider PP cannot be used to satisfy the ammo PP reduction (C10.13) of any Gun that requires a T# (C10.1) to be transported; therefore, a dm 76-82mm mortar is transported by Riders at its normal five PP cost."

Thus, if the 10-3s would have made a difference, one of them could have been placed as a Rider on the US halftrack laden so far with only 4x Hero & 4x LMG, allowing for a fifth Hero+LMG combo as Passenger to use up the 9th and 10th PP while the 10-3 could have directed everything as a Rider. Placing an extra Hero as a Rider does not change anything because Riders may not use SW and only fire within their normal range.

In case the challenge would have been set at range 4 (and not 6), then the extra Hero Rider with his LMG would have added 1.5FP (halved for being a Rider) and an extra -1 Heroic DRM.


So it seems that the final score is settled at

443FP @ -102.


von Marwitz

And with my luck, I'd roll a 6,6
 

von Marwitz

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And with my luck, I'd roll a 6,6
Well, the good news is that it wouldn't matter.

Even if a dozen or two .50 Cals would break, Cowering does not affect fire by SMC (A7.9).
But even if you would Cower from 443FP to the 'next column' of 36FP (call that a drop...), the -102 DRM would do in Major von Schoff with a 7 KIA even if he were a cat. So Oberst von Marwitz's bottle of Château Lafite-Rothschild is safe.

Obviously, he has forseen the possibility of boxcars and taken precautions.

von Marwitz
 

von Marwitz

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But what if Random Selection is boxcars all the way down?
Doesn't matter. As per A9.7, the attack is still resolved when you roll boxcars.

Besides, Oberst von Marwitz argues, that the chance for all 51 .50 Cals to malfunction despite being Captured Equipment is only (a still surprising) 1,2%.

Edit:
The percentage of all 51 .50Cals breaking with a 1,2% chance is so surprising, because the calculation is royally gacked... 🤣

von Marwitz
 
Last edited:

von Marwitz

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Oberst von Marwitz cannot understand at all why anyone would like that obnoxious Major von Schoff to survive in the first place!!

But as some here entertain the thought (and maybe even von Schoff himself), I challenge you with the follwing:

What are Major von Schoff's chances to survive the attack as outlined above?
What are Major von Schoff's chances to survive broken the attack as outlined above?
What are Major von Schoff's chances to survive pinned the attack as outlined above?
What are Major von Schoff's chances to survive unscathed the attack (oh, the scandal!) as outlined above?


Note, that Major von Schoff is posing as a Green US Army Halfsquad (fat bugger that he is...).

Cowering, ROF, Captured Use, Malfunction, losing negative Hero DRM due to .50 Cals malfunctioning and thus getting the Heroes out of 'normal range', etc. all have to be taken into account.
And for the ones who are really fanatic: Take the chances of von Schoff Battle Hardening in the various ways into account.

Now, that should do you in while Oberst von Marwitz is sipping his crystal chalice of Château Lafite-Rothschild.


von Marwitz
 

stuh42asl

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I have one ultimate FP demo........................it only needs one shell......................playing Maxim Gorky and hit a stack of russians with a single shell from Dora............Sure it is not German finesse.............but a 800mm shell trumps everything....all you get is a huge, deep shellhole... Firing at a target 6 hexes away would be like swatting a fly with a 2000 pound sledge hammer. Sorry but my Monty Python sense of humor just could not resist.........
 

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jimfer

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Did anyone take into account fortified Buildings with something like the M51 Multiple .50-cal MG Carriage? It is small so can go into a building and has 20 attack. Just saying other than the 800mm gorilla in the room, the Holy handgrenade and the occasional backpack nuke this has been an interesting thread.
 

von Marwitz

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Did anyone take into account fortified Buildings with something like the M51 Multiple .50-cal MG Carriage? It is small so can go into a building and has 20 attack. Just saying other than the 800mm gorilla in the room, the Holy handgrenade and the occasional backpack nuke this has been an interesting thread.
If limbered, the M51 even has 24 FP. However, together with a crew it would count as 1 squad-equivalent for stacking purposes. 10 Heroes with 5x .50 Cals give you 40FP @ -10 DRM, while you "only" get 24 FP @ 0 DRM for the M51.

The thing just ain't gettin' the lead out... :LOL:

von Marwitz
 

jimfer

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If limbered, the M51 even has 24 FP. However, together with a crew it would count as 1 squad-equivalent for stacking purposes. 10 Heroes with 5x .50 Cals give you 40FP @ -10 DRM, while you "only" get 24 FP @ 0 DRM for the M51.

The thing just ain't gettin' the lead out... :LOL:

von Marwitz
I'm just trying to find away to make that unit plausible. :whistle:
 

Paul M. Weir

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But even if you would Cower from 443FP to the 'next column' of 36FP (call that a drop...), the -102 DRM would do in Major von Schoff with a 7 KIA even if he were a cat.
Ahem, it needs a 9 KIA to kill a cat.
Besides, Oberst von Marwitz argues, that the chance for all 51 .50 Cals to malfunction despite being Captured Equipment is only (a still surprising) 1,2%.
A B12 captured weapon becomes effectively a B10, not a B11.

