December 09, 2013, 08:10:44 pm
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 Author Topic: Odd calculator for a single round of battle  (Read 1095 times)
Anonymous
Guest
 « on: January 08, 2003, 07:04:43 am » 0

hi there

just wonder if there exists one. It would be useful to calculate odds of attacks, that are just made 1 or 2 Combat rounds to weaken the enemy and/or get a good retreat position (ie G1 attacking Karelia including troops from Finland hoping for a good roll to get the tank out).

thanks&greetz
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Attilla the Hun
A&A.org New Conscript

Posts: 4

 « Reply #1 on: January 09, 2003, 03:43:33 pm » 0

Of course there is.  I rely on a tiny microchip that was implanted into my brain when I was a child.  It's specifically designed to calculate the odds of winning Axis&Allies battles.   It works on a simple principle: addition of fractions.  Ha!

Please excuse the sarcasm , but the calculations are pretty simple.  Men have a 1/6 chance of scoring a hit (they attack at one, thus 1/the six possible sides of the dice).  Multiply the # of men times that fraction.  Tanks and fighters have a 1/2 chance of scoring a hit.  Planes have a 2/3 chance.  Just add up all of those factors, depending upon what and how many units you're attacking with, and that is the average number of hits that you should score.  Decide which men your opponent will likely put behind the casualty line.

Then take into consideration the number of hits that the defense will likely score.  Same thing, except defensive men and tanks=1/3, defensive planes =2/3, defensive bombers =1/6.  Take that estimate of hits, and decide which units of yours would die if you did attack.  Keep doing this (in your head) until it is clear who will win the battle.  Perfect odds rarely pan out, so it pays to calculate a few extra hits toward yourself or round up when the defender's hits add to a fraction.

OK, you're right, probably quicker just to find some odds calculator.  But it's so much more fun to use the left side of our brain  Good Luck.
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F_alk
Guest
 « Reply #2 on: January 10, 2003, 02:21:01 am » 0

Quote from: Attilla the Hun
Of course there is.  ...   It works on a simple principle: addition of fractions.

Which of course is an approximation to first order. It ususally is sufficient though, unless you go into battles that probably last a few rounds or has many units involved, then the reliability of your precalculated results will be pretty low (due to rounding effects, not taking into account "multiple hits", e.g. your two attacking infantry have a chance of 1 in 36 to hit twice).
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DarthMaximus
Moderator
A&A.org Heavy Bomber

Posts: 20268

 « Reply #3 on: January 10, 2003, 10:18:56 am » 0

Attilla's method is a common way to get a quick overview of how things "should" turn out, however it does not give the odd's of winning a battle (or round of combat) other then the general "Oh, should be able to take it, or not".
An example, 6 inf attack 2 inf.  Using Attilla's method you should win and worst case you take with maybe 3 inf.  That might lead you into a false sense of security thinking "this is a battle I can't lose", but there is still the statistical chance that the 2 inf will defeat you, albeit a small one but there is still the chance.  And I believe that's what a proper odd's calculator should calculate.  It should tell you that there is a xx% chance of 6 inf beating 2 inf.
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Poobaloo
A&A.org Mechanized Infantry

Posts: 60

 « Reply #4 on: January 15, 2003, 02:47:57 pm » 0

That is by far the best method to use to get a good feel for how a battle round will turn out.  (adding up the fractions).  If you want an overall percentage of who should win, then you have to go into permutations.

Example, take a very simple battle:  1 sub attacking 1 sub defending.

Sub A hit: 1/3
Sub A miss: 2/3

So 1/3 of the time, the attacking sub will win right off.  Now that doesn't mean it's a 1/3 matchup!

Now you get into the IF-THENs...

If Sub A hits, (1/3) then it's battle-win for A.
If Sub A misses, then B gets to return fire...

Sub B return fire hit - 1/3
Sub B return fire miss - 2/3

Now sub B only fires IF sub A missed, so they're dependent results, and you have to multiply the probability of A missing, by each of B's prob's of hitting or missing on the return.

So we have...
A hits... 1/3 (or 3/9 for ease of comparison, later)
A misses (2/3) and B hits (1/3)... 2/3 x 1/3 = 2/9
A misses (2/3) and B misses (2/3)... 1/3 x 2/3 = 4/9

So,
3/9 of the time, A will win.
2/9 of the time, B will win.
4/9 of the time neither will, and the battle will reroll.

So for decisive outcomes, you have 3/9 A win, vs 2/9 B win, or 3:2 odds.  3:2 odds means that 3/5 times A will win, and 2/5 times B will win.

So it is 60% A, 40% B.

That's a very good battle for A, as 60% of the time, the final outcome is likely to be a win, provided neither side retreats, and there is some result eventually.  (it could in theory go on forever in eternity, but assuming there is some result eventually, it'll be 60% / 40%.
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MajorC
A&A.org Infantry

Posts: 26

 « Reply #5 on: January 27, 2003, 04:28:19 pm » 0

my head hurts. I gots the brains pains coach!
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