I went when I was really young. I think I mainly visited resorts and amusement parks.
DAAK
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By the way, frood, to do a purely mathematical calculator for battle outcomes, rather than a simulator, the task would be incredibly complicated if you were going to do ADS.
You’d have to keep an array of all the individual units in the battle, and calculate the probabilities that each one would get a hit or miss. Then do the same for the defending army. Then recursively consider the probabilities for each of the possible resulting outcomes. You perhaps could save a bit of calculations by using a Pascal’s triangle or combination table. It’s probably easier if I gave you an example:
Consider 2 tanks 3 infantry (attacking)versus 5 infantry (defending).
For each attacking armor, there is a (1/2) probability of hitting or missing, and for each attacking infantry, there is a a (1/6) probability of hitting and a (5/6) probability of missing. For each defending unit, there is a (1/3) chance of hitting and (2/3) chance of missing.
Considering the defender, the respective probabilities are as follows:
5 hits: (1/3)^5
4 hits: (2/3) * (1/3)^4 * 5
3 hits: (2/3)^2 * (1/3)^3 * 10
2 hits: (2/3)^3 * (1/3)^2 * 10
1 hit: (2/3)^4 * (1/3) * 5
0 hits: (2/3)^5(because there are 5 ways that 4 units can hit and 1 miss, 10 ways that 3 can hit and 2 miss, etc)
Then considering the attacker, things get a bit more complicated:
5 hits: (1/2)^2 * (1/6)^3
4 hits: (2 * (1/2)*(1/2) * (1/6)^3) + ((1/2)^2 * 3 (1/6)^2 * (5/6))
3 hits: ((1/2)^2 * (1/6)^3) + (2 * (1/2)(1/2) * 3 (1/6)^2 * (5/6)) + ((1/2)^2 * 3 * (1/6) * (5/6)^2)
2 hits: ((1/2)^2 * 3 (1/6)^2 * (5/6)) + (2 * (1/2)(1/2) * 3 * (1/6) * (5/6)^2) + ((1/2)^2 * (5/6)^3)
1 hit : ((1/2)^2 * 3 * (1/6) * (5/6)^2) + (2 * (1/2)(1/2) * (5/6)^3)
0 hits: ((1/2)^2 * (5/6)^3)And you know what if you can’t figure out how I came up with that last part I don’t think I’m going to bother to explain it because you need to have some background in probability and combinatorics and I don’t have time to give you a course. The brief explanation is, for 5 hits it was just the chances that both tanks would hit and all 3 infantry would hit, then for 4 hits it was the chance that both tanks would hit and 2 infantry would hit PLUS the probability that only one tank would hit but all 3 infantry would hit, and so on.
The coefficients being used in all cases are the values of the combinations which can be pulled from Pascal’s triangle as I mentioned above. It’s the number of ways of picking k distinct combinations from n items. The standard formula for that is n!/((n-k)!k!) [! = factorial]. So for instance the number of ways 6 units can get 3 hits is 6!/(3!*3!)=20.
Ok, so then anyway, you calculate all those values for the number of hits for each army, then you multiply all that together to get the probabilities for all the outcomes on that round. So for instance, the probability that the outcome of the first round will be 0 units remaining on both sides is the product that both sides will get 5 hits: (1/3)^5 * (1/2)^2 * (1/6)^3.
Now that you have all the respective probabilities of the results of that round, you take those outcomes that don’t result in the end of the battle and you recursively analyze them as if the battle were starting with that. But you see that this is beginning to get extremely computationally intensive, since there are already 36 different outcomes from the first round, only 3 of which result in the battle ending, so you have 33 different things to analyze recursively!
A further problem occurs when you consider the case that one of those 36 outcomes is that no one got any hits. This will always be a possible outcome of every position you analyze, so we face the prospect that our recursion will NEVER terminate. It is always possible that even if we have had 500 rounds of battle and no one has had any hits yet, that for the next round of battle no one will get any hits again.
To force our program to always terminate, we will then have to “eliminate” the case that no one gets any hits, because the situation ends up identical to the previous situation. So what you do is that for all the other 35 outcomes given above, divide the probability of their occurence with 1 minus the probability of no hits. So the “adjusted” probability then for 5 hits on both sides then is:
((1/3)^5 * (1/2)^2 * (1/6)^3) / (1 - ((1/2)^2 * (5/6)^3 * (2/3)^5) )
Got all that?
Anyway, even ignoring all the calculations about probabilities, which is simple enough to do for a computer once you’ve got it programmed, the recursion is still exponential and that is a huge problem. Calculating the outcome of a battle of 5 units against 5 units would probably take several seconds even on a fast computer. Actually probably more, given that after only 5 rounds of battle you already have on the order of 60 million possible threads of recursion, although most of those are probably going to be duplicates. Calculating 10 on 10 would probably take hours. Forget about doing a Karelia vs Eastern Europe scaled battle - it could well take decades for the computation to complete. Such is the nature of exponentially complex functions.
Now if you were trying to do the calculator for LowLuck though, that’s a different issue altogether. That’s easily doable in my opinion, and I made some posts on it a while back on the Axis and Allies forum here in response to someone who was trying to do just that. The reasons that it’s much simpler is that you can ignore the exact composition of units - only the total strength of each army matters. Furthermore, there will always be at most only 4 different outcomes for each round of battle, which shaves the recursion down considerably. Due to the nature of it also, there will be significant overlap after two rounds or more. Large battles also reduce in size much faster since each player is guaranteed to get a certain number of hits.
