I can't figure out how to Google this, so I'm hoping someone here is better at math than I am.
If my opponent is adding X troops per turn, and I am adding Y troops per turn (where Y > X), how long will it take before I have 3 times my opponent's troop number?
Let's say at the time you start adding troops, you have A troops and your opponent has B troops.
After C turns, you will have (C x Y) + A troops, and your opponent will have (C x X) + B troops.
You want C such that:
(C x Y) + A = 3((C x X) + B)
(C x Y) + A = (C x 3X) + 3B
C x Y = (C x 3X) + 3B - A
Y = 3X + (3B/C) - (A/C)
(3B - A)/C = Y - 3X
C = (3B - A)/(Y - 3X)
Now you simply plug in your values for A, B, X, and Y and solve.
sempai
High score: 2200 - July 20, 2015 Game 13890915 - in which I helped clinch the NC4 title for LHDD
sempaispellcheck wrote:Let's say at the time you start adding troops, you have A troops and your opponent has B troops.
After C turns, you will have (C x Y) + A troops, and your opponent will have (C x X) + B troops.
You want C such that:
(C x Y) + A = 3((C x X) + B)
(C x Y) + A = (C x 3X) + 3B
C x Y = (C x 3X) + 3B - A
Y = 3X + (3B/C) - (A/C)
(3B - A)/C = Y - 3X
C = (3B - A)/(Y - 3X)
Now you simply plug in your values for A, B, X, and Y and solve.
sempai
seriously... i wouldn't have even bothered my 5th grader for this answer... jeez...-Jésus noir
Thorthoth,"Cloaking one's C&A fetish with moral authority and righteous indignation
makes it ever so much more erotically thrilling"
a= your opponent's starting troop number
b= your starting troop number
A = total opponent troops
B = your total troops
X= your opponent's troops per turn
Y= your troops per turn
T = turns
nietzsche wrote:
This equation will only work when Y > 3X.
Which means that you can only ever get to three times your opponent's troops if you're earning more than three times as many troops as him. Possibly non-intuitive; naively it seems like if you're out-trooping your opponent then your relative strength over them should grow without bound. But it doesn't; asymptotically you can never amass more relative troops than that implied by the number you each get per turn.
nietzsche wrote:
This equation will only work when Y > 3X.
Which means that you can only ever get to three times your opponent's troops if you're earning more than three times as many troops as him. Possibly non-intuitive;
Yes, it did take me a second when I tried to corroborate it with values of 2 to 1.
I am NO math genius to be sure....but if you get out of strictly math and DO get into the game as the post started out....it will depend on where you are located, if you are playing while accumulating troops, and bonus's you get not to mention the troops you AND your opponent lose.
Anything other than that in this game isn't worth the effort seems to me.
AyeTrain wrote:You guys are scholars and gentlemen! Much obliged.
...looks like I might have to settle for smaller odds.
But even though the troop proportion doesn't really change, your odds do continue to improve as the troop count goes up b/c of attacking dice advantage.
Without knowing the situation, 3x your opponent's troops on a single tert is way more than enough. 2x is a safe margin, 1.5x is generally enough, and in a simple 1v1 situation, even having one more troop than the other player gives you an advantage, particularly the more you each have on the tert.
If your opponent has more territories, though, you'll need more troops of your own, generally +1 for each territory your opponent owns.
