Q about taking a lot of 1's

[Geger]

What is the simplest way to calculate, how many 1's can we take (with good probability)? Let's say we start with x+1 troops.

[Dukasaur]

What is the simplest way to calculate, how many 1's can we take (with good probability)? Let's say we start with x+1 troops.

The simplest way is to use the Assault Odds add-on.
http://www.conquerclub.com/forum/viewtopic.php?t=83451

However, if I find myself in a position where I have to do it manually, I just ballpark it as three for every two terts with a 1 on it. (Plus of course a basic 4 to sustain the attack.) So, with seven troops I expect to safely take two terts with 1s on them, with ten troops I expect to take 4, with thirteen 6, and so on.

[Geger]

What is the simplest way to calculate, how many 1's can we take (with good probability)? Let's say we start with x+1 troops.

The simplest way is to use the Assault Odds add-on.
http://www.conquerclub.com/forum/viewtopic.php?t=83451

However, if I find myself in a position where I have to do it manually, I just ballpark it as three for every two terts with a 1 on it. (Plus of course a basic 4 to sustain the attack.) So, with seven troops I expect to safely take two terts with 1s on them, with ten troops I expect to take 4, with thirteen 6, and so on.

Never use that add-on, I think it is like http://gamesbyemail.com/games/gambit/battleodds

But it's not simple to use it, because I'm to lazy to put a lot of 1's. Oh I forgot to say, that x is big (100+). About the 2nd idea, I'll figure out later, have to go now. Thanks. Or maybe is there a simple formula?

[laughingcavalier]

Duke’s assumptions are way too conservative – the attacker’s advantage is much greater than that. For my rule of thumb I work on the assumptions of 1 loss for every 3 singles killed, and do my last attack when I have 3 troops.
So starting with 4 troops I expect to kill 2
With 5 – kill 2
With 6 – kill 3
With 7 – kill 4
With 8 – kill 4
With 9 - kill 5
With 10 – kill 6
With 11 – kill 6
With 12 – kill 7
With 13 – kill 8
Or something like that – and I couldn’t possibly put it into a formula, that’s much too clever for me.

[Kaskavel]

Duke's assumption seems correct to me, although I would reduce his required attackers by one (why 7 for 2 1s Duke? 6 is the expexted number necessairy). But the basic concept that "3 extra troops in the stack will offer 2 more territories" is correct and very useful. Basically, you should expect to lose one troop for every two 1s you conquer.

[Kaskavel]

Duke’s assumptions are way too conservative – the attacker’s advantage is much greater than that. For my rule of thumb I work on the assumptions of 1 loss for every 3 singles killed, and do my last attack when I have 3 troops.
So starting with 4 troops I expect to kill 2
With 5 – kill 2
With 6 – kill 3
With 7 – kill 4
With 8 – kill 4
With 9 - kill 5
With 10 – kill 6
With 11 – kill 6
With 12 – kill 7
With 13 – kill 8
Or something like that – and I couldn’t possibly put it into a formula, that’s much too clever for me.

Strange statement, it seems obvious that low-rated Duke is right. At least for the concept that 3 more troops at the stack, will offer 2 more regions, although he should start with 6 troops, not 7. And we shouldnt add the 3-1 attack, because it is not obvious that we will want to execute that attack in all circumstances

[RiffArt]

To kill a 1 troop territory you have a 34% chance to lose at least one troop (perhaps what laughingcavalier is thinking of?), or to put it another way, 2 out of 3 times you attack a 1 troop territory you won't lose anyone.

However, you can "expect" to lose a little over 0.5 troops for every such territory you capture. (In fact, for each 1000 such territories you will lose 516 troops). The difference between the two comes because sometimes you will lose 2 or more troops before taking a territory. This is where Duke's 3:2 ratio comes in (1 lost troop + 1 left in each territory for each two territories taken).

For a formula, if you have x + 1 troops in your starting territory as OP stated, you can "expect" to take (assuming you don't attack 2v1 at the end, that's up to you):

Terts = ( x - 2 ) * 2/3

So starting with 5 troops, x = 4, you can expect to take 1 territory, 2 with a little luck.
With 6 - kill 2 (6 troops = 3 to occupy territories including your original + 1 killed + 2 extra at the end which are insufficient to attack 3v1)
With 7 - kill 2 (3 with luck)
With 8 - kill 3
With 9 - kill 4 (9 troops = 5 to occupy territories + 2 killed + 2 extra at the end which are insufficient to attack 3v1)

[spoiler][/spoiler]

[Kaskavel]

To kill a 1 troop territory you have a 34% chance to lose at least one troop (perhaps what laughingcavalier is thinking of?), or to put it another way, 2 out of 3 times you attack a 1 troop territory you won't lose anyone.

However, you can "expect" to lose a little over 0.5 troops for every such territory you capture. (In fact, for each 1000 such territories you will lose 516 troops). The difference between the two comes because sometimes you will lose 2 or more troops before taking a territory. This is where Duke's 3:2 ratio comes in (1 lost troop + 1 left in each territory for each two territories taken).

