The ULTIMATE Death Star Challenge

[Sparafucil3]

Aren't you missing a step in there? Once you roll a 6,6 (1/36), you now need to have all them roll the same thing on 6 sided die for random selection. -- jim

Edit to add more: Actually, isn't this all resolved in a single attack? If that's the case, the probability of none breaking (for B12) is 35/36. Once you have rolled the twelve, then you have to have all the weapons involved roll the same dr for Random selection. Perhaps I missed something up thread. I can't be bothered to go back and read it all.

 

[von Marwitz]

Aren't you missing a step in there? Once you roll a 6,6 (1/36), you now need to have all them roll the same thing on 6 sided die for random selection. -- jim

Edit to add more: Actually, isn't this all resolved in a single attack? If that's the case, the probability of none breaking (for B12) is 35/36. Once you have rolled the twelve, then you have to have all the weapons involved roll the same dr for Random selection. Perhaps I missed something up thread. I can't be bothered to go back and read it all.

Right. But it is not a 6,6 you need to roll. As the .50 Cals are 'captured use', the B# is lowered by 2 (per A21.11, A21.12), and thus a 10 or more is sufficient.

Chances to roll 10+ for malf/breaking are 8.3%. And then it is RS for all .50Cals, which I reckon is 1/6th i.e. 16.7%^51 (3 hexes x 17 MGs = 51).

von Marwitz

 

[Paul M. Weir]

The last one I did would be applicable for independent fire attempts. I was trying to figure out what I was thinking in the first calculation.

Single fire attempt. Any B10 breakdown needs a 6/36 (as per vonM). For all 51 to break they need to all have the same RS dr. So the probability of all the same is 1/6^51 multiplied by the number of possible values which is 6 leaving 1/6^50. Another way to look at it is the 2nd to 51st RS has to exactly match the 1st, regardless of whatever dr the 1st got (1/6^50).

So 6/36 * 1/6^50 = 1/6^51 = 2.062e-40 = 2.062e-38% for a B10. For B11 and B12 replace the 6/36 with 3/36 and 1/36 respectively.

 

[Sparafucil3]

Right. But it is not a 6,6 you need to roll. As the .50 Cals are 'captured use', the B# is lowered by 2 (per A21.11, A21.12), and thus a 10 or more is sufficient.

Chances to roll 10+ for malf/breaking are 8.3%. And then it is RS for all .50Cals, which I reckon is 1/6th i.e. 16.7%^51 (3 hexes x 17 MGs = 51).

von Marwitz

I am setting aside a lower breakdown number for the moment. My point is, you don't break any unless there is a 6,6. I think (and I am no math wizard so it is a guess for sure) that this PA given PB. So what is the chance of all of them breaking given that a breakdown number is rolled. First you must roll the break down number, then all the other math kicks in. No breakdown number, no need to go any further. -- jim

 

[Russ Isaia]

There are two probabilities to be calculated here to determine whether all break and the result is the product of the two. The probability of any MG breaking and the probability that if any break, then all will be randomly selected to break.

The probability of any MG breaking depends on the B#: at 12 it is 1/36, at 11 it is 3/36 and at 10 it is 6/36.

The probability of all being randomly selected if any are is: 6 * (1/6 ^ 51). Translated, six and only six combinations of 51 dr rolls meet the criterion (the same one of the six numbers on each of the 51 dr rolls).

Hence, for all to break:

B12: 1/36 * (6* (1/6 ^ 51)) = some impossibly small number (to use the technical term).
B11: 3/36 * (6* (1/6 ^ 51)) = three times that impossibly small number.
B10: 6/36 * (6* (1/6 ^ 51)) = six times that impossibly small number.

And the possibility of none breaking? Only if the DR roll is under the B#:

B12: 1 - 1/36 = 97.222%
B11: 1 - 3/36 = 91.667%
B10: 1 - 6/36 = 83.334%

Miss anything?

 

[von Marwitz]

You wouldn't believe it...

After much brainwhacking, we came up with 17x captured .50Cals manned by 34 Heroes each per level.

Apparently, there have been escapist nerds in reality, too. Look at this:




von Marwitz

 

[Ric of The LBC]

You guys need to find a hobby.

 

[holdit]

You wouldn't believe it...

After much brainwhacking, we came up with 17x captured .50Cals manned by 34 Heroes each per level.

Apparently, there have been escapist nerds in reality, too. Look at this:

von Marwitz

Talk about recycling... I wonder where the chassis came from.

I hope they had truckloads of captured ammunition to go with it those MGs.

"It worked perfectly comrade, but we only got off a couple of five-second bursts before the ammunition ran out..."

 

[Michael Dorosh]

You wouldn't believe it...

After much brainwhacking, we came up with 17x captured .50Cals manned by 34 Heroes each per level.

Apparently, there have been escapist nerds in reality, too. Look at this:

View attachment 13633


von Marwitz

"It worked perfectly comrade, but we only got off a couple of five-second bursts before the ammunition ran out..."

Worse than that, the use of the assault drum would limit the device to about 3.6 seconds of usefullness before it had to be reloaded, assuming the guns fired at the same time. Without hydraulic assistance, I suppose the point of the gun is to put out a (brief) volume of fire and pray a German airplane flies into its path since I can't imagine any human operator being able to track a moving target with this contraption.

 

[Futbol]

but as fate would have it the attack rolls a 12 resulting in random selection which breaks every single MG gotta love every possible outcome...