Pretending you get that hit, and way to ensure that only 1 fighter dies would dramatically increase the odds of bombers and tactical bombers dieing.
I disagree, you forget to take into account the chances of an extra hit are minimal. Let’s redo the example, but with assuring maximum one type of aircraft survives, for instance, simply remove the fighters/bmrs from the pool after you assign the first hit.
So redoing the example with 999 fighters + 3 bmrs. First assume no extra hit: you allocate the sure hit by rolling a 1002 sided die, if it’s <1000, a fighter dies, if it’s >=1000, a bmr dies. With extra hit: a fighter and bomber dies (this follows from rolling a 1002 sided die for the sure hit first, remove the type of casualty, and roll another die for the other type, which by then is the only type left, so will for sure be hit). Now let’s analyze chances:
The chance you hit a fighter is: the chance the sure hit is a fighter + the chance the sure hit isn’t a fighter (which removes the bmrs from the group) and the extra hit hits. Mathematically: 999/1002 + 3/10022/1000 = 99,7%. The chance you hit a bmr is the chance the first hit is a bmr + the chance the first hit is not a bmr and the extra hit hits. Mathematically: 3/1002 + 999/10022/1000 = 0,5%. (Note these chances do not amount to 1 since on average more than one unit is hit. Moreover, on average, 1,002 units are hit, which is exactly the sum of the chances ) So in almost all battles, a fighter will be killed, but never more than one, and almost never a bomber, and never more than one. So no dramatical increases of bombers dieing with this approach, a very slight increase at best. Very much within the margins of LL imho 8-)
This reminds me of a method of handling AA I invented before, struggling with the same problem in AAR LL. I forgot it in my previous post :roll: It’s fairly elegant, but has (in little cases) a chance of “both bmrs dieing”. Here’s how it goes:
So, 9 fighters, 5 bombers, 7 tacs, 8 lancasters, 1 zeppelin and 3 ufo’s are being attacked by an AA. First remove groups of 6’s and remove these sure hits, just like you said. Then line them up on one long row. So 3 ftr, 5 bmr, 1 tac, 2 lnc, 1 zep, 3 ufo. Now divide them in ordered groups of 6: 3 ftr + 3 bmr, 2 bmr + 1 tac + 2 lnc + 1 zep, 3 ufo (= rest), and roll a die for each group, with the number of the die appointing the aircraft hit. For instance, you roll 4, 3, 4. So in the first group a bmr (the 4th aircraft) is hit, in the second a tac (3rd), and in the third nothing (you needed a 3- to hit, the 4th aircraft is “empty”). This is my preferred way of solving LL AA: no more hits than needed, no deep calculations, easy way of handling the extra hits. Actually, it’s a generalization of the “groups of 6 are sure hits”-rule: a group of 6 is nothing more than 6 aircraft for which a die is thrown, but since all the aircraft are of the same type, and since the group is exactly 6 large, the number on the die will always point to an aircraft of the type of which the group consists.
However, there is a problem with this approach: suppose the second die in the example was a 1, then two bmrs would have been hit, which you want to avoid. My advice: order the groups as good as possible (with as little types in two groups as possible). In the example: 5 bmr + 1 zep, 3 ftr + 3 ufo, 1 tac + 2 lnc. As you can see, no aircraft type is in danger of getting two hits. This however is not always possible, but even in those rare cases, the type to split will contain at least 4 members (always split the largest type), so it is acceptable that 2 of those group can die instead of one. I deem this also very much within the approximative margin of LL to ADS.
For instance, applying this method to your example (999 ftr + 3 bmr): divide in 2 groups, with the ftrs split (they are the largest type). So 3 bmr + 997 ftr, 2 ftr. Now the chances of a bmr being hit are 3/1000 = 0,3% , the chances of (minimum) one ftr being hit are 997/1000 + 2/1000 = 99,9% These are actually the exact average chances of ADS! And compared to your method (which has the advantage of no double bmrs hit), it’s pretty close I think.
To summarize, use my method, with the “split only if needed, split to never more than two, and split the largest types first”-rule, and you won’t need to worry about chances or double bombers. Happy rule crunching :mrgreen: