News flash: AXIS & ALLIES ANNIVERSARY EDITION due out oct 23 08

Agreed, need and want are different. From what I gather, they only waited about 3 years between 1942.1 and 1942.2.

It’s nice to not have to house rule too much - even using a bid implies some additional play testing would have been beneficial. But to be able to discuss strategies here, etc, requires us to have the same rules. :)

@SS:

Ok then. Axis player what is the best way to play it , with or without tech and NO’s?

I would preface this with my opinions about these options.

National Objectives

Usually these are more often lost and hard to (re-)obtain for the allies.  The axis are usually advancing and can more easily achieve their national objectives, with the exception of Italy, whom the allies can deny the Italian NOs more easily.

Advantage with National Objectives:  to the Axis

Weapons Development

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I am not a big fan of the randomness that is built into the achieving of technologies.  Plus the techs are not even in their strengths:  Super Subs for Russia?  practically worthless

But if you play with tech, I think that favors the Axis.  Why?  Most techs are offensive in nature, which is the Axis main strategy/strength, especially early.  It would be nice if Jet Fighters were still a ‘5’ on defense instead of a 4 on offense (no longer a defensive tech).  Similiarly, advanced art could be a defensive aid as well (inf defense to a ‘3’).

Advantage with Weapons Development:  to the Axis

Again, merely my opinion, but with the reason(s) I why I believe so.

Long live tech!

@Krieghund:

@P@nther:

@Krieghund:

Since Jet Fighters increases fighters’ attack values to equal their defense values in normal combat, I’d say it would do the same in SBRs, allowing attacking fighters to roll on a 2. However, Larry will need to confirm that.

But this is still not confirmed by Larry as I think. Or is it?

Yes, it’s just been confirmed.  It will be in the next version of the FAQ (soon, I hope).

Great, thank you :-)

@shifty:

What kind of “program” are you writing?  What equation/algorithm are you using?

It’s written in BlitzMax. The source below should (hopefully :)) not be too hard to understand - if you have questions, feel free to ask:

' Compares 2nd ed. A&A tech dev cost/rate with Anniv. ed. A&A ' 20 turns to test; record turn number that a total of 2 breakthroughs were achieved.  Test 1 & 2 rolls/turn. ' If two breakthroughs aren't achieved, "turn 0" is recorded (ignored for graphing & turn average) ' (When going for two breakthroughs, two aren't achieved in about 2% for both 2nd and Anniv.) Strict Local breakthroughs, totalCost, turn, totalTurns Local answer[21], prefix$, graph$ Local breakthroughLimit = 2 For Local rolls = 1 To 2 ' actual rolls for 2nd Ed, Researchers for Anniv. Print;Print; Print rolls + " Tech Roll(s)/Researcher(s) per Turn:" Print; Print "2nd Edition" totalCost = 0 totalTurns = 0 For Local c = 0 To 20 answer[c] = 0 Next For Local test = 1 To 100000 ' a hundred thousand tests for accuracy breakthroughs = 0 turn = 0 While breakthroughs < breakthroughLimit And turn <= 20 turn :+ 1 totalCost :+ rolls * 5 ' each roll costs 5 IPCs For Local c = 1 To rolls If Rand(6) = 6 Then breakthroughs :+ 1 Next Wend totalTurns :+ turn If turn > 20 Then answer[0] :+ 1 Else answer[turn] :+ 1 EndIf Next ' display results For Local c = 1 To 20 graph = "" ' divide results by 1000 so each asterisk represents one percent For Local cc = 1 To Int(answer[c]/1000.0 + .5) graph :+"*" Next If c < 10 Then prefix = " " Else prefix ="" ' extra space for single digits Print prefix + c + " " + graph Next Print "Avg turn = " + totalTurns / 100000.0 Print "Avg cost = " + totalCost / 100000.0 Print; Print "Anniversary Edition" totalCost = 0 totalTurns = 0 For Local c = 0 To 20 answer[c] = 0 Next Local researchers ' researchers accumulate from turn to turn For Local test = 1 To 100000 researchers = 0 breakthroughs = 0 turn = 0 While breakthroughs < breakthroughLimit And turn <= 20 turn :+ 1 totalCost :+ rolls * 5 ' each researcher costs 5 IPCs researchers :+ 1 * rolls ' accumulate 1 researcher per "roll" Local success For Local c = 1 To researchers If Rand(6) = 6 Then success = True ' only one regardless of # of sixes rolled Next If success Then researchers = 0 ' eliminate accumulated researchers when successful breakthroughs :+ 1 EndIf Wend totalTurns :+ turn If turn > 20 Then answer[0] :+ 1 Else answer[turn] :+ 1 EndIf Next ' display results For Local c = 1 To 20 graph = "" ' divide results by 1000 so each asterisk represents one percent For Local cc = 1 To Int(answer[c]/1000.0 + .5) graph :+"*" Next If c < 10 Then prefix = " " Else prefix ="" ' extra space for single digits Print prefix + c + " " + graph Next Print "Avg turn = " + totalTurns / 100000.0 Print "Avg cost = " + totalCost / 100000.0 Next

