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[NT] How do you think through this?

BlueGray

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

So then was I right? :s

Well all three are random but also dependent on each other as they must add up to 1.
 

Phoenix_400

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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)

Okay, now you're starting to make sense. THIS can be worked with.
 

Arthur Schopenhauer

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Whoever posted this:

Here is the message that has just been posted:
***************
I think sending 1/2 to two of the fronts and leaving the remaining front alone is the most effective strategy.
***************

The answer to that is no.
 

William K

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If you call the 3 battlefronts A, B, C,

Then your probability of winning is the sum of :-
P(Win A&B) + P(Win A&C) + P(Win B&C)

What happens when the number of soldiers on each side is exactly the same? A draw?

Edit : Ok, just saw that you win a tie as well... so you need to add P(Win A&B&C) where you tie all 3 battlefronts
 

forzen

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If you call the 3 battlefronts A, B, C,

Then your probability of winning is the sum of :-
P(Win A&B) + P(Win A&C) + P(Win B&C)

What happens when the number of soldiers on each side is exactly the same? A draw?

Ok, I don't get it. Can you explain it so a non-math genius like me can understand? I'm not familiar with probability so I don't see where you're coming from.

I was under the impression that you can divide your troops by

1/3 | 1/2 | 1/6

to get the majority.
 

Little_Sticks

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I don't know how to explain this, but if you consider that their are only two battlegrounds and you go through each possibility you will see that they always tie in a sort of balancing act where if one player wins one battleground then they must win the other. So when two opponents send the same amount to one battleground you automatically have a tie. Then you want to analyze the changes in that and find a set of ratios with a minimal set.

Considering that first battle, you want to win it but with as little people as possible. The higher your winning number than your opponent, the worse your chances of winning another battle.

Starting from this, let's assume we do 1/3:1/3:1/3.

From the first battle: 66/100 can beat you. Of those 66, only (66-33) 33 possibilities have a chance of winning another battle.

Let's now assume we do 1/2:1/4:1/4.

From the first battle: 50/100 can beat you. Of those 50, only (75-50) 25 possibilities have a chance of winning another battle.

Let's now assume we do 2/3:1/6:1/6.

From the first battle: 33/100 can beat you. Of those 33, only (83-66) 17 possibilities have a chance of winning another battle.

So 33/66 and 25/50 and 17/33 and ignoring my crude math we see the same chance (1/2) of winning another battle.

So it appears completely random so they all have the same chance of winning...I think???? Is this right?

Edit: Nevermind. I'm wrong. I'll wait for the answer.
 

forzen

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I don't know how to explain this, but if you consider that their are only two battlegrounds and you go through each possibility you will see that they always tie in a sort of balancing act where if one player wins one battleground then they must win the other. So when two opponents send the same amount to one battleground you automatically have a tie. Then you want to analyze the changes in that and find a set of ratios with a minimal set.

Considering that first battle, you want to win it but with as little people as possible. The higher your winning number than your opponent, the worse your chances of winning another battle.

Starting from this, let's assume we do 1/3:1/3:1/3.

From the first battle: 66/100 can beat you. Of those 66, only (66-33) 33 possibilities have a chance of winning another battle.

Let's now assume we do 1/2:1/4:1/4.

From the first battle: 50/100 can beat you. Of those 50, only (75-50) 25 possibilities have a chance of winning another battle.

Let's now assume we do 2/3:1/6:1/6.

From the first battle: 33/100 can beat you. Of those 33, only (83-66) 17 possibilities have a chance of winning another battle.

So 33/66 and 25/50 and 17/33 and ignoring my crude math we see the same chance (1/2) of winning another battle.

So it appears completely random so they all have the same chance of winning...I think???? Is this right?

You're looking to allocate your troops so you can win 2/3. A tie is a win.

1/3 : 1/3 own |1/2 win |1/2 win |1 owned | 1/3 tie
1/2 : 1/3 win |1/2 tie |1/4 win |n/a win | 1/2 tie
1/6 : 1/3 own | n/a win |1/4 owned |n/a win | 1/6 tie

This is how I got my answer.

I made a mistake on the 2/3 lol. But you still get majority, majority of the time.
 

Snoopy22

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1/2 - 1/2 - 0, and hope randomness works to you advantage.
 

William K

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Ok, I don't get it. Can you explain it so a non-math genius like me can understand? I'm not familiar with probability so I don't see where you're coming from.

I was under the impression that you can divide your troops by

2/3 | 1/2 | 1/6

to get the majority.

Err.. 2/3 + 1/2 + 1/6 > 1...

No Maths scholar, but here's what I remember from my high school lessons :)

Ok, let's assume you spread your army 1/4 : 1/4 : 1/2 is A,B,C

You win A if your opponent's army there is <= 1/4
You win B if your opponent's army there is <= 1/4
You win C if your opponent's army there is <= 1/2

So...the probability that you win A & B is the probability that your opponent's army at A <= 1/4 and B <=1/4

Since your opponent is random, the probability of each combination he chooses is equal...that is the probability of him setting his army 1/4:1/4:1/2 is the same as the probability of 1/3:1/3:1/3. In other words, 1/(total possible combinations). From there you should be able to calculate the total combinations where A <= 1/4 and B <= 1/4. And from there, the probabilities
 

forzen

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Err.. 2/3 + 1/2 + 1/6 > 1...

