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

@Taewakan

I have moved the topic to the Anniversary-category. @barnee is correct.
This question is part of the official FAQ:

Q. Are sea zone 1 and Western Canada adjacent?
A. No. The gold map border obscures the fact that the Hudson Bay area of sea zone 1 doesn’t quite
reach all the way to the western edge of Eastern Canada.

@warTorn:

@vodot:

Guys, use 7zip!

Seriously, freaking WinRAR!? For what- archiving the album of .mp3s you downloaded from napster over the course of an entire evening with your state-of-the-art, 38880 baud, dial-up modem?

I kid, but seriously: www.7zip.org.

EDIT: If you’re on a mac, use this instead (it’s a 7zip port) www.kekaosx.com/

YOU ARE A SAVIOR!!!

:-)

Well yes. If you get an account with mediafire, you wont get that. Everyone should have at least a dropbox account and probably Mediafire

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.

@axis_roll:

Don’t get me wrong, I like to KJF, it’s more fun, since it IS harder to execute.  I like a challenge.  Probably why I like to play the Axis… they’re the underdog.

So KJF is an underdog type of allied strategy.   :wink:

KJF is harder of master, not of execute. When you master KJF, it’s more powerful than a mastered KGF. First, because very few people know how counter KJF and many start to do strange things. Second, because it’s easier for axis reach economic parity in KGF than in KGF. Third, for league games, I found it’s easier reach 9 VCs with KJF (Manila, Shangai, Paris). Fourth, a KGF strat let Japan try invade american mainland (a strat I still don’t master, but seems very nasty in a initial KGF game)

No prob dude!

Even then you may have two first place winners.  For example the Allies may have won as a team but Germany won as a player.

LT