I release it into the public domain (modify it as desired/convert it to the language of your choice).  breakthroughLimit is initially set for 2, which generated the first set of graphs in this thread (I hand-rounded the accuracy to a single decimal place when copy/pasting results in the thread). Setting it to 1 (i.e. stopping after getting one breakthrough) produced the second set above. I also made another, slightly tweaked version to generate the “only buy Researchers on turn 1” version for wodan46.

:-)

This thread appears to be long dead, but I never get tired of math questions!  :-)

To answer the original question: The statistics of when a bomber gets shot down are determined by something called the geometric distribution, which you can find plenty of info about at http://en.wikipedia.org/wiki/Geometric_distribution.  The geometric distribution describes the first occurrence of an event when you make repeated, independent trials.  For example, the number of times you have to flip a coin before getting heads; the number of times you have to be dealt a poker hand before getting a royal flush; and the number of times you have to roll a die before getting a 1 are all described by geometric distributions.  Once you know the probability of success in a single event, the geometric distribution tells you how long you have to wait before seeing the first success.

Some of the information for the geometric distribution:
–-The probability of getting your first success (in our example “success” is the bomber getting shot down!) on the nth turn is (1/6)*(5/6)^(n-1).  This is because, in order to be shot down on the nth turn, the bomber must first survive the preceding n-1 turns, which has a probability of (5/6)^(n-1), and must then be shot down on turn n, which has a probability of 1/6.
–-The probability of getting your first success by the nth turn is 1 - (5/6)^n.  This is just 1 minus the probability of the bomber surviving the first n turns.  (It’s also the sum of the probabilities of getting shot down on turn 1, 2, …, up to n, but that’s the long way to do it!)

Given this, a number of other facts can be calculated (I’ll spare you the details…)  The question “When will the bomber, statistically speaking, be shot down?” has three different answers: the mean, median, and mode.  Here’s what each of those is:
—The mode is the single turn on which the bomber is most likely to be shot down.  This may be surprising, but it’s the very first turn!  The probability of getting shot down is the same (1/6) on all turns, provided the bomber makes it to that turn , but first it must survive all the turns before it.  Hence, the first turn is the most likely turn-of-death since the bomber doesn’t have to survive any prior turns in order to reach it.
–-The median is the first turn for which the bomber has a 50% chance of getting shot down before it.  This is the fourth turn, because the probability of getting shot down on one of the first four turns is 52% whereas the probability of surviving the first four turns is 48%.  So it’s about 50-50 whether the bomber survives at least 4 turns.
–-The mean , also known as the expected value, is the average number of turns that the bomber will survive.  This is 6 turns.  It turns out that you can actually calculate this by taking the reciprocal of the 1/6 probability of getting shot down on a given turn, although the reason why that works is slightly less obvious than it might seem.

So, which of the mean, median, and mode is the most useful?  When should we expect to lose the bomber?  Well, the mode is the least useful for answering this sort of question.  One can make a case for either the mean or the median, but for the types of calculation that people usually have in mind, the mean is the way to go.  For example, if you’re trying to figure out the average amount of economic damage inflicted minus damage received from each bombing raid, then the statistic you want is the average, aka the mean.

Of course, the question of whether bombing raids are worth doing involves a lot more than calculating the average IPC’s lost and destroyed.  My test: If you want to decide whether bombing a given country with your country is useful, ask yourself, “Would I destroy $10 of mine if I also got to destroy $10 of theirs?”  If the answer is “Heck yes!” then SBR’s are useful.  For example, Japan bombing Russia generally makes sense, as does US or UK bombing Germany.  A weaker power bombing a stronger power is usually not so smart.