# Thread: How do you think through this?

1. Originally Posted by tcda
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.

2. Originally Posted by 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)
Okay, now you're starting to make sense. THIS can be worked with.

3. 1/3,1/2,1/6

4. 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.

5. Originally Posted by MagnificentMind
Whoever posted this:

The answer to that is no.

6. 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

7. Originally Posted by William K
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.

8. 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.

9. Originally Posted by Little_Sticks
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.

10. 1/2 - 1/2 - 0, and hope randomness works to you advantage.

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