three on two teritories of 1 odds

by **jonka** on Fri Apr 03, 2009 12:54 am

Can somebody help me with the odds that I will successfully keep the attacker with 4 troops on their country (3 attacking) out of my country if there is only 1 country with 1 troop separating them? (one troop on the territory that I hope to defend) I got the odds at about 39% but I'm not so sure thats right.

by **oVo** on Fri Apr 03, 2009 2:42 am

I don't know about the 39%, but it would seem that the odds are in the attackers favor for the first roll.
When it's 3 armies vs 1 does the attacker still get to roll three dice?

by **jonka** on Fri Apr 03, 2009 3:12 am

(exhibit a)3 dice vs 1 die 66% chance defender is killed in the first turn
(1)if the defender is killed-2 verse 1 die 75% chance they take the territory
(2)if the defender is not killed- Have to win the 2 vs. 1 and 1 vs. 1 (58% x 42%)

(.9166666...) x 2/3 + 1/3 x (.24)
attacker's odds #1 Probability of #1

Thats probably wrong because with the unsimplified form I'm getting 99.78% which seems a little much.

by **oVo** on Fri Apr 03, 2009 3:24 am

The defender wins all ties... so is there a way of inserting that into the equation?
It seems that would be a major factor to consider when attempting to calculate
or project winning odds.

by **jonka** on Fri Apr 03, 2009 4:12 am

They already are, the percents that I named were of the possible outcomes of the die, for example the 66% the first one I see. Its a little easier to approach it from the defenders side-
1/6th of the time they will roll a 6 and win, then you go into the other options- its long and boring so I'll just give an example with 5
(5) - 1 1 1
(5) - 1 1 2...
(5) - 1 1 5
5 - 1 1 (6)...
4...
3...
2...
1... There are 1296 different combinations. I just get my basic odds off the internet, and plug them into equations.

by **Timminz** on Fri Apr 03, 2009 4:52 am

http://gamesbyemail.com/Games/Gambit/BattleOdds

for any odds-based questions.

by **oVo** on Fri Apr 03, 2009 6:15 am

Another thing you have to account for is SOME DAYS THE DICE JUST SUCK BIGTIME as I just attempted to crack open a big bonus doorway by nibbling on an 18 fort with 12 armies
and managed to sacrifice 11 to kill off a mere 3. 8-[

A FREAKIN DOUBLE DIE DISASTER!

by **Georgerx7di** on Mon Jun 15, 2009 5:05 pm

I did this on paper and also got 66.0% for attacker to win, and 34.0% for defender to win. Accurate to 3 sig figs.

Edit: Btw, these odds are only for the first roll. If the attacker looses he will still have 3 armies, and can attack 2 dice to 1. I'll work out the odds for the second and third roll, and the total when I have a few minutes free, and update this post.

Edit2: I read the question as what are the odds of success if you attack 4v1. That means a territory with a number 4 on it attacking a territory with 1 army on it.

by **iancanton** on Mon Jun 22, 2009 12:21 pm

i think it's a bit less than 66%. to break a bonus, attacking 2 regions of 1 each is normally more difficult than 1 region of 2.

ian.

by **AceArtemis** on Wed Jun 24, 2009 2:02 am

Assault Odds Calculator says the chance of the attacker winning a 4 v. 1,1 is 0.5796 or 57.96%.

Assault Odds can be found here: http://www.conquerclub.com/forum/viewtopic.php?&t=83451

by **tryagain** on Mon Jul 06, 2009 12:26 am

4 vs 1 Chance of NOT winning = .3403
3 vs 1 Chance of NOT winning = .4213
2 vs 1 Chance of NOT winning = .5833

Therefore the Chance of NOT winning the whole battle = 0.3403 x 0.4213 x 0.5833 = .0836
Therefore the Chance of winning the whole battle = 1 - 0.0836 = 0.9164 =

91.64%

There's your answer.

by **peacemakers** on Sun Aug 02, 2009 7:25 pm

tryagain wrote:4 vs 1 Chance of NOT winning = .3403
3 vs 1 Chance of NOT winning = .4213
2 vs 1 Chance of NOT winning = .5833

Therefore the Chance of NOT winning the whole battle = 0.3403 x 0.4213 x 0.5833 = .0836
Therefore the Chance of winning the whole battle = 1 - 0.0836 = 0.9164 =

91.64%

There's your answer.

correct

by **AAFitz** on Sun Aug 02, 2009 7:34 pm

peacemakers wrote:
tryagain wrote:4 vs 1 Chance of NOT winning = .3403
3 vs 1 Chance of NOT winning = .4213
2 vs 1 Chance of NOT winning = .5833

Therefore the Chance of NOT winning the whole battle = 0.3403 x 0.4213 x 0.5833 = .0836
Therefore the Chance of winning the whole battle = 1 - 0.0836 = 0.9164 =

91.64%

There's your answer.

correct

Isnt that the odds of just attacking a one. The Op was asking about attacking two ones. Im almost certain a 4 does not have a 91% chance of killing 2 ones.