No Maths scholar, but here's what I remember from my high school lessons :)

Ok, let's assume you spread your army 1/4 : 1/4 : 1/2 is A,B,C

You win A if your opponent's army there is <= 1/4
You win B if your opponent's army there is <= 1/4
You win C if your opponent's army there is <= 1/2

So...the probability that you win A & B is the probability that your opponent's army at A <= 1/4 and B <=1/4

Since your opponent is random, the probability of each combination he chooses is equal...that is the probability of him setting his army 1/4:1/4:1/2 is the same as the probability of 1/3:1/3:1/3. In other words, 1/(total possible combinations). From there you should be able to calculate the total combinations where A <= 1/4 and B <= 1/4. And from there, the probabilities

Haha, yah i saw that after I saw my original answer on top. I was thinking you have to win 2/3 of the time and wrote that instead of 1/3 :doh:. Damn, I don't even remember what I did in math in high school.

Ok I get where your coming from now, the equation kind of confused me. Domo arigato gozaimasu.
 

William K

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Haven't done any calculations, but my gut feel is that 0:1/2:1/2 would give the best results...
 

Arthur Schopenhauer

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Alright, so I've spent some time running this puzzle through my mind, and I've found a definite, 100%, win/win scenario.

Firstly, let's review the errors in logic:

- THE ERROR OF 'NOT REALLY INFINITY BUT STILL KINDA LIKE IT'

You are the general of an army of essentially infinite numbers. The opposing general has an army equally as massive.

a) The first error and most obvious error, is found in the OPs use of the term 'essentially infinite'.

- Infinite numbers cannot be divided, otherwise the number divided will be infinite as well.

- It is impossible to subtract anything from infinity and come to something less than infinity; e.g., if you take one banana away from an infinite number of bananas, the number of remaining bananas is infinite.

>>> Thus, dividing a certain amount of troops to conquer and having all of your troops killed is logically impossible; all possibilities result in an infinite battle. Of course, the OP tried to correct the flaw, but it failed harder than Bill Clinton on his wedding night.

b) It's motherfucking infinite.

- It's infinite, motherfucker.


- THE ERROR OF RANDONIMOTYINOMITY

The opposing general uses no strategy and distributes his troops randomly.

a) It is obvious that you're fighting some imbecile of a warlord, as he randomly sends his troops out to die. (not that their deaths matter, as there is an infinite number) This raises a few questions:

- How did this dumbshit of a warlord come into power?

- How does he keep the economy and people under his control?

- Where do they live, as the universe might not be able to contain an infinte amount of people.

- Etc.

-----

Yet, EVEN IN THE MIDST OF SUCH OVERWHELMING ODDS I've arrived at a DEFINITE CONCLUSION:

bacon-05.jpg


IT'S MOTHERFUCKING DELICIOUS!!!!
 

Provoker

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

Aha! That's a sensible lad. Now, might I bring our attention to something much more meaningful to the mathematics, philosophy, and physics communities? This is the disjunction that infinity and quantitative distinction cannot be postulated to both exist at the same time in any objective sense; for the objective validity of the one would negate that of the other and vice versa. The sequence of reasoning is something like the following.

If the universe is infinitely expanding, then dividing it into quantitatively distinct parts is illusory for the same reason it makes little sense to divide troops if they are infinite. Since such an endeavor would preclude quanta from being discovered and distinguisheable, it follows that quanta is the necessary creation of a limited earthbound mind, which breaks the world up into discrete parts for convenience, but is not an accurate representation of the world as it is. If, on the other hand, quantitative distinctions of the universe are objective, then this necessitates that the universe is finite rather than infinite; for only in a finite world could quantity take on meaning. In that case, it is infinity that is necessarily created by the mind.

It is no small coincidence that maintaining that the universe is infinite lends to the Copenhagen Interpretation of quantum mechanics and its intense focus on the observer--i.e. creator. For quantum mechanics without the observer means the universe is not infinite and that would contradict many of the long-esteemed beliefs of our pop scientists.

Now, should quantum physicists get out of their dogmatic slumber and one day get it right by embracing Karl Popper's notion of quantum mechanics without the observer, and stop misconstruing Heisenberg's principle as precluding the possibility of measuring movement and position rather than as a statistical scatter, this, it seems, would not be entirely without consequence.
 
R

ReflecTcelfeR

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Throwing the first fight seems logical and then dividing into half your armies for the last two, because even if the opponent throws one opponent into the first fight you automatically win, but this is only strategically speaking, in other words... I agree with others before this post.
 

forzen

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Infinity divided by infinity would equal to one, oh snap! Besides the concept of infinity in the problem is an inference which successfully caused MM's mind to asplode. We know that it's stating people and there is a limited amount of people in the world. So all we have to do is divide the total amount of people in the world in half (if it's odd, that unlucky soul would be first blood and would be used to rationalize the war) and use that number to compute the data. Unfortunately, that sounds like too much work. But, if you treat this problem like an abstract question, infinity would own your ass.
 

forzen

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Infinity works a little bit different from numbers such as 3 or 4.

Check this out: WikiAnswers - Does infinity divided by infinity equal

Yes, I know what infinity stands for, obviously if you have infinite amount of people you would not even have a battle field to even work with. Infinity is an abstract idea in math. Trying to intergrate it in real life concept such as the probability question in this thread would give you no solution. However, it is used quite often in the concept of limits as a way to replace infinitesimal.

It is also the reason why I wonder how DC was able to reduce an infinite amount of universes in "Crisis on Infinite Earths" storyline to just 52 universes.

"You are the general of an army of essentially infinite numbers. The opposing general has an army equally as massive."

"Since infinity is not a set number, you cannot assume that infinity divided by infinity would equal one. Infinity could be 1 or it could be 9 trillion, the number is not set in stone."

:cheese:
 
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