This is an issue that has cropped up in numerous A&A games. The box board photos are often taken by marketing department people who simply want to make the picture look pretty and who aren’t overly concerned about depicting things accurately.
Re: Statistical odds of AA guns + standard deviations associated low/high sample
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By that logic, you can’t say that there’s a 16.6% chance of a die rolling a one
Their is only if you got a large sample. In terms of 7 total rolls you are correct. Its not 16.6% either.
To get a true average you need a large sample.
If you just played the game once in your life and out of that ONE GAME you only rolled the dice 7 total times for AA capability. I can say the % of what will occur is anywhere from 0% to 100% with anything in between possible.
I suppose if you took every AA game ever played, of which nobody was part of then you might infer a proper % because you got a large sample. But you could personally never see that sample unless you spent the rest of your life rolling dice.
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@Imperious:
By that logic, you can’t say that there’s a 16.6% chance of a die rolling a one
Their is only if you got a large sample. In terms of 7 total rolls you are correct. Its not 16.6% either.
To get a true average you need a large sample.
If you just played the game once in your life and out of that ONE GAME you only rolled the dice 7 total times for AA capability. I can say the % of what will occur is anywhere from 0% to 100% with anything in between possible.
I suppose if you took every AA game ever played, of which nobody was part of then you might infer a proper % because you got a large sample. But you could personally never see that sample unless you spent the rest of your life rolling dice.
Again, I’m talking theoretically. You’re talking experimentally.
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The best way to stop the attack on an Italian fleet (if you feel you need to save it to play Italy well) is to take a very big chance and attack the Carrier, DD and Tac with 2 German subs.
You are likely to lose this fight or worse, give the Tac better range, but since we are taking gambles to save 3 Italian ships, success would end a likely successful attack on the Italian fleet.
Even so, I argue and have seen played out in one game thus far: UK in German hands is still a close game. Our German player ended up fighting Russia in German territories after seizing UK, giving Russia extra NO income. Eventually, with the boost in income, Russia was forced to back up away from Germany. However, as Russia got closer to its production center: Moscow, it was able to slow the German conquest of Moscow well past mattering. This was done by not engaging in a big land battle as the outcome could cost Russia the game, rather when the threat became large enough, Russia withdrew 1 space at a time towards its reinforcements, and since all German units did not move 2 spaces (i.e. Italy can opener did not matter), it was enough distance to insure Germany could not crush the stack as each space moved back, moved into 5-8 additional units.
(for those Russian players needing advice: the route of retreat was Slovakia => E. Poland => W. Ukraine => Bryansk => Russia. Also note many Russians lose Moscow by depleting units from their stack by trying to hold Novgorad. Units stationed there are too slow to race to Moscow if by-passed. It is better to stand off in Belarus and create that city as a kill zone, thus conserving force inventory for Moscow when it matters later in the game.)
Maybe the outcome will differ based on player abilities, but if its still a close game after many more attempts, then the game would not be “broken”. Who would have thought a strategy game, might have many strategies to choose from? And maybe some work better than others, or work better against other strategies. Hmm, sounds like a fun game to me.
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@Imperious:
The rolls of each sample is still 16.6%. Playing 7 times the same game only gives you more chances and to get a 67% you need a larger sample. AA games are not a foundation to demonstrate these samples and why the 67% is a fail.
You’re right, but about the wrong thing.
The chance of shooting down plane 1 is exactly the same as shooting down plane number 2 is exactly the same as shooting down plane number 3 … etc etc etc … is exactly the same as shooting down plane number 100. Because, as you so rightly point out, the dice rolls do not influence eachother at all. Every time the chance is 16.6%.
So the chance of shooting down a specific plane in a single roll will always be 16.6% … no doubt about that.
That’s not the case here though. We’re talking about the chance of shooting down at least one plane in a larger number of rolls. So instead of needing a specific die to show a 1 for a success, I can roll a handful and if even a single 1 shows (no matter which die) it will be a success. This also means that the more dice I roll, the greater the chances of getting one or more 1’s.
So the chances of shooting down at least one plane in a large number of rolls will grow with the number of planes in the attack (= the number of AA rolls made).
It’s the same with land battles. Would you rather attack 1 defending infantry with 2 infantry of your own or with 6 infantry of your own? The answer is of course 6 infantry … because they have a much bigger chance of scoring that hit, even though individually all those infantry have exactly the same chance of hitting as the infantry from the 2 high stack.
