[NT] How do you think through this?

[BlueGray]

I'm curious how people work through this problem.

You are the general of an army of essentially infinite numbers. The opposing general has an army equally as massive. There are three battlefronts that will decide the war. A battlefront is won if you send as many or more troops than the opposing general and lost if you send less. To win the war you must win two of the three battlefronts. You sent your troops to the battlefronts in the magnitudes of 1/4 : 1/4 : 1/2 of your total army. The opposing general uses no strategy and distributes his troops randomly. What is the probability that you win the war?

 

[Arthur Schopenhauer]

I'm curious how people work through this problem.

You are the general of an army of essentially infinite numbers. The opposing general has an army equally as massive. There are three battlefronts that will decide the war. A battlefront is won if you send as many or more troops than the opposing general and lost if you send less. To win the war you must win two of the three battlefronts. You sent your troops to the battlefronts in the magnitudes of 1/4 : 1/4 : 1/2 of your total army. The opposing general uses no strategy and distributes his troops randomly. What is the probability that you win the war?

This makes no sense... If the troops are infinite, how can you divide them?

 

[INTP]

i solve this by escaping from the army and let the idiots kill each others

 

[BlueGray]

I said essentially infinite. Basically there are too many for a single soldier to matter but still not infinite.

 

[Arthur Schopenhauer]

I said essentially infinite. Basically there are too many for a single soldier to matter but still not infinite.

It's either infinite or it's not. There is no 'almost infinite'.

 

[Phoenix_400]

Logic is flawed. Can't work with 'near-infinite'. Now if you want to say "Both forces have equal resources and manpower", then we're getting somewhere.

Now:
If we know that there are 3 battlefronts, and the opposing force is randomly distributed, do we have intelligence as to the enemy distribution across the battlefronts?

Recon is vital in a war. So is strategy. Also, need lay of the land. If the enemy has the high ground, they'll expend less troops maintaining it than the opposing force would to capture it.

Given two equal forces, the force with the superior strategy should be victorious. Wars of attrition should be avoided. Personally, I want to see a layout of the war-zone, to include topology, enemy strongholds, and distribution of enemy forces, supply lines, etc....Anybody got a RISK board handy?

Couple of different possible strategies. My fav's for playing something like Risk are:
1.) Flank- Attack the enemy's weak point w/ 1 force. When they move to reinforce, move your flanking unit in to attack them from the back or the new weak point they've opened up.
2.) Spearhead- Concentrate your forces and run a wedge straight up the middle of the enemy. Keep reinforcing your 'blades' from the center. Drop units from center-rear to engage disjointed forces as you break the line and divide their forces.

If ya give me more to go on, I could pick the brains of a few retired officers, SF, and some door-kickers I know.

Also, needs more Sun Tzu and Musashi

 

[Stanton Moore]

What is the stregic value of each of the three battlefields? Are they all equal in value? If so, you can assume that the other side will distribute 1/3 of troop total to each front. If not, you can use that to your advantage, if you know in advantage what the values are.
but you still need to know more rules of the game to win it.

 

[BlueGray]

The point of saying near infinite is that the numbers become incredibly messy when dealing with a finite size but they do approach simpler numbers. It's basically asking for the limit as the number of troops approach infinity.

 

[BlueGray]

What is the stregic value of each of the three battlefields? Are they all equal in value? If so, you can assume that the other side will distribute 1/3 of troop total to each front. If not, you can use that to your advantage, if you know in advantage what the values are.
but you still need to know more rules of the game to win it.

The strategic value of each battlefield is the same. Win any two and the war is won. You have already sent your troops. The other side uses no system of placing troops. Basically they are just as likely to send all of their troops to 1 battle field as they are to send 1/3 of their army to each battlefield. The question is asking you to evaluate how effective a 1/4:1/4:1/2 strategy is when your opponent uses no logical strategy. A tougher question could be, find the most effective strategy.

 

[Shimmy]

My brain tells me there is a way to solve this. Checking back when I figure it out!

 

[Phoenix_400]

The point of saying near infinite is that the numbers become incredibly messy when dealing with a finite size but they do approach simpler numbers. It's basically asking for the limit as the number of troops approach infinity.

The strategic value of each battlefield is the same. Win any two and the war is won. You have already sent your troops. The other side uses no system of placing troops. Basically they are just as likely to send all of their troops to 1 battle field as they are to send 1/3 of their army to each battlefield. The question is asking you to evaluate how effective a 1/4:1/4:1/2 strategy is when your opponent uses no logical strategy. A tougher question could be, find the most effective strategy.

Crappiest....Sit-rep...Ever. Tell me you didn't come up with this on your own.

EDIT:
I was going to give it a shot, but how the hell do you develop a strategy or know where to send your greater forces w/o knowing the layout of enemy forces? Even if you don't know their strategy you should at least know where to concentrate forces. Methinks this problem wasn't developed by someone with experience in military history.

If you know the enemy and know yourself, you need not fear the result of a hundred battles. If you know yourself but not the enemy, for every victory gained you will also suffer a defeat. If you know neither the enemy nor yourself, you will succumb in every battle.
- Sun Tzu

 

[Stanton Moore]

The strategic value of each battlefield is the same. Win any two and the war is won. You have already sent your troops. The other side uses no system of placing troops. Basically they are just as likely to send all of their troops to 1 battle field as they are to send 1/3 of their army to each battlefield. The question is asking you to evaluate how effective a 1/4:1/4:1/2 strategy is when your opponent uses no logical strategy. A tougher question could be, find the most effective strategy.

