"People in glass houses shouldn't use Windex when living near bird sanctuaries."- myself
"We are never alone my friend. We are constantly in the company of victories, losses, strengths and weaknesses. Make no mistake, life is war...and war is hell. Those who fight the hardest will suffer the most...but that's what you have to do: Fight. As long as you're feeling pain, then there's hope...because only the dead do not suffer." -RD Metcalf
1/3,1/2,1/6
This post grammatical errors had been intentionally left uncorrected.
Whoever posted this:
The answer to that is no.Here is the message that has just been posted:
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I think sending 1/2 to two of the fronts and leaving the remaining front alone is the most effective strategy.
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INTJ | 5w4 - Sp/Sx/So | 5-4-(9/1) | RLoEI | Melancholic-Choleric | Johari & Nohari
This will not end well...
But it will at least be poetic, I suppose...
Hmm... But what if it does end well?
Then I suppose it will be a different sort of poetry, a preferable sort...
A sort I could become accustomed to...
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
4w5, Fi>Ne>Ti>Si>Ni>Fe>Te>Se, sp > so > sx
appreciates being appreciated, conflicted over conflicts, afraid of being afraid, bad at being bad, predictably unpredictable, consistently inconsistent, remarkably unremarkable...
I may not agree with what you are feeling, but I will defend to death your right to have a good cry over it
The whole problem with the world is that fools & fanatics are always so certain of themselves, and wiser people so full of doubts. ~ Bertrand Russell
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.
gone
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.
This post grammatical errors had been intentionally left uncorrected.