8-)
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@Imperious:
Sealion is not a requirement, rather stopping the Italian fleet attack is. The only way to guarantee that is to use my proposal and sink the Labrador DD/AP, which forces UK to leave Italy alone and hole up in UK. because i get 55% on that attack given what we have left. UK wont take a chance.
It’s a bit hard to figure out exactly what pre-Sealion moves we’re now talking about, but if they’re (nearly) the same as the one we previously discussed … it’s nowhere near a guarantee at all.
The chances of success drops dramatically if even one of the following happens:
- the UK transport at SZ106 survives
- one or more planes are shot down during the sea battles around the UK
- one or more planes are shot down by AA during Sealion
Placing a cruiser in SZ104 does not remedy this, it only makes the initial attacks on the UK fleets weaker (= higher chance of losing a plane). By weakening the initial attacks and making the chances of critical losses higher, it actually lowers the chances of a successful Sealion. (It does protect the Kriegsmarine from a direct counter naval attack by the UK … not that that helps the Italians, but still.)
What makes it even better for the UK is that they will know the exact outcome all the pre-Sealion battles before they have to make any decisions about what to buy and what to attack.
That actually makes the G3 Sealion operation from Jim a lot more interesting, because that requires the UK to buy the correct units before Germany starts buying transports.
8-)
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at least one plane in a larger number of rolls.
And we can always assume a sample of 7 rolls cannot indicate anything in terms of establishing 67% of hitting a plane.
Thats all i am saying and what your agreeing with anyway.
Placing a cruiser in SZ104 does not remedy this, it only makes the initial attacks on the UK fleets weaker (= higher chance of losing a plane).
Well no. The BB must assist in the G1 attacks and that places it in harms way. TO stop this the CA blocks the UK naval from being used to kill the main German fleet. The purpose of the CA block is to stop naval fodder from hitting with 4 planes.
SZ 104 is the only way to stop this.
That’s not the case here though. We’re talking about the chance of shooting down at least one plane in a larger number of rolls. So instead of needing a specific die to show a 1 for a success, I can roll a handful and if even a single 1 shows (no matter which die) it will be a success. This also means that the more dice I roll, the greater the chances of getting one or more 1’s.
So the chances of shooting down at least one plane in a large number of rolls will grow with the number of planes in the attack (= the number of AA rolls made).
Still the problem is its not 67% or anywhere near this. The sample is too small to field that result. A large number of rolls is not 7. The variation of swings in results of these few events cannot predict an outcome of 67%.
Also, when you roll a tank, its hitting on a 1,2,3 each time it rolls. If you got 3 tanks you got 9 rolls each round. So essentially your playing more numbers, but AA guns still hit at one.
AA guns are one roll hit on one. Again the sample is too small to estimate with accuracy. I have played most games where i have escaped many times without a plane loss and also flew over 6+ times. NO where except perhaps a few times i even got close to losing 2 out of 3 times one plane in 6+ SBR runs.
Also, where exactly are Jim’s moves…Item by Item, move, combat result, buy turn by turn.
I don’t see anywhere thats its written down. If you know the moves point out the post please.
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@Imperious:
As of now, I’d say Sealion is not close to breaking the game.
What’s more a threat to balance is the UK sinking the Italian fleet, and possibly the US buying a major IC in Norway.Sealion is not a requirement, rather stopping the Italian fleet attack is. The only way to guarantee that is to use my proposal and sink the Labrador DD/AP, which forces UK to leave Italy alone and hole up in UK. because i get 55% on that attack given what we have left. UK wont take a chance.
The other point in my plan is to protect my main fleet from his carrier group from being used as soakers with his planes against my BB, CV and 2 planes with 2 AP.
The CA block stops that. No matter what UK does, she cant get the flexibility in being able to attack and sink either my fleet or the Italian fleet. This is a key provision, that Jim’s plan fails at.
Well, as pointed out, without sinking the LAB trn UK can go after the Italian navy. (and sinking the trn with 1 sub is not a 55% chance)
And even if the LAB trn is sunk, UK can use its Tac to destroy the Italian navy and still have better odds. Allthoug not by much, making this a risky move.
So all in all, since the strategy depends, well not even on a coin flip, since the chances are not 50/50, in LAB, combined with the need for lucky rolls in the England attack (even without the TAC), I wouldn’t use it.@Imperious:
And btw, IL, I’d still like to know the following, What is, in your opinion is the probability of:
An AA hitting one plane?
An AA hitting at least on of six attacking planes?This was answered in the last post. You don’t got a sample of “large numbers” with 7 rolls. Your concept is flawed.
