Re: Statistical odds of AA guns + standard deviations associated low/high sample

[Lydian]

Besides the question how probability works, what remains:

Trying to find a Sealion strategy that’ll work more often than not

Or a strategy that threatens Sealion, thus forcing the UK to leave the Italian navy alone and forcing the UK to a defencive buy, thus stalling its navy build up and buying time for Germany, but with which it is also viable to go after Russia if England reacts accordingly, and thus making Sealion a (far) less than 50% winner.

This is the bummer with Jim’s strategy, a G3 Sealion gives the UK the chance to kill the Italian fleet UK 1.
A G2 hasn’t really got real chances of success, but it could lead the UK Player to leave the Italian navy alone. But he doesn’t really need the Tac in England to defend, especially not, if the transporter off of Labrador survived.
(Question IL, you propose to attack England in G2 with 3 inf 3 arm 3 ftr 3 Tac 1 bom 1 CV 1 BB right?)

So in my eyes, both Sealion strategies do not prevent the sinking of the Italian fleet, and even Jim’s doesn’t let you win 50% of the time. It may be worth it though, to force the UK to buy land units, thus pushing their fleet build up back.

Questions that need to be addressed:

Does a successful Sealion even effectively end the game, or will Russia come strong and the US take back England eventually (or Russia be on the steps of Berlin if Germany chooses to defend England massively)?

Can the Italian navy be saved?

[Imperious Leader]

Well yes, the probability of success for each die doesn’t go up, but with the number of attempts I increase the % that I’ll, at least once, get the desired result.

Well you finally get it. The number of chances does give you more opportunities even if the 16.666% figure wont be changing.

These dice are each reset because they are rolled not together, but individually against each plane.

If i roll 6 dice and you ascribe some ‘combined’ result that i should get…say 70% of rolling a one. I

More units in A&A will usually get more hits won’t they? The probability of 4 arm making at least one hit is far greater than that of one wouldn’t you agree? to be precise 93,75% compared to 50%.
This all is true in the case of independent events, such as throwing dice.

Well sort of. There is something called the Law of Large Numbers (or the Law of Averages) which states that if you repeat a random experiment, such as tossing a coin or rolling a die, a very large number of times, your outcomes should on average be equal to (or very close to) the theoretical average.

Suppose we roll three dice and get no 6’s, then roll them again and still get no 6’s, then roll them a third time and STILL get no 6’s. (This is the equivalent of rolling nine dice at once and getting no
6’s, there’s only a 19.38% chance of this happening.) The Law of Large Numbers says that if we roll them 500 more times, we should get at least one 6 (in the 3 dice) about 212 times out of the 503 rolls (.4213 * 503 = 211.9).

This is not because the probability increases in later rolls, but rather, over the next 500 rolls, there’s a chance that we’ll get a “hot streak,” where we might roll at least one 6 on three or more consecutive rolls. In the long run (and that’s the key - we’re talking about a VERY long run), it will average out.

In the case of AA rolls, This is a short run of potential rolls. The chances of this average wont likely approach anything of the sort that you claim with such a small sample. That is why your numbers are bunk.

There is also something called the Gambler’s Fallacy, which is the mistaken belief that the probability on the next roll changes because a particular outcome is “due.” In the example above, the probability of rolling at least one 6 in the next roll of the three dice (after three rolls with no 6’s) is still 42.13%. A (non-mathematician) gambler might think that the dice are “due,” that in order to get the long-term average back up to 42%, the probability of the next roll getting at least one 6 must be higher than 42%. This is wrong, and hence it’s called “the Gambler’s Fallacy.”

http://en.wikipedia.org/wiki/Gambler's_fallacy

( you can also bring sub to SZ 109 and get better odds, but take the bomber to SZ 111)

Did you miss this attack? I can bring the Sub over to help out in SZ 109 and forget Labrador. giving it a fighter and sub against a DD, the bomber can now replace the fighter elsewhere to boost the odds elsewhere.

