Imagine that you have a single enemy territory bordering two of your territories. The enemy has 10 armies, and your territories have say, 9 and 14. If you can be sure that neither you nor the enemy will deploy in the area, and no transfers/attacks will be made to/from the enemy territory, which territory should you attack from first (the 8 or 13) to have the best chance of taking the territory?

TLDR: Is it better to attack with a larger number, then a smaller number, or vice-versa?

i always thought large then small but im not sure

Let's do the basic math...

If the 13 attack first, they take out 7.8, leaving 2.2 for the 8 to mop up. You are left with 7 on your first territory, 1 in your second and 6.6 in the new one.

If the 8 attacks first, they take out 4.8 leaving 5.2 for the 13 to mop up. You are left with 2 on your first territory, 1 on your second and 9.5 in the new one.

End result of option 1 - Victory with 10 killed and 8.4 lost and 14.6 left alive.
End result of option 2 - Victory with 10 killed and 10.5 lost and 12.5 left alive.

The first option is superior in most instances... the only advantage to the second is if you want the larger force occupying the newly taken territory.

In that example, The Larger is better, I believe. Best case scenario is that the larger amount weakens it enough for your second army to take it, and the larger army wont lose far more armies than the defende either way. Worst case is that both forces fail to take it, but with less armies lost in the process.

As far as I know, the smaller number is unlikely to reduce the armies significantly, and will probably end up destroying itself against the defence.

it might also depend on luck settings

I just picked some random numbers... say for instance you had attacking groups of 7 and 9... then which option would be better? Is it possible to make a generalisation for the better option?

you're always more likely to take a territory if you attack with the larger number of armies first.

i don't know if it's worth making a new thread over, but i have a statistical question: is it certain that you will take a neutral (of 2) if you attack it with 2 and then 2 again from a different territory? (on 16% luck and 0%) if not, are you more likely to take it than if you only attack with 3?

depending if u use straight rounding or not

yeah, that too.

With the standard rounding (weighted random) the 2x2 vs 2 attack always works even on 16% luck - when you attack with 2 armies you're always guaranteed to kill at least 1 army. A very useful trick to know!

On 16% luck, the worst that can happen is (16%*0)+(84%*1.2) = 1.008 armies, ensuring a kill on one army.

With 0% luck, you're going to take an average of 1.2 armies.

However, with straight round, a 3v2 will always work on 16%, or, if approaching from 2 sides, a 1x2 vs. 2 will also work if the 1 comes first. Therefore, a 2v2 attack is only wasting an army that could be used elsewhere.

The answer might be clear to the OP already, but nobody specifically stated that the chance to take the territory is equal no matter with which territory you attack first. The only difference is the amount of troops lost on your side. However, in normal games, you can expect transfers/income, and you generally don't want to lose more troops than necessary.

Actually it's a very good tactic sometimes.

Answering your question x - yes, you have a 100% chance of takeing the territory with 2 x 2 v 2 attacks.

In straight round you have a 100% chance of takeing a territory with - 1 v 2 + 2 v 2 attack.

In both cases you need the same amount of armies globaly (4 in weighted round, 3 in straight round) to ensure capturing a region. BUT: if you have a situation where, you have 1 army in 1 territory and 1 army in another and they both border the territory you want to capture, then this tactic actually saves you an army, which can be used elswhere. Its pointless to deploy to two diffrent territories, if you dont already have troops there, as you will propobly loose more and dont transfer all armies to the captured territory.

To x:

yeah, I did the math on that, 2+2vs2 is pretty much always better than 3vs2.

To the primary question:

z defending troops - x,y are your troops, x>y Also for the sake of argument let us suppose 0.6(x+y-2)>=z so that you take over the territory (if that is not the case ut's even more obvious then to attack with smaller stacks first). Also everything with 0% luck.

Option 1 - attack with bigger first:
You kill z troops total, you lose (min(x,0.7z)+min(y,0.7(z-0,6(x-1)))

Option 2 - attack with smaller first:
You kill z troops total, you lose (min(y,0.7z)+min(x,0.7(z-0,6(y-1)))

Since we assumed that 0.6(x+y-2)>=z and x>y, thus
min(x,0.7(z-0,6(y-1)))=0.7(z-0,6(y-1))
min(x,0.7z)=0.7z

So now we only have to compare 0.7z+min(y,0.7(z-0,6(x-1))) with (min(y,0.7z)+0.7(z-0,6(y-1)))