For a formula, if you have x + 1 troops in your starting territory as OP stated, you can "expect" to take (assuming you don't attack 2v1 at the end, that's up to you):

Terts = ( x - 2 ) * 2/3

So starting with 5 troops, x = 4, you can expect to take 1 territory, 2 with a little luck.
With 6 - kill 2 (6 troops = 3 to occupy territories including your original + 1 killed + 2 extra at the end which are insufficient to attack 3v1)
With 7 - kill 2 (3 with luck)
With 8 - kill 3
With 9 - kill 4 (9 troops = 5 to occupy territories + 2 killed + 2 extra at the end which are insufficient to attack 3v1)

Second paragraph is wrong. Think about it a little, it doesn't work that way. Sometimes we lose 10 troops to win 1 region, sometimes we win 20 regions and lose nothing. We dont care, we will make a lot of 4 vs 1 attacks and we will overall lose 1 for every 2 regions.
But last paragraph is absolutely correct and those are the exact numbers. In fact they are Duke's numbers corrected towards 1 lower.

[AslanTheKing]

Duke’s assumptions are way too conservative – the attacker’s advantage is much greater than that. For my rule of thumb I work on the assumptions of 1 loss for every 3 singles killed, and do my last attack when I have 3 troops.
So starting with 4 troops I expect to kill 2
With 5 – kill 2
With 6 – kill 3
With 7 – kill 4
With 8 – kill 4
With 9 - kill 5
With 10 – kill 6
With 11 – kill 6
With 12 – kill 7
With 13 – kill 8
Or something like that – and I couldn’t possibly put it into a formula, that’s much too clever for me.

this looks very logical to me,
but it doesnt work ( at least not for me, look at my rank and look at yours, so for u it works)

out of 20 times when i attack 6 vs3, i fail 19 times- no joke
i dont know why
and 6 vs 3 is the MOST important Attack on this site, so whoever can manage a 6 vs3 -be very happy
5 vs2 works for me easily

butt just yesterday in a crucial moment ( a gameturner)
i lost 9 vs 2

and with my dice, if i have 10 troops i can take 5 singles down, that makes a 50 percent take down
with 9 - i take down 4 singles
with 7+ 8- i take down 3 singles

look at your own battle outcome in dicestats
everybody here who has green in attack, or green in defense is bloody lucky
or if your attack and defense is balanced ( choose the last 25 battles to get more accurate numbers)
mine is minus 12 ( green + 4 in attack, and red - minus 16 in defense) so i lose more than i can win
there is no formula
i could be now a conspiracy clown
but i choose to shut up

[FreeFalling123]

Look guys I'm not trying to be an asshole but this rough estimate shit is donkey business. If we are analyzing this properly it's simply +.66 on each extra attacker
4v1s = .66 kill on average
5v1s = 1.33
6v1s = 2
7v1s = 2.66
It's not rocket science. If you want to add in 3v1s and 2v1s to
Those calculations you are talking about +.58 and +.42 to the average kill ratio so of you have 5 stacks 3v1 to try for 1s then on average you will get .58x5or 2.9 stacks

[Dukasaur]

Look guys I'm not trying to be an asshole but this rough estimate shit is donkey business. If we are analyzing this properly it's simply +.66 on each extra attacker
4v1s = .66 kill on average
5v1s = 1.33
6v1s = 2
7v1s = 2.66
It's not rocket science. If you want to add in 3v1s and 2v1s to
Those calculations you are talking about +.58 and +.42 to the average kill ratio so of you have 5 stacks 3v1 to try for 1s then on average you will get .58x5or 2.9 stacks

Not even sure what you're trying to say. For instance, 6v1 = 2? Does that mean if you have 6 on tert you expect to kill 2 terts with it? After your first kill, one attacker advances, so at best you have a 5v1 on your second attack, and that's if you didn't lose any on the first attack. There's no way you can comfortably expect to kill 2 1s with only a 6. If you didn't have to advance after the first attack, I could see you taking a chance. That advance throws your expectations right off.

Unless I'm misunderstanding what you're saying.

[D4 Damager]

Look guys I'm not trying to be an asshole but this rough estimate shit is donkey business. If we are analyzing this properly it's simply +.66 on each extra attacker
4v1s = .66 kill on average
5v1s = 1.33
6v1s = 2
7v1s = 2.66
It's not rocket science. If you want to add in 3v1s and 2v1s to
Those calculations you are talking about +.58 and +.42 to the average kill ratio so of you have 5 stacks 3v1 to try for 1s then on average you will get .58x5or 2.9 stacks

Not even sure what you're trying to say. For instance, 6v1 = 2? Does that mean if you have 6 on tert you expect to kill 2 terts with it? After your first kill, one attacker advances, so at best you have a 5v1 on your second attack, and that's if you didn't lose any on the first attack. There's no way you can comfortably expect to kill 2 1s with only a 6. If you didn't have to advance after the first attack, I could see you taking a chance. That advance throws your expectations right off.

Unless I'm misunderstanding what you're saying.

He appears to be inadvertently agreeing with your original suggestion: add 3 troops to your stack for each 2 extra 1s that you want to conquer. Except he is expressing it as a fraction instead

(well, decimal actually, but it's 2/3 in my head)

[Kaskavel]

Look guys I'm not trying to be an asshole but this rough estimate shit is donkey business. If we are analyzing this properly it's simply +.66 on each extra attacker
4v1s = .66 kill on average
5v1s = 1.33
6v1s = 2
7v1s = 2.66
It's not rocket science. If you want to add in 3v1s and 2v1s to
Those calculations you are talking about +.58 and +.42 to the average kill ratio so of you have 5 stacks 3v1 to try for 1s then on average you will get .58x5or 2.9 stacks

That is also accurate and agrees with Duke's and mine's numbers.
6vs 1 is expected to win 2. 9 vs1 is expected to kill 3 etc. I am not sure what we are debating here.