Closed Week #42 2018-02-19

[LadyHecate]

I really doubt this. If a ghost guild is a satellite of some other large guild, allowing it to take whatever they just ripped off their rivals, ghost guild wins, because attrition will be big guild vs big guild, and Traz lv80 would supply ghost guild fighters with expendable troops to defeat a sector off the rival guild, and then they just don't set up defense to not waste their small treasury.

If the ghost guild has no intentions of keeping a footprint on the map then I'd agree. However, if the larger guild keeps its footprint away from landing sites (as most aim to do) then it's easier for the larger guild to take the ghost guild off the map thereby making it harder for them to re-establish contact with them. Pretty much all the GvG action involving ghost guilds that I've seen is NOT involved on the AA map and thus the guilds involved are relying on more junior members for the the Treasury goods and troops; AF/OF players will only have a finite amount of lower Era goods and troops to support a campaign on the lower Era maps unless giving up city space to rebuild barracks of the lower Era.
And level 80 Traz's are not too common! There are none on En11!!!

 

[DeletedUser104719]

It sounds like the intricacies of GvG might be really interesting to get into; unfortunately I don't have the time to play that heavily these days, so I'm a casual fighter at best; I negotiate GE as soon as I start having to put actual thought into the battles

 

[DeletedUser]

I quote my response from Week #31:

Oh, I get it now. The cost for placing sieges and all that is proportional to the number of sectors controlled.

 

[LadyHecate]

Oh, I get it now. The cost for placing sieges and all that is proportional to the number of sectors controlled.

Exactly! The cost for a ghost guild to place a siege that currently holds no sectors is only 5 of each respective good for that map's Era. The top guilds on a GvG map will have at least 20 if not 30 sectors. Placing a siege to recapture a sector when you already hold 25 sectors costs 1215 of each good; over 2000 of each good when 30+ sectors are held.
Added to that is the DA unlocking costs to fill all the DAs; only the first two are free! Each additional two DA slots cost an increasing amount of goods to unlock.

 

[DeletedUser]

Thanks for the information. That helps me understand GvG a lot better.

 

[DeletedUser104719]

Thanks @LadyHecate; although I was aware of the general principal, I don't think I'd ever appreciated the scale of increase involved!

 

[LadyHecate]

http://forgeofempires.wikia.com/wiki/Siege_Costs covers the lot.

 

[DeletedUser]

But it doesn't give you the chance to win stuff.

 

[LadyHecate]

But it doesn't give you the chance to win stuff.

What GvG?
Apart from GvG players generally winning the weekly PvP Towers and hence medals, prestige within the 'hood, and the general appreciation of the guild!
What more could you want?

 

[DeletedUser]

What GvG?
Apart from GvG players generally winning the weekly PvP Towers and hence medals, prestige within the 'hood, and the general appreciation of the guild!
What more could you want?

Oh no, sorry I realise I wasn't very clear. I meant to say that reading the wikia doesn't give you the chance of winning anything, as opposed to asking questions in this thread. And since you get about the same information from the wikia and from Hecate, well the preferable option is clear.

 

[DeletedUser111866]

Aaawww, another full set of tickets lost vs 5% at double. Nahh, too much oddities, and 0.9^11 seems to be too frikking big to ever try, let alone 0.95^11.

 

[DeletedUser]

Sorry to hear that...

 

[DeletedUser110179]

Aaawww, another full set of tickets lost vs 5% at double. Nahh, too much oddities, and 0.9^11 seems to be too frikking big to ever try, let alone 0.95^11.

I found the neighbourhood pot at 560 ... and won in two tickets with the double chance.
It's probably not worth it at lower levels (when you're forced to spend your ticket(s) at 200).

7 hours ago by Mr.Quib:
" I think the double chance is a waste of florins. For each loss at the 5% game you gain 15 florins, but spend 20.
The best chance for the 5% is collecting at least 50 tickets before playing.
No guarantees, it has cost me around 130 tickets before I won and I started when it was at 460. "

 

[DeletedUser]

Tickets are overrated.

 

[AussieAlex]

(Tickets to most things in RL are overpriced!)

 

[DeletedUser111866]

I think the double chance is a waste of florins. For each loss at the 5% game you gain 15 florins, but spend 20.

Provided there is a pool in progress, losing 20 for adding 15 to pot is justifiable, provided you eventually win. Same with double chance in 15% game, while you lose more (10 per ticket) if you don't spend more than starting pool/10 tickts, you're still in the positive. But... if that kid spent 130 tickets at 5% without a single win... it's 1 in 830 chance, quite unlucky. This means I should recalculate some odds and starting winning pots given quite big chance of failure.

Hmm. If X is starting pot, and you have 11 tickets at 5%, to break even on average you need X to be greater than... (1-0.95^11)*X>=110, X>=256, and if spending 20 for double, (1-0.9^11)*X>=330, X>=480. This does not count increasing pot, though, actual values should be about 5 pot increases lower, that is, 180 for naked game and 405 for increased stakes. Hmmmm...

 

[DeletedUser]

Ih8regin, teach me. I've always wanted to learn how to calculate probability.

 

[DeletedUser111866]

This depends on what do you actually want to calculate. Assuming each try is independent to be a win or a loss from all the othr tries, there are simple formulae to get chances of 0 wins, at least 1 win or all wins out of N. These involve the main probability law of two independent events. Say A and B are some events that may happen and are deemed independent, and P(A) and P(B) are their respective probabilities. Then the chance of them both happening is the product of individual chances, P(A and B)=P(A)*P(B). So, to win all N events if a single win chance is X will have a probability of X^N, to lose all of em (1-X)^N and to win at least one (1-(1-X)^N), because if you won at least once, you exactly did not lost all of them, so you subtract the probability for losing all from 1 (there's another basic law, P(A)+P(not A)=1, meaning that an event either happens or does not happen). Then, as you desire to calculate a probability of a certain event, you need to split the set of all possible events into independent elementary events, calculate their probabilities, then sum up all probabilities for these elementary events that make your desired event A happen, you then get its exact probability.

 

[DeletedUser111866]

A set of elementary events is a set of all posible outcomes, say for a d6 being thrown it consists of an event "it comes with 1 up", "2 up" and so on until "6 up", and for a lottery say 6 in 49 it consists of an event "numbers are 1,2,3,4,5,6" through "44,45,46,47,48,49". In case of an honest die and honest lottery these elementary events are assumed to have equal probability, so you just need to know the number of them to get probability of a single event. This makes theory of probability border combinatorics and some big numbers of C(m,n) and P(m,n) which are (n!)/(m!*(n-m)!) And (n!/(n-m)!) respectively. Say for a lottery 6 in 49 the number of outcomes is C(6,49) = 13983816, so your odds at guessing all 6 numbers are slightly better than 1 in 14 million.

 

[DeletedUser111866]

When it comes to your ordinary applying of probability calculatikns, combinatorics is normally enough, as you can almost always enumerate all possible outcomes, differentiate them, classify them and combine a more complex event from them, such as guessing two numbers right out of 6 rolled out of 49. Then the random numbers start being less discrete and occupy a set of values with a given distribution function over that set, be it finite or infinite, countable or uncountable. There you get some familiar distributions like flat in an interval, "normal", binomial and so on, there are quite a lot of them known by names, some as weird as their names suggest. There you need to take integrals of these functions over a subset to determine the desired probability, which end up not always available in a functional form, or reference a non-analytical functions such as Ф(x) or erf(x) (these are related), and this is what makes the maths-deep probability theory too complex to fit in here. However, binomial distributions are pretty common in everyday life, so get those formulae and crunch the numbers.