@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.