For 51 SW with an effective B#, the chances of not all breaking, all breaking
B12: 76.23%, 23.77%
B11: 98.82%, 1.18%
B10: 99.9908, 0.0092%
 

von Marwitz

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Ahem, it needs a 9 KIA to kill a cat.
It depends on whether the hex in question is located on the Anglo-American or German turf.
On the former, cats are said to have nine lives, on the latter only seven.

I have wondered more than once what the 'opinion' on this is in other European countries. Not unlikely, it might be one of these efforts of going great lengths to be different: Driving on the wrong side, no metric system, not abolishing the mounted lance before 31. December 1927... Probably, you need two extra lives as a cat to put up with all that.;)

von Marwitz
 

Gordon

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It depends on whether the hex in question is located on the Anglo-American or German turf.
On the former, cats are said to have nine lives, on the latter only seven.

I have wondered more than once what the 'opinion' on this is in other European countries. Not unlikely, it might be one of these efforts of going great lengths to be different: Driving on the wrong side, no metric system, not abolishing the mounted lance before 31. December 1927... Probably, you need two extra lives as a cat to put up with all that.;)

von Marwitz
REALLY?!? I'd have thought Germans would have engineered a superior cat with at LEAST 10 lives. 😁
 

Paul M. Weir

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REALLY?!? I'd have thought Germans would have engineered a superior cat with at LEAST 10 lives. 😁
Yeah. It had long fur, fangs and claws but it only produces one or occasionally two kittens per litter, is grossly overweight, can't cross soft ground, its legs frequently give way, eats trice as much as an alley cat and only capable of walking to the end of the back yard. Oh and the vet bills!
 

von Marwitz

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Yeah. It had long fur, fangs and claws but it only produces one or occasionally two kittens per litter, is grossly overweight, can't cross soft ground, its legs frequently give way, eats trice as much as an alley cat and only capable of walking to the end of the back yard. Oh and the vet bills!
But if it hits you, it gorges on your scattered entrails like a 'King Tiger' for a week or so...

von Marwitz
 

Gwinnell

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Ahem, it needs a 9 KIA to kill a cat.

A B12 captured weapon becomes effectively a B10, not a B11.

For 51 SW with an effective B#, the chances of not all breaking, all breaking
B12: 76.23%, 23.77%
B11: 98.82%, 1.18%
B10: 99.9908, 0.0092%
Paul
I am struggling with this.
How can it be more likely that a B12 weapon will all break than a B10 weapon?
Also a 23.77% chance of all breaking seems unlikely as you would need 51 identical dice rolls on the RS dice.

Maybe I'm just missing something.

Gavin
 

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Ahem, it needs a 9 KIA to kill a cat.

A B12 captured weapon becomes effectively a B10, not a B11.

For 51 SW with an effective B#, the chances of not all breaking, all breaking
B12: 76.23%, 23.77%
B11: 98.82%, 1.18%
B10: 99.9908, 0.0092%
Paul
I am struggling with this.
How can it be more likely that a B12 weapon will all break than a B10 weapon?
Also a 23.77% chance of all breaking seems unlikely as you would need 51 identical dice rolls on the RS dice.

Maybe I'm just missing something.

Gavin
I think he may have mistyped (i.e., reversed) the B10 and B12 results, for exactly the reason you suggest.

But there is also something about the formula he is using for all breaking (not all breaking is just derived by subtraction). I am not sure how to calculate the likelihood of rolling the same number on one die 51 straight times for each of the six possible numbers but whatever it is, the result should be the same for every B#. I think the rest of the formula should be likelihood of breaking down times that result of random selection. If (a big if) I have that right, the rate of change should be linear with the rate of change in the cumulative probability of rolling the breakdown number or worse. A 3-fold increase from B12 to B11, a further two fold increase from B11 to B10. But not in his figures.

Now I'll step back and wait for the adults in the room to explain my errors.
 

Paul M. Weir

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I think it's best I restart my calculation and figure out what values I did calculate and what they should be.

B12: Probability of 1 breaking = 1/36 (PB1), not breaking (PNB1) = 35/36. All 51 breaking = PB1^51 = 4.252e-80 ~ 0%, none breaking = PNB1^51 = 0.2377 = 23.77%. It looks like what I posted was the probability of at least 1 breaking and none breaking. Lets do the other two.

B11: PB1 = 3/36, PNB1 = 33/36. All 51 breaking = PB1^51 = 9.157e-56 ~ 0%, none breaking = PNB1^51 = 0.01182 = 1.182%

B10: PB1 = 6/36, PNB1 = 30/36. All 51 breaking = PB1^51 = 2.062e-40 ~ 0%, none breaking = PNB1^51 = 9.157e-5 = 0.009157%

So I got the pre calculation logic wrong and posted at least 1 breaking and none breaking.

For all B12, B11, B10 the probability of all breaking is less than 2e-40 or less than 2e-38%.
 
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