Its like a guy playing black on roulette, thinking incorrectly that if he keeps doubling on black eventually its bound to “hit”
This too falls under the gamblers fallacy. Because simply the sample is 7 events and that can’t be enough to justify the results of whatever % you want to assign to rolling one die. Its not enough chances to get that result.
If it were, then back when i was 18 and in Vegas my strategy of doubling the red or black, hoping that eventually i will hit paydirt within the stated number of turns would work. It didnt back then and i learned alot about it because it cost me thousands at the time. Vegas has a table limit for THIS VERY REASON.
The table limit is your 7 rolls just like it was mine where i could only double 7-9 times before i hit the maximum bet.
This is where the law of Large Numbers plays out and 7 is not it.
I can only say as i said before its 16.6% each event and the aggregate of 7 rolls total cannot sustain the % you posted no matter what you say.
Thats not how math works. Its not 67% or whatever you posted.
Wrong, the % of an event not taking place at all is the combined % of the negative event. Other way around, the % of a thing happening is 1-combined% of the negative event. This is a basic formula and will be found in every math book about the topic.
Large number of dice will make it sure, that the numbers 1-6 are distributed evenly, I need not determine the % of that happening, it just will, if the sample is large enough.
What are the chances of 1 2 3 4 5 6 coming up with 6 dice rolls? 1,5%
So don’t bet on that. What are the odds of at least one 1? 67%
What are the odds of throwing 6 dice twince and getting no 1? 11%
It’s not magic, it is science. If you’d read a book about it, maybe you’d understand.A common fallacy is, to see 11% and think, it won’t happen. Yes it will, 1 in 10.
So if you throw 6 dice twice and get no one, that is that 1 in 10.
Another mistake would be, and you agree on this, to think if I just had no 1, then the next time I have to get one. Wrong, 11% that that’ll happen again if it just happened now.
But before you throw any dice, the chances that throwing 6 dice twice two times and gettin no one is 1,25%. Unlikely, but far from impossible. So if you do that, and come up with no 1, you haven’t proven math wrong, you are just one of that 100 cases where this happens.With low numbers, you can very well determine the chances of the possible outcomes.
The formula for hitting at least on plane with an AA is valid. It says 67%, that doesn’t mean it will happen every time, but in 7 of 10 cases. Your sample shows 1 in 3. While that is an unlikely outcome, it is not an impossible one.To roulette:
If you can double up infinitely, you will hit black sooner or later.
Now since you can’t, chances you’ll win are there, but it won’t be much. (You invest 1+2+4+8+16+24+48 = 103 if you loose, e.g. red comes 7 times. If you win, no matter at which point, you win 1 (if black comes up first, invest 1 get 2, win 1. If black comes up after three times, invest 1+2+4=7 get 8, win 1). The problem is, if you are unlucky, and 7 times no black comes up more than it should, then you loose big time.
And since you’re only rolling a few dice, you have no guarantee that the numbers will even out, but you can determine the chances of them doing so.@Imperious:
Still the problem is its not 67% or anywhere near this. The sample is too small to field that result. A large number of rolls is not 7. The variation of swings in results of these few events cannot predict an outcome of 67%.
Nope, the sample is large enough to have a 72% (67% is 6 dice) of hitting at least one plane.
Now if you throw dice, you’d need to throw them infinitely to have a 72% of the results showing at least one 1.
But still, each time you throw 7 dice, the chances are 72% that at least one 1 will show up.
Now if you throw them 100 times, will 72 of them have at least one 1? Probably not, that would be pure luck. But if you throw them 100 times, the most likely result is, that around 72 of them will show at least one 1. It is unlikely that 90 or only 20 do so, but not impossible, thus it can very well happen in an A&A game.
But that does not chance the 72% each throw of seven dices has.The concept of large numbers only say, that you will only get the % right when throwing a lot of dice. If you do not throw large numbers, chances are great, that not 72 of 100 cases will yield a result which probability is 72%.
But it doens’t say, that the 72% of an event happening is not true. -
@Imperious:
AA guns are one roll hit on one. Again the sample is too small to estimate with accuracy. I have played most games where i have escaped many times without a plane loss and also flew over 6+ times. NO where except perhaps a few times i even got close to losing 2 out of 3 times one plane in 6+ SBR runs.
Wooooooow …. Hold on to your horses there!
That’s a gigantic misunderstanding we have right there. I never claimed you would lose anywhere near to 67% of your attacking aircraft to AA fire. That calculated 67% chance is the attacker losing 1 or more planes planes … of which the biggest chance would be just losing one, a much smaller chance of losing 2 and a very very tiny chance of losing 3 etc etc.
