@wanguskhan
They may reprint it again, preserving all the errors and the illogicially ordered setup charts.
Comp for WW1 is $350-600 now.
That’s more a statistics question then a mathematical one. It’s based on probability, the binomial equation I think.
Statistically speaking you have a 17% chance to shoot down a solitary bomber each round. Many of the users here will tell you to add that up and expect to lose a bomber every 6 rounds. I don’t believe that is the correct method since the odds do not magically increase each round. (The results are independent, not dependent.)
You could take the number of rounds you want the bomber to survive and work out the equation as follows:
Odds of Survival: 5/6
Odds of Death: 1/6
Let’s say you want 5 rounds and to be shot down on the 6th Round:
(5/6)(5/6)(5/6)(5/6)(5/6)*(1/6) or 7% of the time your bomber will live for 5 rounds and be shot down on the 6th round.
The most common method for determining the longevity of a bomber vs an AA gun is to just assume it will die every 6 shots so you’ll get 5 attacks before you die.
Obviously, the numbers shift dramatically with Radar since it’s a 1/3 chance of being shot down instead of 1/6 now!
hiya,
i’m not that deep into math but that’s how i would figure it out:
first you must find out the odds. this is very basic and i think you got it yet. a normal aa has a hit probability of 1/6 while a radar aa has a hit probability of 1/3.
to get the average hits you simply add up the odds: in two rounds you score 0,333 hits (standard aa: 1/6 + 1/6) / 0,666 hits (radar: 1/3 + 1/3).
but i would rather suggest that you calculate the probability for a bomber to be hit in x rounds. therefor you multiply the odds that a bomber is not hit ^ number of rounds and substract this from 1:
(example given for standard aa)
prob to be hit in 1 round: 1- 5/6 = 1/6
prob to be hit in any one of 2 rounds: 1 - (5/6 * 5/6) = 11/36 = 30,5%
prob to be hit in any one of 3 rounds: 1 - (5/6 * 5/6 * 5/6) = 671/1296 = 51,8%
and so on.
hope that helps.
p.s.: aas fire once per combat - you got that, didn’t you?
p.s.: aas fire once per combat - you got that, didn’t you?
Which is the problem with using Frood’s site and bombers vs x number of transports. the transports would get to hit hundreds of times, the AA Gun could only hit once.
Also, to clarify, 1-(5/6) is a way to determine the odds of something happening.
Odds of Something Happen = 1 MINUS the odds of something NOT happening.
Standard statistical principle.
Can any of you guy explain some mathematical basics to me?
I would like to know in which round a bomber would be statistically shot down on normal AA-rolls and radar-improved AA-rolls? How do you calculate this?
THX!
And you should remember one thing: stats have no memory. After the 5th successful bombing the bomber should be shot down according to theory. Probably it won’t. Or the bomber gets shot down on the 1st bombing; and you send a 2nd bomber considering that now the odds are 1/11 that it will be shot down. But it does get shot down by the AA right away.
On the long run (and that can be millions or zillions of tries) the stats work out. On the short one…
And you should remember one thing: stats have no memory. After the 5th successful bombing the bomber should be shot down according to theory. Probably it won’t. Or the bomber gets shot down on the 1st bombing; and you send a 2nd bomber considering that now the odds are 1/11 that it will be shot down. But it does get shot down by the AA right away.
On the long run (and that can be millions or zillions of tries) the stats work out. On the short one…
yo. the odds for a single event are not influenced by the events that happened before. otherwise you could become a rich man by throwing a fair (!) coin secretly until you get a very very very long row of heads, and then bet a huge amount on tails :-D
Ok, I’ve applied your statistics created a diagramme going up to ten passes (attachment)
Now the most interresting point I’ve found is, that the effectivity of SBR should indeed rise with higher numbers of participating aircraft. That is because with each additional aircraft the number of bombers that might survive multiple turns
rises. Just watch yourselves
Try using the search function sometime. :wink:
This thread explains everything.
Exactly, that’s why I don’t believe in Low Luck where you say the bomber WILL be shot down every 6 rolls. Perhaps it will on average, say, if you rolled 6 AA Gun shots 100,000,000 times you should average 1 hit in 6 rolls…but that does not really apply to the instance of one AA Gun shot at your bomber that is currently conducting it’s 6th SBR mission. (The chance there is still 1 in 6, it is not 6 in 6 because the gun already missed 5 times.)
@Cmdr:
Exactly, that’s why I don’t believe in Low Luck where you say the bomber WILL be shot down every 6 rolls. Perhaps it will on average, say, if you rolled 6 AA Gun shots 100,000,000 times you should average 1 hit in 6 rolls…but that does not really apply to the instance of one AA Gun shot at your bomber that is currently conducting it’s 6th SBR mission. (The chance there is still 1 in 6, it is not 6 in 6 because the gun already missed 5 times.)
Low Luck doesn’t do that. It takes the average amount of damage per run(3.5) and then takes the average value of loss per bomber per bombing run(1/6th of the bombers value[15]=2.5) and since you can’t give or take 0.5 IPC’s, it rolls a die for high or low for each value. The bomber either does 3 or 4 points of damage and the bombing run costs the bombers owner 2 or 3 IPC’s.
It’s the same principle. You are assuming that every 6 runs you WILL lose a bomber to the AA Gun, so you completely nerf the damage from SBR making it a kind of stupid move to make. (Gee, I can lose 2.5 IPC and do 3.5 IPC in damage, or I can just kill one of his infantrymen and probably not lose the bomber.)
BTW, that is the basic idea of LL. You assume that 3 defending infantry WILL get 1 hit. But that’s really a topic well covered a few times in the past. (Search for LL vs ADS)
the triplea low luck just gives the aa GUN a to-hit = to the # of planes
so if you bring 6 planes you lose 1 for sure, the other 5 bomb as normal. If you bring 5 it rolls and on 5 or less they lose 1 bomber.
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.