improve your odds.any board almost all settings

[jac99]

on each roll of the dice the player defending gets the tie breaker advantage but is limited to just 2 die the attacker is limited to 3. but if the attacker gets 3 and the defender only has 1 armie he will lose. so if you don't have a terriotory try placing 2 instead of 1 on each teritory this puts the odds in your favor and even if your opponet has built a large army he will be wary of taken you out. i have found this tactic helpful in 1v1 and multi player games. jac99

[White Moose]

It's better to have as many 4's as you can instead.

[jac99]

do you use 4 for psychological reasons or for better odds.

[The Neon Peon]

do you use 4 for psychological reasons or for better odds.

Better odds

With a 3, you roll 2 dice, with a 4, you roll 3 dice
With a 5, you roll 3 dice.

So why deploy 2 and roll three dice when you can deploy 1 and roll 3 dice?

[Wolffystyle]

Whoa, I think you two are talking about two different things. jac99 writes about defense while Neon writes about offense.

Back to Moose, though, I'm actually wondering though if 4 is better to defend with. I've seen players like King use it on city mogul. In a small 1v1 game, if you have four, you might as well attack, right? However, a wall of four gives a chance to roll 2 dice more than twice as often. I'm making up these numbers actually. Let me get new batteries for my TI89....

[Wolffystyle]

Ok I did the math for one example only. MY CONCLUSION IS DRAWN ONLY FROM ONE EXAMPLE.

Under the assumption that attacker A attacks all the way through to his final roll:

Given: Attacker A has 7 troops bordering a string of three of Defender D's territories (i.e. 7 in China attacking Siam, Indonesia and Western Australia).

Given: Defender D has unlimited reinforcements and wants to know how to reinforce his 6 total defending troops.

If Defender D places all 3 mobile troops in Siam (4-1-1) he has an 11% of keeping at least one troop on the board.
If Defender D places one of each of his three mobile troops in each of the three territories left (2-2-2) he has an 18% of keeping at least one troop on the board.

Of course, this doesn't take into consideration the weight of spoils.

This is one rare situation, but jac99 is right, stacks of two prove a better defense than the alternative...


7 v 4 v 1 v 1
0.138 0.01 0.028 0.00003864
0.164 0.028 0.083 0.000381136
0.179 0.083 0.246 0.003654822
0.14 0.246 0.583 0.02007852
0.083 0.583 1 0.048389
0.039 1 1 0.039
0.111542118


7 2 2
0.371 0.11 0.214 0.00873334
0.222 0.214 0.34 0.01615272
0.184 0.34 0.64 0.0400384
0.09 0.64 0.89 0.051264
0.046 0.89 1 0.04094
0.0197 1 1 0.0197
0.17682846

Wolff

[BaldAdonis]

Wolff, if you're interested, the math supports always attacking when you have 4 or more (provided your objective is to make your opponent lose as many armies as possible) instead of using them for defense. When you are defending, even numbers are better. Two 4s are better than 5,3, but 6,2 is better than both.

[JoshyBoy]

So what I gather from this post is that the dice are not random at all.

Am I correct?

[Seulessliathan]

When you are defending, even numbers are better. Two 4s are better than 5,3, but 6,2 is better than both.

why? any math for this theory?

[Timminz]

When you are defending, even numbers are better. Two 4s are better than 5,3, but 6,2 is better than both.

why? any math for this theory?

Not quite the math involved, but http://www.gamesbyemail.com/Games/Gambit/BattleOdds will show you that those examples are true.

[alster]

Back to Moose, though, I'm actually wondering though if 4 is better to defend with. I've seen players like King use it on city mogul. In a small 1v1 game, if you have four, you might as well attack, right? However, a wall of four gives a chance to roll 2 dice more than twice as often.

As a defense - 2 armies provides a great boost to your defense compared to 1 army.

But, placing any more than 2 armies as a defense does not provide much odds-wise. So, anything above 2 is normally best placed on your attack stacks.

[Seulessliathan]

When you are defending, even numbers are better. Two 4s are better than 5,3, but 6,2 is better than both.

why? any math for this theory?

