Odds of losing a 3 vs 1 attack twice

[Grifter]

Can someone answer the below problem and show the math? Thanks.

The odds of losing a 3 vs 1 attack is about 34%. This is according to what is posted on 'Wikipedia. This is actually slightly higher then losing a 3 vs 2 attack (where attacker loses 2). What are the odds of losing a 3 vs 1 attack twice?

My record for most loses when attacking one army is 5. Needless to say I did not win that game.

[Jeff Hardy]

dunno about the math but i lost a 13vs1 today what were the odds of that?

[The Neon Peon]

The attacker will always have an advantage as long as the attacker rolls 3v2, 3v1 or 2v1.

3v2: 37.2% win, 33.6% tie, 29.3% loss
3v1: 66% win, 34% loss
2v1: 57.9% win, 42.1% loss

two 3dice v 1dice losses in a row are 11.6%

[The Neon Peon]

dunno about the math but i lost a 13vs1 today what were the odds of that?

.11%

So, that converts to 1 in 1000

[Itrade]

dunno about the math but i lost a 13vs1 today what were the odds of that?

.11%

So, that converts to 1 in 1000

Considering how many rolls are made every day, those odds should make 13 vs 1 losses a pretty common occurence!

Either the math is wrong or I have no idea how probabilities work. I suspect it's the latter.

[edbeard]

first I'll answer the OP's question (if I've understood him correctly) which someone else has answered but here's the math

you're asking the chances of losing two separate 3v1 rolls? one roll has nothing to do with the other so we simply multiply the probabilities of each separate event which is

.34 * .34 = .1156



if you understand the logic behind that you can answer the 13 v 1 question

that would entail losing ten "3v1" rolls, one "2v1" roll, and one "1v1" roll

according to the numbers sully posted (quoted by neon) this translates to

.34^10 * .421 * .583 (.583 found on http://www.plainsboro.com/~lemke/risk/ since I don't want to calculate it myself)

which equals about .000005 or about 5 times every million (assuming I didn't screw up the actual calculation thanks to the shitty windows calculator)

[Quicksand_Jesus]

The attacker will always have an advantage as long as the attacker rolls 3v2, 3v1 or 2v1.

3v2: 37.2% win, 33.6% tie, 29.3% loss
3v1: 66% win, 34% loss
2v1: 57.9% win, 42.1% loss

two 3dice v 1dice losses in a row are 11.6%

Are you sure the attacker always have an advantage statisticly? I thought the opposite and tried to prove it by making a program that simply generate one million fights and displays the result. I'm quite new to programming, but this wasn't too hard. What i got was:

result after 1000000 fights with 3 attackers and 2 defenders
attackers killed: 750578
defenders killed: 589426
percent chance of winning if attacking with equal number of forces: 44
percent chance of winning if defending: 56

this would if my program is correct prove that you have a small advantage if defending
Am I right or not?
If not please prove my wrong
/Jesus

[Timminz]

Are you sure the attacker always have an advantage statisticly? I thought the opposite and tried to prove it by making a program that simply generate one million fights and displays the result. I'm quite new to programming, but this wasn't too hard. What i got was:

result after 1000000 fights with 3 attackers and 2 defenders
attackers killed: 750578
defenders killed: 589426
percent chance of winning if attacking with equal number of forces: 44
percent chance of winning if defending: 56

this would if my program is correct prove that you have a small advantage if defending
Am I right or not?
If not please prove my wrong
/Jesus

You should check out Gambit.

[FabledIntegral]

The attacker will always have an advantage as long as the attacker rolls 3v2, 3v1 or 2v1.

3v2: 37.2% win, 33.6% tie, 29.3% loss
3v1: 66% win, 34% loss
2v1: 57.9% win, 42.1% loss

two 3dice v 1dice losses in a row are 11.6%

Are you sure the attacker always have an advantage statisticly? I thought the opposite and tried to prove it by making a program that simply generate one million fights and displays the result. I'm quite new to programming, but this wasn't too hard. What i got was:

result after 1000000 fights with 3 attackers and 2 defenders
attackers killed: 750578
defenders killed: 589426
percent chance of winning if attacking with equal number of forces: 44
percent chance of winning if defending: 56

this would if my program is correct prove that you have a small advantage if defending
Am I right or not?
If not please prove my wrong
/Jesus

You either did something wrong or you have a statistical anomaly. Attackers have an advantage with 3 attackers 2 defenders. Hopefully you didn't do 3v2 in terms or armies, which is 2 attackers 2 defenders where the defender obviously has an advantage.

[lgoasklucyl]

Wouldn't it be impossible to determine the odds, as it's 'entirely random' in the generator?

[The Neon Peon]

My original figures were off. After a few lessons from fellow CCers about statistics, I am showing a 6% chance of losing both.

Chance of winning a 3v1 attack (dice, not troops) = .754
1- .754 = Chance of losing
Chance of losing ^ 2 = 6%

[The Neon Peon]

Wouldn't it be impossible to determine the odds, as it's 'entirely random' in the generator?

