Attention mathematical geniuses...part III

[the liquidator]

So I was wondering--how often will I win a 6 on 3 if a) I'm only willing to lose 3 armies, and b) if I'm willing to take it all the way down to 1?

[the liquidator]

Let me modify this--for a) let's say I'll attack 4 on 3,which changes the question a bit...

[tahitiwahini]

So I was wondering--how often will I win a 6 on 3 if a) I'm only willing to lose 3 armies, and b) if I'm willing to take it all the way down to 1?

a) 63%

b) 77%

[tahitiwahini]

Let me modify this--for a) let's say I'll attack 4 on 3,which changes the question a bit...

Yes it does, quite a bit:

a) 25%

b) 47%

[Aegnor]

Well I'm not a mathematical genius, or a genius at all for that matter, but let me try to take a shot at it:

In a 4vs3 situation You have :
1)37.17% to beat 2 armies (Next attack--> 4vs1)
2)33.58% to win 1 and lose 1 (Next attack- 3vs2)
3)29.26% to lose 2 (Next attack 2v3)

So basically you have 37.17% to meet your condition in one shot.

If you're willing to take more shots, let's see what happens:

3vs2:

1)22.76% to beat 2 armies. (You now took his territory)
2)32.41% to beat 1 and lose 1 (Leaves you at 2vs1)
3)44.83% to lose 2 (now it's 1v2 and you've lost)

Now let's add up the odds !! 37.17% (if you made it in one shot)+[33.58% X (22.76%+32.41%)]=18.52%
37.17%+18.52%=55.69%

So basically if you're willing to risk 3 armies out of your 4, you have 55.69% to bring him down to 1.

Well at least I think so, it might be wrong of course. I know crap about math.

[Aegnor]

Hmm there's also the 2vs3 situation which I didn't even take into account, so you might have even a higher chance to win a 4v3

[the liquidator]

Sorry,I was unclear in my modification--I still meant that I was starting out 6 on 3,but my original question didn't take into account the possibility of losing 2 armies if it got down to 4 on 3...

[oVo]

Sometimes the odds just suck. I've recently had an 18 on 7 where the 7 lost none and then a 12 on 3 where the 3 lost none. Sometimes a territory just doesn't give at all so you have to stop the attack and wait 'til the next round...

[sully800]

Sometimes the odds just suck. I've recently had an 18 on 7 where the 7 lost none and then a 12 on 3 where the 3 lost none. Sometimes a territory just doesn't give at all so you have to stop the attack and wait 'til the next round...

The odds are always the same, they don't just sometimes suck.

And why does waiting till next round improve odds?

[oVo]

I have no idea why some territories won't give up a single army regardless of the odds,
but I've found that when it does happen they tend to roll over on the next turn or
conquering the adjacent country and attacking from there seems to work too.

[Coleman]

You guys know there is an application for this stuff right? I'd tell people which one I use, but I like the advantage of knowing the odds for myself.

[tahitiwahini]

Sorry,I was unclear in my modification--I still meant that I was starting out 6 on 3,but my original question didn't take into account the possibility of losing 2 armies if it got down to 4 on 3...

If this is what your modification meant, then the original answer stands:

What are the chances that a 6v3 army attack will conquer a country if you are willing to lose up to 3 armies in the attack (attacker stops with 3 armies)?

a) 63%

What are the chances that a 6v3 army attack will conquer a country if you are willing to lose up to 5 armies in the attack (attacker stops with 1 army)?

b) 77%

There are two nice battle odds calculators that I use:

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

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

[tahitiwahini]

Well I'm not a mathematical genius, or a genius at all for that matter, but let me try to take a shot at it:

In a 4vs3 situation You have :
1)37.17% to beat 2 armies (Next attack--> 4vs1)
2)33.58% to win 1 and lose 1 (Next attack- 3vs2)
3)29.26% to lose 2 (Next attack 2v3)

So basically you have 37.17% to meet your condition in one shot.

If you're willing to take more shots, let's see what happens:

3vs2:

1)22.76% to beat 2 armies. (You now took his territory)
2)32.41% to beat 1 and lose 1 (Leaves you at 2vs1)
3)44.83% to lose 2 (now it's 1v2 and you've lost)

Now let's add up the odds !! 37.17% (if you made it in one shot)+[33.58% X (22.76%+32.41%)]=18.52%
37.17%+18.52%=55.69%

So basically if you're willing to risk 3 armies out of your 4, you have 55.69% to bring him down to 1.