Not quite the math involved, but http://www.gamesbyemail.com/Games/Gambit/BattleOdds will show you that those examples are true.

no, it doesn´t , that´s why i ask.

i have tried 2 examples, 10v8 and 15v8 , here the results

10v...
6,2 67,34% (chance for attacker to win)
5,3 67,37%
4,4 66,66%
3,5 67,05%
2,6 66,17%

15v...
6,2 93,06%
5,3 93,11%
4,4 92,97%
3,5 93,07%
2,6 92,81%

6,2 seems to be worse than 2,6 for the defender, and 10v6,2 gives the attacker better odds than 10v3,5 ...

i doubt there is any rule about even and uneven numbers

[laughingcavalier]

Great link - does anyone know if the maths behind it are sound?

[susanspy]

hello all,


i am susan here, i am not sure if i uderstand any of the content posted at the link improve your odds.any board almost all settings please explain.




cheers!!!!!!!!!!
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[Bruceswar]

Or if you are Jac99 then to better your odds of winning, you just foe the people who beat you or give you a run for you money...

[BaldAdonis]

why? any math for this theory?
....
i doubt there is any rule about even and uneven numbers

Of course there is. Your numbers aren't large enough, and the bottom 2v2 and 1v2 rolls are skewing results (on the order of 0.05%). Expected losses per defender are maximized at 2, but climb slightly on even numbers and drop on odd numbers, so you best move is to make a bunch of twos and stack the remainder at the front.

The formula for expected losses makes it obvious that there is a rule about even and uneven numbers. It assumes there are enough attackers that you don't have to throw 2v2 or 1v2 (which are the inefficient rolls, where you lose more than you kill).

Losses = 0.85n + 0.22*(-0.53)^n - 0.22

proof

If you throw 3 dice, and they throw 1, call the probability of winning "a". For 3v2, call a win "b", a loss "c" and a draw "d". c = 1- b - d, so I'll make this switch wherever I can. Call the expected loss against n armies x_n.
x_1 = a*0 + (1-a)*(1 + x_1), because if you win, then your loss is 0, and if not, then it's one more than it was before, and you have to try again.
x_1 = 1/a -1 = (1296-855)/855 = 49/95 = 0.516

x_2 = c(2 + x_2) + d(1 + x_1) = (d*x_1 + 2c + d)/(b + d) since b + d = 1 - c
x_2 = 808234/522595 = 1.547

x_3 leads us to the general case, since it is the first value that cannot be taken in one throw. I need the value for initial recurrence conditions, but I'll just write the general equation.
x_n = c*(2+ x_n) + d*(1 + x_(n-1)) + b*x_(n-2)
Since this is an inhomogenous recurrence (it contains 2c + d on the right side, with no x values attached), we convert it to a homogenous recurrence by subtracting x_(n-1) from both sides.
x_n - x_(n-1) = c*(2+ x_n) + d*(1 + x_(n-1)) + b*x_(n-2) - c*(2+ x_(n-1)) - d*(1 + x_(n-2)) - b*x_(n-3) or more simply
x_n = [(b + 2d)*(x_(n-1)) + (b - d)*(x_(n-2)) - b*x_(n-3)]/(b + d) and canceling all the fractions, we get
x_n = [8112*(x_(n-1)) + 279*(x_(n-2)) - 2890*x_(n-3)]/(5501)

We solve this in the usual way, by solving x^n = [8112*(x^(n-1)) + 279*(x^(n-2)) - 2890*x^(n-3)]/(5501), and find that x=1 (twice) or x = -2890/5501. So we look for a general solution of the form x_n = A + Bn + C(-2890/5501)^n

A little bit of leg work (and using x_0 = 0) gives us A = -(2a - 2b - d)(b + d)/[a(2b+d)^2], C = (2a - 2b - d)(b + d)/[a(2b+d)^2], and B = (2 - 2b -d)/(2b + d), or A = -0.2213412, C = 0.2213412, and B = 0.8534144

Here are some papers for you to read. The first is simpler and explains things without requiring much knowledge of probability. The second would make more sense if you know what a stochastic process is.
http://www.rose-hulman.edu/mathjournal/ ... n2-3pd.pdf
http://www.recreationalmath.com/Risk/RiskPaper.doc
There's another article by Tan that might help you, but I assume you don't have any MAA subscriptions, and I don't know where to find it online.