And so:

Attack SZ 109 1 Fighter and 1 SS vs. 1 DD , 5 vs. 2. 92.5%

Attack SZ 110 2 subs, 1 tactical, 1 fighter, 1 bomber vs. 1 BB, 1 DD ( should win) 99.5%

Attack SZ 111 2 subs, 1 tactical, 1 Fighter, vs. 1 BB, 1 CA ( should win) 85.7%

Attack SZ 112 1 BB, 1 tactical, 1 Fighter vs 2 CA ( should win hit on BB) 89%

Key Move: CA blocks UK Gibraltar fleet in SZ 104!

92.5%
99.5%
85.7%
89.0%
–--------
what is the combined odds now?

(Question IL, you propose to attack England in G2 with 3 inf 3 arm 3 ftr 3 Tac 1 bom 1 CV 1 BB right?)

No. I propose the cause UK to leave Italy alone because i “could do that” This plan has always been about stopping the UK attack on Italy. Nothing else.

UK will have as a result of this new attack:

12 Inf, 1 tank, 3-4 planes against 3 inf 3 arm 3 ftr 3 Tac 1 bom 1 CV 1 BB, which is 30% or 12.4% if AA gun hits

However, the best that Germany can do is take a chance on that Labrador DD/AP and make the odds:

55.2% success vs. 41.5% for UK  ( these results are if UK still goes after Italian fleet with tactical bomber, which they need to win that other battle.)

[Lydian]

I know my math, no lecture required…

@Imperious:

Well sort of. There is something called the Law of Large Numbers (or the Law of Averages) which states that if you repeat a random experiment, such as tossing a coin or rolling a die, a very large number of times, your outcomes should on average be equal to (or very close to) the theoretical average.

Of course, if you’ve thrown infinite dice, each number will have come up 1/6 (or something really close to it) of the time.
But if you throw only 6 dice, only in 33,5% of all cases will you have not even one result of 1.
The math behind that is true no matter how many dice you throw.
The only thing is, knowing you have 99,99% chances of success means you can still loose. So you can determine what your chances are, but even if very probable, it can still all go wrong.

@Imperious:

And so:

Attack SZ 109 1 Fighter and 1 SS vs. 1 DD , 5 vs. 2. 92.5%

Attack SZ 110 2 subs, 1 tactical, 1 fighter, 1 bomber vs. 1 BB, 1 DD ( should win) 99.5%

Attack SZ 111 2 subs, 1 tactical, 1 Fighter, vs. 1 BB, 1 CA ( should win) 85.7%

Attack SZ 112 1 BB, 1 tactical, 1 Fighter vs 2 CA ( should win hit on BB) 89%

Key Move: CA blocks UK Gibraltar fleet in SZ 104!

92.5%
99.5%
85.7%
89.0%
–--------
what is the combined odds now?

70% success rate.
But without Labrador it doesn’t matter, since then UK can use its Tac for the Italian navy and still be safe.

@Imperious:

(Question IL, you propose to attack England in G2 with 3 inf 3 arm 3 ftr 3 Tac 1 bom 1 CV 1 BB right?)

No. I propose the cause UK to leave Italy alone because i “could do that” This plan has always been about stopping the UK attack on Italy. Nothing else.

UK will have as a result of this new attack:

12 Inf, 1 tank, 3-4 planes against 3 inf 3 arm 3 ftr 3 Tac 1 bom 1 CV 1 BB, which is 30% or 12.4% if AA gun hits

However, the best that Germany can do is take a chance on that Labrador DD/AP and make the odds:

55.2% success vs. 41.5% for UK  ( these results are if UK still goes after Italian fleet with tactical bomber, which they need to win that other battle.)

Now we’re talking. But as demonstrated in a post above, including the Labrador fight lets the succesrate of G1 drop to roughly 1 in 4, chances I wouldn’t like to take.
But if you don’t successfully sink the trn of Labrador, England can ignore a G2 Sealion (below 20%, taking the AA into account, without the Tac) and still take out the Italian fleet.
Even if the Labrador trn is sunk, you could take your chances as UK, but if the AA misses (unlikely, 28%) you’d loose England (well, at least that would be the most probable outcome).
UK at least has to buy inf only, thus stalling its fleet build up, but I’m not sure if that’s worth it.
It would be interesting though, to see a Sealionstrat that actually saves the Italian navy. That would be worth it I think.

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.

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?

[Imperious Leader]

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.

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.

[calvinhobbesliker]

@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.