Remember, we were talking about combined odds.
8-)
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@Imperious:
Also, where exactly are Jim’s moves…Item by Item, move, combat result, buy turn by turn.
I don’t see anywhere thats its written down. If you know the moves point out the post please.
http://www.axisandallies.org/forums/index.php?topic=20231.0
Those are the G1 moves, the rest should follow when he’s back (though G2 and G3 are not that hard to figure out once you know the opening moves).
8-)
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And since you’re only rolling a few dice, you have no guarantee that the numbers will even out
This is my main point of which confirmed exactly what i have been saying.
That’s a gigantic misunderstanding we have right there. I never claimed you would lose anywhere near to 67% of your attacking aircraft to AA fire. That calculated 67% chance is the attacker losing 1 or more planes planes … of which the biggest chance would be just losing one, a much smaller chance of losing 2 and a very very tiny chance of losing 3 etc etc.
MY comment does not say 67% of your planes are lost. It says this:
"AA guns are one roll hit on one. Again the sample is too small to estimate with accuracy. I have played most games where i have escaped many times without a plane loss and also flew over 6+ times. NO where except perhaps a few times i even got close to losing 2 out of 3 times one plane in 6+ SBR runs."
The sample is too small 7 rolls cannot establish the 67% accuracy.
So all in all, since the strategy depends, well not even on a coin flip, since the chances are not 50/50, in LAB, combined with the need for lucky rolls in the England attack (even without the TAC), I wouldn’t use it.
A sub attacking at 2 vs. a Destroyer defending at 2 is not 55% IN THE DEFENDERS ADVANTAGE. Math does not support that number.
And those so called “lucky rolls” of something north of 85% winning and a combined 70% win if you leave out the 50% attack in Labrador.
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Loose means not tight
Lose means the opposite of winning.
So we dont “loose” battles, we “lose” them.
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@Imperious:
The sample is too small 7 rolls cannot establish the 67% accuracy.
The size of the sample is irrelevant to determine the odds (if you do it mathematically. If you throw dice and write the result down, you need a larger sample). I mean come on, how often must you throw a coin to figure out that heads and tails each have 50% chance of showing. Not once I would believe. Now a coin is easy, since you only have two events. If you take one die, it is still easy, you have 6 events so a chance of 1/6.
With two dice you have more events, 36 to be exact (6x6).
If you say an AA rolls twice, out of these 36 combinations, the following mean a hit was scored: 1,1; 1,2; 1,3; 1,4; 1,5; 1,6; 6,1; 5,1; 4,1; 3,1; 2,1;
So, 11 of the 36 events (which are all equaly probable) mean a hit was scored, thus 11/36 (30,55%).
Or you could just take the odds of throwing no one (5/6), mulitply them, and deduct that number from 1 (=100%), giving you the 30,55%.
So you can easily find out the probability of throwing 2 dice and getting at least one 1 (30,55%).The law of large number comes in an another point:
When you throw two dice infinite times, 30% will show at least one 1.
When you throw two dice ten times (like in a game of A&A for example) it is not sure, that the dice will show at least one 1 three times out of the ten.
You can get lucky and hit more than you should, are less, or exactly. And each of these cases has a certain probability which you can figure out. But the outcomes near the 3 in ten will be the most likely.
All that doesn’t change the probability of each two dice roll to be 30% of showing one 1, though.(The same with flipping a coin. Heads or Tails is 50%. If you throw infinite times, 50% of the tosses will show each. If you throw only ten times you do not have that guarantee. But, the possibilites around the 5:5 distribution are still the most likely. But of course you can get tails ten times, it is just unlikely (0,01%) But if you have thrown tails nine times already, chances are 50% that it’ll come up a tenth time)
Back to the AA gun.
When throwing 7 dice, the chances of at least one 1 are 72%. You can figure this probability out the same way as with the 2 dice above. You can either draw a table and put in the possible combinations, then count the % of beneficiary outcomes, or just use the formula.
That is a fact, just as tossing a coin and having a 50% chance is fact. That are the same kind probabilities.
If taken to the test, things can go a lot different of course. 0 hits, 1 hit, 2 hits, 3 hits….