Well at least I think so, it might be wrong of course. I know crap about math.

these are all the possibilities that lead to the attacker winning (that is, when the defender has 0 armies left):

4v3-->4v1-->4v0: 0.3717*0.6597=0.245210490
4v3-->4v1-->3v1-->3v0: 0.3717*0.3403*0.5787=0.073199479
4v3-->4v1-->3v1-->2v1-->2v0: 0.3717*0.3403*0.4213*0.4167=0.022205956
4v3-->3v2-->3v0: 0.3358*0.2276=0.076428080
4v3-->3v2-->2v1-->2v0: 0.3358*0.3241*0.4167=0.045350619
4v3-->2v3-->2v2-->2v1-->2v0: 0.2926*0.2246*0.2546*0.4167=0.007903412

The sum of all possibilities is: 0.470298037

You don't want to go through that calculation so I highly recommend the Gambit Battle Odds calculator which gives as a result: 0.4702511

There's a 0.000046937 discrepancy between the Gambit and the probability I calculated and I'm not sure where to attribute the error. The probabilities I used came from the Dice Analyzer.

[Aegnor]

Thanks tahiti, when I realized there are more probabilities to calculate I pretty much stopped where I stopped , I'm kinda lazy

[Georgerx7di]

I still am not sure about tahitis answers, I will comment more on this when I have time, but I think that those numbers are wrong.

[Bean_]

With the probabilities only at 4 significant figures, I think you can only rely on your hand calculated results to 4 significant figures, so 47.03%.

[tahitiwahini]

With the probabilities only at 4 significant figures, I think you can only rely on your hand calculated results to 4 significant figures, so 47.03%.

You're right Bean_, there goes the discrepancy.

[demigod]

so if i were to attack someone with 1,000 armies and they had [say] 5,000 armies, how much damage could i expect to do?

[Bean_]

Your question blows well past the risk calculators available. What, you playing in a build game?

I hypothesize the answer along the following lines: In a 3 die vs 2 die attack, the attacker loses about 0.921 armies per turn, the defender loses about 1.079 armies per turn. If you have 1000 armies and make 1080 attacks, you would have an expected loss of about 995 armies and the defender would have an expected loss of about 1166 armies.

However, this is not necessarily the actual expectation because you could run into zero armies a lot sooner (i.e., your expected number of attacks may be less than 1080, as opposed to a situation where you start with 100,000 armies against 100,000 and are willing to lose up to 1000).

[tahitiwahini]

so if i were to attack someone with 1,000 armies and they had [say] 5,000 armies, how much damage could i expect to do?

The first thing to notice is that the overwhelming majority of the attacks will be made at 3v2 dice odds. Consequently I will make the simplifying assumption that all attacks will occur at 3v2 dice odds, which will have the effect of making the answer slightly exaggerate the loses suffered by the defender.

Let n be the number of armies lost by the defender.
Let a be the number of attacks in which the defender lost two armies.
Let b be the number of attacks in which the attacker lost an army and the defender lost an army.
Let c be the number of attacks in which the attacker lost two armies.

Since we know the probabilities of each outcome of a 3v2 dice attack we obtain the following three equations:

(i) a = 0.3717 (a + b + c)

That is, we expect the defender to lose 2 armies 37.17% of the time in a 3v2 dice attack.

(ii) b = 0.3358 (a + b + c)

We expect the attacker to lose one army and the defender to lose one army 33.58% of the time.

(iii) c = 0.2926 (a + b + c)

We expect the attacker to lose two armies 29.26% of the time in a 3v2 dice attack.

(iv) 999 = b + 2c

We will continue to attack until we are left with a single army and b and 2c represent the number of attacking armies we will lose in our attacks.

(v) n = 2a + b

The defender will lose 2a + b armies from our attacks.

Solving for a, b, and c we obtain:

a = 403
b = 364
c = 317

So the total number of armies lost for the attacker is 999 and the total number of armies lost by the defender is (from equation v) 1170.

So the answer to the question is that on average the attacker could expect to kill less than 1170 of the 5000 defender armies at a cost of 999 of the attacker's 1000 armies.

[demigod]

expected loss of about 1166 armies.

thanks. i was trying to make the numbers simple btw.

the point of the question is to ask that if you are losing in a game and u need some last ditch efforts to try and win - that is, you are attacking a defender who has more armies - what would be the minimum number of armies you statistically need to defeat that person and perhaps break that important continent.

For example if you had 20 armies, by your stats one could expect to kill 1.166 armies for every attacking army. so if the defender has 23.32 armies you would win??

thanks also to tahitiwahini

[Bean_]

It depends. If you have 20 armies and 1 single territory to attack for all the marbles, you have 52.2% to break into a 21 army territory, and 47.2% to break into a 22 army territory. You have a >90% chance to break into a territory with 12 or fewer armies.

But failing to break into even a 12 army territory with 20 armies is well within normal variation!