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.
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?
(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.
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.
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.
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.
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.
Last edited by Georgerx7di on Wed Jun 24, 2009 10:47 pm, edited 1 time in total.
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 =
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 =
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.
I'm Spanking Monkey now....err...I mean I'm a Spanking Monkey now...that shoots milk Too much. I know.