And each of these results is differently probable, which you can figure out without actually throwing dice, but just doing some number crunching. Or looking them up in a table.Again, the law of large numbers says (or rather the one of low numbers actually), if you throw 6 dice a hundred times, you probably won’t get at least one 1 exactly 72 times. But each possible outcome from rolling at least one 1 once to 100 times has a certain probability, which you can figure out (or look up).
check out the following (the first one actually uses AA as an example)
http://www.edcollins.com/backgammon/diceprob.htm
http://wizardofodds.com/gambling/dice.html
http://gwydir.demon.co.uk/jo/probability/calcdice.htmNow I know they’re meant for beginners, but you seem to have gotten something wrong from the start.
If you check wiki, they have the real formulars behind all this.
It is really pretty simple actually, no magic, just High School Math, at least by us.@Imperious:
So all in all, since the strategy depends, well not even on a coin flip, since the chances are not 50/50, in LAB, combined with the need for lucky rolls in the England attack (even without the TAC), I wouldn’t use it.
A sub attacking at 2 vs. a Destroyer defending at 2 is not 55% IN THE DEFENDERS ADVANTAGE. Math does not support that number.
And those so called “lucky rolls” of something north of 85% winning and a combined 70% win if you leave out the 50% attack in Labrador.
You’re right, it’s not 55%, it’s 60%. Remember, the trn survives if the the Des and Sub take each other out.
The odds are:
40% sub wins
20% both get destroyed
40% des winsAnd the next misunderstanding:
The “lucky rolls” I was talking about are not the fleet attacks with the exception of LAB, but the attack on England itself, which is under 50%, taking the AA into account (which your dice sim should actualy be able to do) and without the TAC.
So even if LAB is succesful (defined as sinking the trn, which is more unlikely than likely, even if not by far) the TAC can be used to sink the Italian fleet, which lets the odd for the England attack go up, but your still over 50% of holding it.
But lets say, the odds of LAB and UK where exactly 50% (which they’re not, they’re below that), it would still only be 25% of both succeding.
That is why I wouldn’t base my strategy on that.
Another thing is, if G1 moves are good in itself and work for Germany even if you don’t actually plan on attacking England, then you can always wait if LAB is succesfull and the attack UK (which I still wouldn’t since chances of winning are far bellow 50%). But then, I would rather have a strategy that actually saves the Italian fleet. -
@Imperious:
MY comment does not say 67% of your planes are lost. It says this:
"AA guns are one roll hit on one. Again the sample is too small to estimate with accuracy. I have played most games where i have escaped many times without a plane loss and also flew over 6+ times. NO where except perhaps a few times i even got close to losing 2 out of 3 times one plane in 6+ SBR runs."
The sample is too small 7 rolls cannot establish the 67% accuracy.
Yeah, I misread that … my bad.
Since we obviously have very different views on how probability calculations work, I’m going to try a different route: start rolling them dice!
In this case the difference between both positions is so big (roughly 1/6 versus roughly 4/6), that even with relatively few rolls the trend will become clear very soon.
So I hereby invite everybody who is interested/curious (or just bored) to grab six dice and start rolling. Every roll of six dice that contains one or more 1’s is a success, every roll of six dice without any 1’s is a fail. You don’t need to do this hundreds of times to get an idea which of the two points of view is more accurate … after 20-30 rolls it already starts to show.
A sub attacking at 2 vs. a Destroyer defending at 2 is not 55% IN THE DEFENDERS ADVANTAGE. Math does not support that number.
Don’t forget that the defender also has a transport that happens to be the whole point of the attack … if the transport stays afloat, the attack is a failure. I’m not going to calculate the odds since we disagree on that anyway, I’ll just show you why the odds aren’t even.
Possible outcomes:
G hit + UK miss = sunk transport = G victory
G hit + UK hit = intact transport = UK victory
G miss + UK miss = no result = next round
G miss + UK hit = intact transport = UK victoryThe Submarine has to hit before the Destroyer hits for a victory while the Destroyer only has to hit … that’s why it’s easier to get a UK victory result and why the odds are not 50-50.
8-)
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Since we obviously have very different views on how probability calculations work, I’m going to try a different route: start rolling them dice!
In this case the difference between both positions is so big (roughly 1/6 versus roughly 4/6), that even with relatively few rolls the trend will become clear very soon.
So I hereby invite everybody who is interested/curious (or just bored) to grab six dice and start rolling. Every roll of six dice that contains one or more 1’s is a success, every roll of six dice without any 1’s is a fail. You don’t need to do this hundreds of times to get an idea which of the two points of view is more accurate … after 20-30 rolls it already starts to show.
Oh well, I’m not going to calculate the standard deviation, but if one person rolls the dice 30 times, chances are live he will be nearer the 1/6 (thogh very unlikely).