With fixed x,z we can have 3 cases:
i)y<= 0.7(z-0,6(x-1))
ii)0.7(z-0,6(x-1))<y<0.7z
iii)y>0.7z

i)
We have to compare
0.7z+y with y+0.7(z-0,6(y-1))) which is quite obvious since y>0 (option 2 wins)
ii)
We have to compare
0.7z+0.7(z-0,6(x-1)) with y+0.7(z-0,6(y-1))) which means we have to compare
0.7z-y with 0.42(x-y)
So for our option 2 to be better we need 0.58y<0.7z-0.42x which is very often satisfied. In small words we want 0.7z-y (which is the net loss 1st attack in taking option 1) to be relatively "small"
iii)y>0.7z
We have to compare
0.7z+y with y+0.7(z-0,6(y-1))) which again is obvious (option 2 wins)


So to sum up always attack with smaller first (the only scenarios where it won't be good is when the first attack will save you some troops and it hapens when 0.7z is close to y), but since most of the time those attack happen late order anyway and opponent may counter, it'll at least 90% of the time lie in range of i) and iii) scenarios. Of course that is without saying anything about strategy, for example if you have slight income/troop advantage and yuo defend bonus like that, I can often recommend putting 2 in first territory, rest in second territory and hit 1st order the big stack and last order the small one, but that's beside the point. Anyone instinct should tell you that, hit with smaller first ;)

what will happen to all the math lessons when you retired?

Think you missed a few things. First of all 0.7(z-0,6(x-1)) is negative when you would takeover with x. So you should have for option 1 max(0.7z,0.7z+min(y,0.7(z-0,6(x-1)))). Then it would be pretty much always good to chose option 1, except of course with staight round, which gives a total other dimention to it.

Round up really could have a big impact on this theory. So to be sure, calculate it a little.
For when option 1 is useful, we have equation 0.58y>0.7z-0.42x to be satisfied, which means that 0.7z-y (don't get it how this should be net loss) is small compared to x-0.7z. So kinda like saying the neglacted loses if you first attack with y are relative small compared to the survivors from when you attack with x first. This just shows how narrow it is which option you should chose. Just calculate it out before you do it, this is probably best advice. Doing it blindly could cost you a lot of losses.

A simple example of the difference it can give. When your enemy has 10 armies. You have x=13 and y=7. First option 1, you attack with 12 which kills 7.2. Then enemy has 2.8 left and 7 losses. 2nd attack gives 2.8*0.7= 1.96. So total loses 8.96. Now option 2, gives 3.6 kills and 6 losses. So 6.4 left, which gives 4.48, total of 10.48 loses. Big difference, but if we just shift the armies with x=16 and y=4. Option 1 gives 9.7 losses and option 2 gives 7.63 losses. So it is really close which option you should chose.

This is why I'm not getting any better at this game. Should've paid more attention in math class.

Lol, I thought it was obvious from how the question was stated, obviously the question is asked when the condition z>0.6(x-1), otherwise the question is trivial...

Also regarding the rest of the post:

You conveniently chose the worst case scenario... Anyway please read the last paragraph carefully, most in-game situations don't have the level of control we have creating abstract models, thus we should opt for the solution that provides greater coverage and since option 2 has much greater range it seems like a very simple decision for me... Anyway when option 1 is chosen it is always for other reasons (mainly defensive as I said earlier) and what is most important we can see that when option 1 is preferable the net losses in comparison to option 2 can be easily bounded:

with case ii) we have net income loss from picking option 2 0.42x+y-0.7z when 0.7(z-0,6(x-1))<y<0.7z so obviously that varies from -0.18x to 0.42x in worst case scenario. On the other hand in cases i) and iii) the variation of the net income is equal to 0 and it is always 0.6(y-1). It is quite hard to predict expected value of net income in case 2 not knowing the distribution of y in that case, but we can observe a few things:
- in most cases X+Y=income (or some other constant) and by getting into case ii) with our deployment we make x smaller, so on one hand raising y to a certain point gets us closer to the worst case scenario, on the other hand in practice it also makes it smaller by deploying more to Y with the cost of deploying less to X
- since the range of cases i) and iii) is much wider if we count the expected total value we have to weight the cases with certain probabilities
- if we try some sort of symmetrical distribution on Y on (0.7(z-0,6(x-1)),0.7z) (which is not that far-fetched considering in practice we don't have that much control over specific values/enemy deployment, but we can estimate expected value) then the expected net losses on case 2 would be 0.12x

I actually don't have time to go into detail into that, but I could present a model explaining my reasoning, but i think the general idea is presented in a pretty easy to understand way.

Head hurts

Advice don't read when drunk ... infact probably don't read when sober, unless you know you get wood over formulas.

ps thank you Opera for inbuilt spell check.