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Or just run the sim in your head and allow for some variations. If you get a set of rolls that are abnormal, deal with it :-P
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Well he was the one asking how you would calculate the odds mathematically. My last post was in response to frood’s query:
The question is, does anyone know how to calculate those odds? Even for a single round of 8 armor v. 8 infantry, as a moderately complex example? What’s the formula?
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I am sending Dan a few bucks to help maintian the site… I am not going to buy him a Cray to run the calcs :-D
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@ncscswitch:
Or just run the sim in your head and allow for some variations. If you get a set of rolls that are abnormal, deal with it :-P
Or just assume you’ll get 20% of your expected hits, the defender will get 160% of his expected hits and plan accordingly.
(Don’t assume a win unless your odds show 97% or better, works too.)
(Always bring 4:1 odds, works well too!)
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(Always bring 4:1 odds, works well too!)
Is there any other way? (unless playing LL?)
You ALWAYS use 3-1 odds as a minimum, 4-1 prefered. And when you can get 20-1, you bring ALL of it.
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America’s all about compromise, right? Lol how about a mix of LL and ADS, based on luck? At the beginning of every battle, roll a die. If it’s 1-3, you do it LL. If it’s 4-6, you do it ADS :-D :mrgreen: :wink: :roll:
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no ll for me, i like ads, allows for the games to be very different instead of just repetitive……ive also noticed bombers get shot down a lot when playing ftf
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:-) From the department of beating a dead horse :-)
If you don’t like dice :cry: play Chess, Go, Checkers, et al. If you don’t want a random element beyond your opponent play a game without one. :-)
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Well duh. If you can get 200:1 odds you do it! Unless you need them for other battles.
Better to kill all enemies in one round of combat then to deal with two rounds of return fire. :)
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ADS is what gives you hair on your chest. Some people will look at ADS and chicken out when they see the possibility of 1 infantry beating 127 tanks. Or it may be more reasonable such as chickening out when sending 4 tanks against 2 tanks. Sure, losing those battles does happen, but on the other hand, astouding success also happens. Your 4 tanks could suffer no injuries. Your 127 tanks could fight 127 infantry in separate battles and never take a loss. In the long run, those who do their math the best and stick to it are the ones who win. It’s just like in poker, you have to play the odds consistently with the math every time, or you will end up losing in the long run in spite of short term hot streaks. It is hard, and separates the men from the boys. Overcompensating too much for luck is as bad as bad luck.
But then again, I do see the use of LL. If you want a shortcut to the long term average, if you want a game that does not have such highly varying degrees of luck, if you want to test a strategy without having to factor in screwy dice, then go ahead and use it. Just remember that the risk and challenge of ADS makes it a different and possibly more exciting game.
And by all this talk of hair on chests and boys and men, I didn’t mean to exclude you, Jennifer 8-)
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Actually, no luck has it’s benefits too. When DAAK is down you can still play!
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What exactly is no luck? o_O
Does 1 infantry even get a roll, or what? Do you do something weird like decide every 6th infantry you roll will hit or something?
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If I am not mistaken…
Like LL, but if the remainder is 1-3, it misses, 4-6 it hits. -
Yea. Though, I believe it is 3-6 it hits, 1, 2 it misses. Gives you benefit of the doubt.
So yes, 1 attacking infantry misses 100% of the time. 3 attacking infantry hits once, every time. Though, not typically accurate given that most times 2 attacking infantry are better then 1 defending infantry.
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Hmmm… I’ve never played a no-luck game actually. I’d be interested in trying one some time.
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The euphemism for no-luck is “A&A Chess”
The result of every battle is a pre-determined attacker victory,the only way an attacker loses a battle is to screw up what they send to it.
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That’s certainly not true - an attacker might decide to lose a battle on purpose. I can certainly think of situations where doing so might be to his advantage. Not common situations, but it’s not inconceivable.
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OK, then add “or by choice” to my statement.
Either way, the battle results ARE 100% pre-determined.
For example, in a battle of 49 INF, and 49 ART against 99 INF:
In No-Luck the defedner will always have 8 INF left.
In Low-Luck, the range is from winning the fight with 7 ART surviving to losing with 16 INF left
In ADS the attacker has a 43% chance of victory and a chance at up to 37 ART alive. The defender has a 57% chance of winning, with up to 41 INF left, and the percentage chance for 37 INF left is exactly the same as to have 8 INF left: 2 %Another Example: 6 ART attacking 6 INF.
In No Luck it is an Unresolvable Battle. Both sides get into a perpetual miss scenario with 1 ART attacker, 1 INF defender left. Attacker MUST retreat 1 ART leaving the Defender victorious.
In Low Luck the odds are 46% attacker, 47% defender, 7% mutual destruction, with an equal percentage chance, 22% of 1 ART attacker or 1 INF Defender
In ADS, it is a 48% attacker, 51% defender, 2% mutual destruction… a coin toss.Hell of a big difference between 50/50 and a guaranteed win.
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Yea, but let’s not forget that in ADS 1 submarine defending against 5 destroyers and 20 battleships has a chance to win. In No luck and Low Luck that submarines toast, even the particles left over after the shells are done falling on that submarine are damaged, they’ll never be the same again. At best, that submarine has a small chance of actually killing the cook on the destroyer with an errant deck gun shot…but it’s a small chance.
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