So have 30 persons roll the dice 30 times, then we’re pretty much safe, as safe as being safe from being hit by a meteor or something like that. I guess, haven’t done the figures. -
Oh well, I’m not going to calculate the standard deviation, but if one person rolls the dice 30 times, chances are live he will be nearer the 1/6 (thogh very unlikely).
So have 30 persons roll the dice 30 times, then we’re pretty much safe, as safe as being safe from being hit by a meteor or something like that. I guess, haven’t done the figures.Exactly! This is the point you agree with? It happens to be mine. IN ACTUAL GAMES, the sample closer to the type in a game is the first one. The second is where you employ the concept of large numbers of which i posted earlier. AA is only a few rolls in a game, thus the % of results is not predictable at any rate approaching 67% kill out of 7 events. It can swing either much higher or much lower and everything in between. IN the field of numbers and math and calculations of same and of which i have never once argued against, it is correct that the % increases with more chances.
IN so few chances AND A FINITE NUMBER OF ROLLS ( remember people don’t live forever and play AA, they just play a few times) the accuracy of that 67% is entirely questionable.
This is why Vegas stays alive because they have deeper pockets and place finite limits on play. IN Vegas your not allowed to keep doubling up your bets. They have a ceiling on maximum bet on all numbered games. This is because they understand the concept of Large numbers very well. If they didn’t Vegas would go bust. Thats why i lost in Vegas when i kept doubling my bet…. i reached the maximum and the standard variation did not. :roll:
Since we obviously have very different views on how probability calculations work, I’m going to try a different route: start rolling them dice!
I did that in another post. IN three samples it yielded:
2 ones in first run of 7
0 ones in second
0 ones in thirdIt was either greater than 67% or 0%
You see the sample is too small to predict, because your playing a small sample. Thats why the odds are only correct if you use a large sample.
In actual games the sample is too small to make any determination. I know i keep saying this but its true. And you can keep hiding from that fact and i will continue to bring that monster out of its closet.
Imperious writes:
A sub attacking at 2 vs. a Destroyer defending at 2 is not 55% IN THE DEFENDERS ADVANTAGE. Math does not support that number.
37.6 % SS wins 41.65 % DD wins. tie is 20.8%
Well perhaps sending 2 subs over and shorting another battle could work… looking at numbers
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@Imperious:
I did that in another post. IN three samples it yielded:
2 ones in first run of 7
0 ones in second
0 ones in thirdIt was either greater than 67% or 0%
I’m not sure how and what exactly you roll each time, but it should look something like this:
- Take 6 dice
- Roll all 6 dice at the same time
- Repeat this 10 times
- Count how many of these 10 rolls show one or more 1’s
To make it more accurate, roll another few series of 10x6 dice, counting each time how many of the 10 rolls show one or more 1’s.
My test batch (success = one or more 1’s , fail = no 1’s)
10 rolls (5x success, 5x fail)
10 rolls (6x success, 4x fail)
10 rolls (6x success, 4x fail)
10 rolls (7x success, 3x fail)
10 rolls (4x success, 6x fail)A total of 50 rolls were made, each of 6 dice.
Of those rolls, 28 contained one or more 1’s, 22 contained no 1’s.8-)
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@Imperious:
Exactly! This is the point you agree with? It happens to be mine. IN ACTUAL GAMES, the sample closer to the type in a game is the first one. The second is where you employ the concept of large numbers of which i posted earlier. AA is only a few rolls in a game, thus the % of results is not predictable at any rate approaching 67% kill out of 7 events. It can swing either much higher or much lower and everything in between.
Nope, read my longer post above and check the links.
The % of 7 AA dice netting at least one hit is 72%. Large or low numbers have nothing to do with it. That just is the probability of that event.
It’s the same with tossing a coin, the probability is 50% for tails.Where the large, or rather low, numbers come into effect is, that if you toss a coin a few times, you’ll not neccessarily get tails 50% of the time, but that doesn’t change the probability of it coming up each toss being 50%.
The exact same is true for the 7 AA dice rolls. It’s not likely that while your AA career you end up with 72% of your 7 AA rolls making at least one hit, but that still is the probability of that happening each time you throw 7 dice.
Saying 7 dice AA is not 72% is like saying tosing a coin is not 50% tails.@Imperious:
IN the field of numbers and math and calculations of same and of which i have never once argued against, it is correct that the % increases with more chances.
