Low Luck with AA guns and anti-aircraft fire

  • Customizer

    Ok, I have a question regarding how to use Low Luck with aa guns’ anti-aircraft fire.

    lets say i attack somewhere that has an aa gun, with 4 fighters.
    in low luck, you would roll a 4, to see if you take a single fighter as a casualty.
    this, i understand fine.

    now what if i attack with 6 fighters and 6 bombers?
    that would be 1 fighter dead, and 1 bomber dead?  correct?

    now what if I attack with 2 fighters and 2 bombers?
    I say, you would roll at 2 to take a fighter as casualty, then roll at 2 to take a bomber as casualty.
    My friend says that with low luck, you would roll at 4 to take a single casualty.  Then roll another dice to determine if it is a fighter or a bomber (ie: roll a die, if it is 3 or less fighter dies, if it is 4 or more, bomber dies).

    Which is correct?

    We both agree that if you attack with 3 fighters, and 6 bombers, you would lose a single bomber, and roll at 3 to determine if you lose a fighter or not.
    But he says, if you, say, attacked with 1 fighter and 5 bombers, you would lose 1 aircraft only.  you would roll a dice, if it hits at 1, you lose the fighter, if it hits at anything else (2-6), you lose a bomber.

    So what is the correct way?

    thx,
    veqryn

    edit:

    actually, i was wrong in what he thought was the system:

    here is the system he thought Low Luck was:

    you attack with 9 fighters and 3 bombers
    it means 2 definite hits
    so you randomly choose 2 hits
    which means you could get 2 bombers dead

  • '16 '15 '10

    I think something like this is subjective…there is no ‘correct’ way.

    While it would be nice to have a separate roll for figs and bombs, I think the principle that takes priority ought to be 6 planes=1 guaranteed hit.  I don’t mind so much if the plane hit is determined by random selection…

    Sounds like a matter of preference in the end.

  • Customizer

    comon…. i know lots of people do Low Luck here…
    is there a consensus?


  • Remember, he needs this for TripleA. He needs a definite answer.


  • @Veqryn:

    comon…. i know lots of people do Low Luck here…
    is there a consensus?

    In my opinion, bombers and fighters are kept completely seperate, so you would always roll based on the totals.  If 9 fighters and 3 bombers, you autokill one fighter and roll a 3 for the remaining fighters and roll a 3 for remaining bombers.

    But that’s my opinion, and part of that is based on my feeling that low luck limits randomness and having to roll to see what you hit (a fighter or a bomber, in a stack of 6) seems counter to that.  And that’s also why I don’t care for low luck, because you can specifically design an attack for a minimum needed win in every attack, rather than truly needing a safety factor.


  • Hmm, it depends. Both ways are LL, but one is with less luck than the other. It also depends on the complexity you want to add in the rules.

    version 1: you roll for fighters and bombers separately. This means 1 fighter and 1 bomber attacking an AA would be the same as ADS => this is the “high” LL variant, but it’s also the most simple one.

    version 2: you roll a “to hit” die for fighters and bombers separately, and roll another die to determine if the casualty is a bomber or fighter. This is “real” low luck, but it’s very complicated. For instance, 5 fighters and 4 bombers are attacking a territory, what dice to roll? 5+4 = 9, so this is one sure hit and one die @3. Suppose no extra hit. It’s clear it’s very complicated deciding which aircraft is hit. One solution is: roll 5 dice for the fighters, 4 for the bombers, and keep the die with the highest score (make sure to remember which dice were what aircraft!). In case of a tie, reroll the dice with the tie scores. Keep doing this untill there are as many dice with a highest score as there are casualties to make. To keep in touch with our example: 5 dice for fighters, rolled at 2, 2, 3, 4, 4, and 4 dice for bombers, rolled at 1, 3, 3, 4. Reroll the 4’s: ftr: 1, 6, bmr: 6. Reroll again: ftr 3, bmr 2. Conclusion: the fighter is the one casualty to take. This would get even more complicated if the die @3 from the beginning was also a hit!

    So in my opionion, version 1 is to be preferred. If your friend opposes this view, ask him how he would handle the 5ftr+4bmr attack 😉

  • Customizer

    it gets even more confusing if you play games with more than just 2 types of aircraft.  For example, A&A Pacific 1940 has 3 types of aircraft, and New World Order has 4 types (early fighter, modern fighter, bomber, and a special aircraft for some powers, like the Lancaster, or the B29, etc.)

    Because in games like this, or games with say 6 types of air, you are basically playing dice now Low Luck.

    The way we are settling on for now is this:

    First, remove any groups of 6 for each type.  So if you are attacking with 17 bombers, 1 fighter, 6 bi-planes, and 4 lancasters, you would remove 12 bombers (2 die, 10 live), and 6 bi-planes (1 die, 5 live).
    That leaves you with 5 bombers, 1 fighter, and 4 lancasters.
    Thats 10 total.
    So thats 1 more hit, plus a 4/6 chance for another hit.
    Roll at 4 to determine if there is an extra hit or not.  We will pretend it is a hit.
    Now you have 2 hits to distribute.
    Line up all the aircraft, and randomly select 2 of them to die.  (this can be done by randomly generating 2 numbers between 1 and X, where X is the total number of aircraft.  if the second number is the same, you can loop until it is not.)

    The only problem with this method is that you can technically get both hits on bombers.

    We have been investigating other methods to ensure that the hits will be distributed, so that you can not get 2 hits on 1 aircraft type in the final example, but all of the methods we have thought of increase the odds of getting a hit on a less represented aircraft.
    For example, pretend you were playing a game with 1000 sided dice.  You attack with 999 fighters and 1 bomber, and 2 tactical bombers.
    This means 1 definite hit and a chance of a 2nd hit if you can roll at 2/1000.  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.


  • 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:

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