If there were a finite number of troops, then you could solve it, but with an infinite number (nearly infinite here just means undefinable) then there is no way to do it, since you can't take 1/4:1/4:1/2 of near-infinity.

 

[Phoenix_400]

If there were a finite number of troops, then you could solve it, but with an infinite no (nearly infinite here just means undefinable) then there is no way to do it, since you can't take 1/4:1/4:1/2 of near-infinity.

I'm choosing to use my interpretation of "both forces have equal resources and manpower". If we do it that way, then you can concentrate solely on the ratios.

Sounds like the problem has you stuck at a 25/25/50 troop distribution across the 3 battlefields. That's where I'm stuck at. If we're stuck at that ratio, do we have to send all 3 forces to all 3 battlefields? Can we send 2 forces to one battlefield? But then its not a 25/25/50, technically its a 50/50 split. There's a dozen other questions I could ask here.

This problem is far too vague to be of any use as a strategy problem. The only use I see for this is to calculate probability of defeat if stuck at 25/25/50 ratio when compared to the enemy distribution ratio.

Basically if any 2 of the enemy forces is under 25% or 1 is under 25% and a second under 50%, you win.

This is still crap for coming up with a 'strategy'. There's just not info intel to work with.

 

[BlueGray]

There are two possible problems you can do. Find the probability that a 1/4:1/4:1/2 split wins or find the best strategy possible. The first is what the original question asked the second is an expansion upon the problem.

This was originally a math problem.

 

[tcda]

So basically

army=1

each of the 3 battlefields can be represented by anything < 1.

presumably 1/3 or 0.33% is the "mid-point" as we are dividing by 3, so the probability is that he is going to get more than 1/3 on one battlefield, less than 1/3 on another, and the last one is random.

So essentially if I distribute my troops 1/4, 1/4 and 1/2 there is probaility of something above 0.5 that he will win (you would have to work out the maths then by calculating difference between 1/4 and 1/3, which I can't be fucked to do).

Or is that complete bullshit?

 

[BlueGray]

Actually the first two are random the third is not. The enemy does use all his troops so the third is all remaining troops.

 

[Phoenix_400]

There are two possible problems you can do. Find the probability that a 1/4:1/4:1/2 split wins or find the best strategy possible. The first is what the original question asked the second is an expansion upon the problem.

That's the thing though, there's not enough info here for any kind of 'strategy'. I can see how you can work this as a 'probability' problem. I don't know of a single military strategist who could effectively work this as a 'strategy' problem the way its written, and I DO know people who've forgotten more about battle tactics and troop movements than most civilians will ever learn.

I think its misleading to claim this as anything other than a statistics problem.

I still want to know where you came up with this problem. It really is vague and poorly worded.

Actually the first two are random the third is not. The enemy does use all his troops so the third is all remaining troops.

Yes, the 3rd is all remaining troops. HOWEVER, since the size of the 3rd is a function of the 1st and 2nd, and the 1st and 2nd are random, the 3rd is still random by design. Its ratio to the whole is not fixed as long as 1 and 2 are capable of fluctuation.

 

[forzen]

Why not put the 1/2 in the middle so it would look like 1/4:1/2:1/4 so you can back up the two 1/4s as you gain more informations on how the enemiy distributed its troops. The 1/2 can easilly support the two 1/4s while you can put either 1/4s to the 1/2 if the enemy is concentrated there. Basically, put the largest so it can support the two smaller group.

 

[BlueGray]

I paraphrased the problem as I hadn't seen it in a while.
You are the general of an army. You and the opposing general both have an equal number of troops to distribute among three battleelds. Whoever has more troops on a battlfield always wins (you win ties). An order is an ordered triple of non-negative real numbers (x, y, z) such that x+y +z = 1, and corresponds to sending a fraction x of the troops to the rst eld, y to the second, and z to the third. Suppose that you give the order (1/4,1/4,1/2) and that the other general issues an order chosen uniformly at random from all possible orders. What is the probability that you win two out of the three battles?
Here is the original wording of the problem.
It is a probability problem. By strategy I meant a division of your troops that has the highest probability of winning. (The 1/4, 1/4, 1/2 split has a higher than .5 chance of winning)

 

[tcda]

I paraphrased the problem as I hadn't seen it in a while.
You are the general of an army. You and the opposing general both have an equal number of troops to distribute among three battleelds. Whoever has more troops on a battlfield always wins (you win ties). An order is an ordered triple of non-negative real numbers (x, y, z) such that x+y +z = 1, and corresponds to sending a fraction x of the troops to the rst eld, y to the second, and z to the third. Suppose that you give the order (1/4,1/4,1/2) and that the other general issues an order chosen uniformly at random from all possible orders. What is the probability that you win two out of the three battles?
Here is the original wording of the problem.
It is a probability problem. By strategy I meant a division of your troops that has the highest probability of winning. (The 1/4, 1/4, 1/2 split has a higher than .5 chance of winning)

So he doesn't then get to choose the last one; the last one is in fact random?

So then was I right? :s