IN so few chances AND A FINITE NUMBER OF ROLLS ( remember people don’t live forever and play AA, they just play a few times) the accuracy of that 67% is entirely questionable.
Well, not only in theory but also when actually throwing dice your odds of making more hits grow with the number of dice you throw. And again, that increase can be determined.
And in the case of the AA example, that would be 72% for 7 dice.
The accuracy of that 72% is not questionable, that is the probability of that event.
What is questionable is whether you actually get that result 72% of the time in your AA career. As said, probably not, but the most likely results you’ll get are around that 72%.
So if planing a strategy, the 72% are what you should take into account.@Imperious:
This is why Vegas stays alive because they have deeper pockets and place finite limits on play. IN Vegas your not allowed to keep doubling up your bets. They have a ceiling on maximum bet on all numbered games. This is because they understand the concept of Large numbers very well. If they didn’t Vegas would go bust. Thats why i lost in Vegas when i kept doubling my bet…. i reached the maximum and the standard variation did not. :roll:
Well, with the numbers we’re talking about in Vegas you are acutally talking large. That is why Vegas wins. The calculate the % and make the rules so that the bank wins. Since that is only true for large numbers, sometimes a player will win because he’s lucky, but overall it evens out and the bank wins.
One player loses 7 times in a row, another one wins 7 times in a row. But as that are the same probabilities it evens out for Vegas, since over the years the Casino results will get the numbers it predicted (if they did their math right) since they operate on a large number basis.
And you can predict the odds with which you’ll go out winning something with all black.
(It actually doesn’t matter if it’s all black or you change around, the chance of winning is 50% each time)
So, if you’d known your math, you could have determined before hand that the all black strategy won’t win you any money.@Imperious:
A sub attacking at 2 vs. a Destroyer defending at 2 is not 55% IN THE DEFENDERS ADVANTAGE. Math does not support that number.
37.6 % SS wins 41.65 % DD wins. tie is 20.8%
How on earth can sub hitting with 2 an a des hitting with 2 have fighting against each other have different odds? Roflmao, that is just so wrong on first sight.
The 40% is correct, you need to set your Simulator on throwing more often, and you’ll see, you’ll get closer to 40%. (A simulator doesn’t do math, it just rolls dice, so it’ll be of a bit of the true odds, unless you do a huge number of rolls (like a million))What is interesting, is that you take the odds by your Simulator and let yourself be guided by them, but according to your flawed logic the % doesn’t count, since in your ganes of A&A that % won’t show.
But if you use you calculator and, for example, say you won’t make an attack if you lose 72% of the time (according to your Simulator), then you shouldn’t make an attack with 7 planes against an AA, if you can’t afford to lose on (or more) planes, because that will happen in 72% of the cases.And when we’re talking about strategies, it makes no sense to base them on the dice results of one game, no matter how probable or improbable they were. The only sensible way is to use the numbers which you sim gives you (and those numvers are only valid if you set the dice rolls high) or figuring them out mathematically. The result will be the same, it is just more convinient to us a simulator.
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Can we stop arguing about the probabilities? It’s getting repetetive and isn’t going anywhere
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How on earth can sub hitting with 2 an a des hitting with 2 have fighting against each other have different odds? Roflmao, that is just so wrong on first sight.
The 40% is correct, you need to set your Simulator on throwing more often, and you’ll see, you’ll get closer to 40%. (A simulator doesn’t do math, it just rolls dice, so it’ll be of a bit of the true odds, unless you do a huge number of rolls (like a million))I posted the wrong numbers.
anyway. The result is like 39.7 to 41.2% with 10,000 events
http://www.dskelly.com/misc/aa/aasim.html
But you see the statistical variation also shows the differences depending on throws.
with 1,000 runs the odds are 40.8 for the sub
with 5,000 runs the odds are 39.9 for the sub
at 10,000 it goes to 39.7This is why with only 7 rolls i don’t see why you can make a claim of 67%+ Not enough numbers.
http://en.wikipedia.org/wiki/Standard_deviation
Only with large numbers can the truth of plane loses be qualified.