That is the only thing that lets us determine the "odds"

The probability that something will happen if it is perfectly random. If it was not random, there would be no statistics of "odds" and we could give you a concrete number.

[lgoasklucyl]

Wouldn't it be impossible to determine the odds, as it's 'entirely random' in the generator?

That is the only thing that lets us determine the "odds"

The probability that something will happen if it is perfectly random. If it was not random, there would be no statistics of "odds" and we could give you a concrete number.

Ah, gotchya.

Me + Math = Nope

[Quicksand_Jesus]

hehe, thanks for all the answers. That Gambit site was great.
However I just found a minor fault in my program. I wasn't reseting a certain variable, also I modified it to display odds of split and so on. Well this is my new results:

result after 1000000 fights with 3 attackers and 2 defenders
attackers killed: 821632
defenders killed: 1178368
percent chance of attackers win all: 47
percent chance of defenders win all: 29
percent chance of split: 24
percent chance of winning if attacking with equal number of forces: 59
percent chance of winning if defending: 41

That does somewhat agree to other odds calculations, except for 10% that should be moved from attackers win all to split.
Anyway I think I rather trust the results from gambit than my own homemade program

P.S. The program is calculating 3 attacking dices vs 2 defending and nothing else to be clear.

[Timminz]

percent chance of winning if attacking with equal number of forces: 59
percent chance of winning if defending: 41

The odds of winning with equal forces changes depending on the size of the force. You're MUCH more likely to win 1000v1000, than you are to win 10v10.

[Quicksand_Jesus]

HA! I knew it. Something is influencing the odds when the number of forces grow.
What is it and how do it work?

btw either the gambit site is incorrect or I am being very unlucky. I am having a "cold war" in Game 3972062.
Two times I have attacked his forces now and calculated the odds with gambit. Both times my surviving forces was far under what was most probable, it was somewhere about 2% chance that that number of forces would survive.
Two times in a row?! Lightning doesn't strike twice at the same place.
Soon I will believe the there ain't no attackers advantage, but a defenders advantage.

[The Neon Peon]

In a 3 dice vs 2 dice attack
The odds of losing 2 are 2890/7776
The odds of killing one and losing one are 2611/7776
The odds of killing 2 are 2275/7776

While the odds of each individual roll remain the same, you will notice that in a 5 vs 5 attack, you can only attack with 4 armies, while in a 1000 vs 1000 attack, you can attack with 999 armies. So since you attack with less men than you start off with, the % of the men you can't attack with is reduced the more troops you have to begin with.

The odds of winning a 10v10 are 48%
The odds of winning a 20v20 are 58%

You will notice that the 10% variation could not have been caused by simply the amount of men you had to start off with. Reason being is that when you are making one 3 dice vs 2 dice roll, one chance determines the outcome. But when you do a million of those rolls, you will find that you succeed more than you fail.

It is not that in a 2000 vs 2000 attack you will kill a noticeably higher percentage of men than in a 1000 vs 1000. It is simply that, the more men you have, the higher the chances that the dice will act according to the figures given at the start.

And P.S. Lightning strikes the Empire state building 7 times when a storm passes over it. Much more than "never hits the same spot twice" don't you think?

[sailorseal]

I lose 3 v 1s all the time

[Quicksand_Jesus]

Me too, that's why I don't trust the odds. Sometimes I roll according to the odds and that's cool, but I never roll better than the odds. Bad luck? I don't know...
Well, thanks all for spending time answering my questions.

and P.S. Lightning doesn't strike by random.

[flexmaster33]

People complain about the dice all the time, but when you think how many rolls go through this site in a single day it shouldn't be surprising to see some long odds come through on a regular basis. And think how boring the game would be if the attacks were all static...every 3 vs. 1 resulted in a win and so on. You need the element of chance to keep the game interesting and strategies evolving.

Plus, in a real war there is plenty left to chance...most things can't be scripted perfectly to how the numbers say they should happen.

This post isn't meant as a rant against anyway, rather simply trying to make a point about the dice.

[e_i_pi]

I just rolled 3v1 twice, and won.

I declare the dice rig theory debunked!

[MarathonMax]

I found this web site to determine the probability of winning a battle based on the number of armies on each side.

Interesting!

http://bartell.org/risk/riskodds.shtml

Max

[Rocketry]

i never play 3 v anything. 3 is just a bad number on CC...

Rocket.

[owenshooter]

i never play 3 v anything. 3 is just a bad number on CC...

the science behind this is amazing, and i soundly agree!!! although 3 is the magic number, it is a bad number on CC to attack with... the black jesus has spoken...-0

[Gregster60]

Odds are just that: ODD! You can talk about the odds all you want, but I can tell you from experience here on CC that the dice here don't seem to follow the odds- lol. I've lost my butt in recent games!