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.

Okay, let’s use a simple example so you can understand. If you flip a coin, the probability of heads is 50%(ideally) and of tails is also 50%. If you flip the coin twice, there is a 75% chance of getting at least one head. here’s why: there are 4 possibilities: HT, HH, TH, TT. 3 of the 4 have at least one heads. To do math without looking at all the possibilities of dice rolling, do this: the chance of an AA missing 1 plane is 5/6. The chance of it missing 2 planes is (5/6)(56), since both planes have to be missed for this condition to be fulfilled. The chance of the AA missing 6 planes is (5/6)^6, which is 33.4%. Thus, there is a 66.6% chance that at least one plane will die

[Imperious Leader]

Okay, let’s use a simple example so you can understand. If you flip a coin, the probability of heads is 50%(ideally) and of tails is also 50%. If you flip the coin twice, there is a 75% chance of getting at least one head. here’s why: there are 4 possibilities: HT, HH, TH, TT. 3 of the 4 have at least one heads. To do math without looking at all the possibilities of dice rolling, do this: the chance of an AA missing 1 plane is 5/6. The chance of it missing 2 planes is (5/6)(56), since both planes have to be missed for this condition to be fulfilled. The chance of the AA missing 6 planes is (5/6)^6, which is 33.4%. Thus, there is a 66.6% chance that at least one plane will die

Well then goto Vegas and try that 67% thing.

If the house edge of playing the same number out of 7 events is so great, Then Kino is your game!

there is a 75% chance of getting at least one head.

This is the gamblers fallacy. The dice have no memory of “getting hot” while rolling cold or vice versa. each roll is independent of the other. Entirely. Also, your looking at a series of a finite number of rolls. Its entirely possible that the only occurrence in the game where AA guns roll in one game could be 7. Your concept may prove sound if the number of casts where of a very large sample, say you will roll 1 million times, Now the averages of all the possible results will demonstrate a more or less even distribution of results.

In 7 events of which you maintain the 67%, go ahead and roll this 70 times in 7 roll segments which is a yield of 10 samples of getting your number.

It will not be 67% I promise.  If you roll out 1 million times for example, yes THEN you got a sample that we can work with.

7 samples is not enough to justify your numbers.  I don’t know what the exact % is but If i could yield 67%. In 7 rolls it could even be worse, like 90% or be about 10% or zero.

2,4,3,1,2,1,6 = game 1

6,2,6,3,4,4,3= game 2

4,3,4,6,6,6,6= game 3

I just rolled these numbers.

One result came up twice in the same game.  I didn’t come up at all in the second two games.

Results:
2/7
0/7
0/7

Now where is your 67%? it didn’t work three times?

WHY?

BECAUSE YOUR SAMPLE IS NOT LARGE ENOUGH!

Its not that hard.

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.

[calvinhobbesliker]

By that logic, you can’t say that there’s a 16.6% chance of a die rolling a one, since it’s possible that it will roll 0 1’s in 6 rolls or more than 1 1 in 6 rolls. For example:
Rolls: 6@1; Total Hits: 06@1: (5, 4, 5, 4, 3, 3)

[Griffin34]

@calvinhobbesliker:

What do you think?

I don’t think so.  I think it’s a good boost for the Axis, but to say it breaks the game is going too far.  A crafty Russian player can still win the game for the Allies.

I mean, the “purpose” of the UK is not to take out Germany by itself right?  It’s there to harass Germany until 65 IPC/turn USA shows up to crack into Europe.  A successful Sealion is painful to Russia because Germany gets the UK’s money and a nice income boost, but I don’t think its a game breaker.

More interesting than the Sealion debate is what German players do as an opener in Russia and where they go after they take Leningrad/Stalingrad.

[Imperious Leader]

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.

[calvinhobbesliker]

@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.

[JamesAleman]

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.

[Latro]

@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-)

[Latro]

@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-)

[Imperious Leader]

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.

[Lydian]

@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.

[Latro]

@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-)

[Latro]

@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-)

[Imperious Leader]

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.

[pusfilledwart]

Loose means not tight

Lose means the opposite of winning.

So we dont “loose” battles, we “lose” them.

[Lydian]

@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.htm

Now 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 wins

And 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.