here is another result: This accounts for 8 different rolls and the odds of killing 0-8 planes
To kill one plane it shows: 18.98% which is much closer to my 16.6% than your 67%
Probability % # units / losses
23.12% 8: 4 Fig, 4 Bom. no units. : 0 IPCs
18.98% 7: 3 Fig, 4 Bom. 1 Fig. : 10 IPCs
18.27% 7: 4 Fig, 3 Bom. 1 Bom. : 15 IPCs
14.52% 6: 3 Fig, 3 Bom. 1 Fig, 1 Bom. : 25 IPCs
5.48% 6: 2 Fig, 4 Bom. 2 Fig. : 20 IPCs
5.42% 6: 4 Fig, 2 Bom. 2 Bom. : 30 IPCs
4.87% 5: 3 Fig, 2 Bom. 1 Fig, 2 Bom. : 40 IPCs
4.67% 5: 2 Fig, 3 Bom. 2 Fig, 1 Bom. : 35 IPCs
0.68% 5: 4 Fig, 1 Bom. 3 Bom. : 45 IPCs
0.8% 5: 1 Fig, 4 Bom. 3 Fig. : 30 IPCs
1.33% 4: 2 Fig, 2 Bom. 2 Fig, 2 Bom. : 50 IPCs
0.65% 4: 3 Fig, 1 Bom. 1 Fig, 3 Bom. : 55 IPCs
0.56% 4: 1 Fig, 3 Bom. 3 Fig, 1 Bom. : 45 IPCs
0.06% 4: 4 Fig. 4 Bom. : 60 IPCs
0.01% 4: 4 Bom. 4 Fig. : 40 IPCs
0.24% 3: 2 Fig, 1 Bom. 2 Fig, 3 Bom. : 65 IPCs
0.23% 3: 1 Fig, 2 Bom. 3 Fig, 2 Bom. : 60 IPCs
0.04% 3: 3 Fig. 1 Fig, 4 Bom. : 70 IPCs
0.04% 3: 3 Bom. 4 Fig, 1 Bom. : 55 IPCs
0.01% 2: 2 Bom. 4 Fig, 2 Bom. : 70 IPCs
0.01% 2: 1 Fig, 1 Bom. 3 Fig, 3 Bom. : 75 IPCs
0.01% 1: 1 Fig. 3 Fig, 4 Bom. : 90 IPCs -
@Imperious:
To kill one plane it shows: 18.98% which is much closer to my 16.6% than your 67%
Um, no, and IL, despite a decent tactical mind and some very nice Illustrator skills, you’re killing me here. For the love of god, you need to take a statistics class (or refresh your statistics notes).
This has never been a calculation of losing ONE plane, and no one has ever argued that that is anything BUT 1 in 6, as that’s the freaking dice roll! A one in six chance of the pip coming up is ALWAYS a 1 in 6 chance! This is a combined odds calculation of the chances of losing ANY plane. The calculation you just posted confirms all this, as muddied as it is. Why on earth would you add in two different types of planes? That just makes the whole thing even more complicated because now we have completely unecessary data with three possible types of hits (on bombers or on fighters or on both). If you had bothered to look at the other instances you posted instead of just copying and pasting (odds of losing 2, odds of losing 3, odds of losing 4) you would notice that it freaking CONFIRMS the 67%. By simply looking at the odds of losing 0 fighters and 0 bombers, you inadvertantly prove what we’re saying. In 8 planes, you have a 76.88% chance of losing SOMETHING, if you have a 23.12% chance of losing NOTHING (because everything adds to 100%).
The general formula for rolling at least one hit in n rolls is 1 - (5/6)^n. Yes, you are correct, that each roll is 1/6 (16%) and yes, you are correct that each roll has no impact on the previous or the next. But bringing standard deviation and statistical variation for number of throws is completely unnecessary, and apparently is confusing you because you appear to think it’s relevant.
In the infinite number of instances where 6 planes are fired at by AA, in 67% of them at least one plane will die. It might be an instance of one plane, but it could be an instance of all 6 planes (which is still at least one plane). It’s the COMBINED odds. Not just the odds of losing ONE plane. and 33% of the time (a whole 1 in 3) you lose NOTHING (which apparently you remember more often than the times where you lose more than one plane - good on you, it’s fun to revel in our enemies’ destruction).
When AA odds calculators do there stuff, and we end up with a calculation that has a win percentage, and that win percentage includes all possible results that equal a win. They do not give you the odds of only one type of win, or only one type of result. Whether it’s the odds of taking the win with only one loss, two losses, etc etc, it’s all in the combined odds. Stop getting hung up on your dangerously incomplete understanding of statistics.
By the by, in Vegas, if you WERE to bet on the same number, eventually you WOULD win. The reason Vegas works is despite the statistical inevitability that eventually, with an infinite number of rolls you will eventually result a win (in fact, an infinite number of wins), no one has infinite money, and the odds are better that you’ll lose more than you can afford to before you get an UNLIKELY win early.
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