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Post by GrannyCarol on Mar 15, 2005 1:44:22 GMT -5
Also, when the warrior goes from regular to vet, doesn't he gain back a hitpoint? So, its possible he lost two hitpoints during the battle - a bit more likely also. I hit streaks like that in single player games - they either suck or they rock, depending which side I am on!
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Post by Don K Hotay on Mar 15, 2005 13:39:13 GMT -5
.... if the archer lost 1 hp, then it would be (.645)^2 * (.355), etc... Actually, that's clearly incorrect because it basically says The archer lost one HP and the warrior lost two HP. For a battle to be won, the loser has to lose all of it's HP . You have to know two things when calculating the probablities: 1.) Chance the defender has lose an hp or the chance the attacker has of losing an hp. Knowing one will imply the other, so in the case of Avo's unlucky archer it had a 2/3.1 chance of taking an hp away from the fortified regular warrior. 2.) The number of hit points the defender and attacker have. This is important because the HPs are where the attack and defense modifiers are being applied. So, Avo first asked what is the probablity of a fortified regular warrior not losing a single HP and defeating a regular archer. We answered that with: (1.1/3.1)^3, which is ~ 4.5%, which is correct. But that's just one way the warrior could've won the battle. It could also have won the battle by losing 1HP, or losing 2HP (obviously if it lost 3HP, then it has lost the battle). So in terms of Hit Points, here are the possibilities relative to the warrior: 0HP lost: WWW 1HP lost: LWWW, WLWW, WWLW 2HP lost: LLWWW, LWLWW, LWWLW, WLLWW, WLWLW, WWLLW So, knowing that the warrior and archer each have three hitpoints, we know that there can be at most 5 HP's between the attacker and defender (the loser will lose 3 and the winner will lose at most 2). The probality of a unit winning or losing a battle, thus, is distributed Negative Binomially, which simply means that there is a formula that can be used to punch in the parameters: P(Winning battle) = [(x-1)!/(r-1)!(x-r)!](p^r)(1-p)^(x-r) -x represents the number of HP that are lost (between the defender and attacker) -r represents the number of HP the loser loses -p represents the chance of the winner keeping an hp So, if we want to know the probality of a fortified regular warrior losing: 0HP: [(3-1)!/(3-1)!(3-3)!](1.1^3)(1- 1.1/3)^0 = .045 1HP: [(4-1)!/(3-1)!(4-3)!](1.1^3)(1- 1.1/3)^1 = .085 2HP: [(5-1)!/(3-1)!(5-3)!](1.1^3)(1- 1.1/3)^2 = .112 Add all three ways for the warrior to win the battle up and you get what Phanman provided, which is from a combat C3C combat calculator I found here, and tells us for any given battle between a regular archer and a fortified regular warrior the overall probality of losing/winning a battle: tokyo.cool.ne.jp/civ3/Civ3ConqCombatCalc.htmSo my question still stands, if all we know is that the warrior has won the battle, what is the probality that it has lost 0HP. And to anyone's next question, yes I am the biggest tool around ;D
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Post by heroray on Mar 15, 2005 14:43:22 GMT -5
Hey Donkey,
Do they have that tool in English?
I cant figure out that Jap wrighting.
Peace Out.
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Post by Don K Hotay on Mar 15, 2005 14:45:34 GMT -5
I googled "C3C combat calculator", and then had google translate the page for me.
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Post by Reptile on Mar 15, 2005 15:41:43 GMT -5
Hmm, so there are japanese people who play Civ, but they havent found the path to mp yet...
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Post by Avogadro on Mar 15, 2005 15:46:48 GMT -5
Untrue, Savant is Japanese
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Post by Reptile on Mar 15, 2005 16:09:54 GMT -5
Interesting, i didnt know that.
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Post by LordPhan on Mar 15, 2005 19:11:08 GMT -5
CivFanatics has the combat calc in english.
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Post by Don K Hotay on Mar 18, 2005 17:21:46 GMT -5
... So my question still stands, if all we know is that the warrior has won the battle, what is the probality that it has lost 0HP. ... No fans of the Reverend Thomas Bayes? The question is asking what is the probality the warrior has lost 0HP given that we know it has won the battle against the archer. The answer: .045/.242 = 18.6% And if all we know is that the warrior has won the battle, the probability that it has lost: 1HP: .085/.242 = 35.1% 2HP: .112/.242 = 46.3% Typical conditional probality/Bayes' Theorem question.
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Post by nuckleralugard on Feb 28, 2006 21:02:48 GMT -5
lol i was savant
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Post by Avogadro on Feb 28, 2006 23:12:00 GMT -5
Yes I know that now, for some reason thought it was the case. Slytime maybe? Whoever had suggested you to LRP had said you were really good but could not speak English, only Japanese. However since then we foudn a huge civ3 mp community in the asian continent. I cant find the link but Yumpak from civ4players belongs to it.
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Post by nuckleralugard on Feb 28, 2006 23:17:47 GMT -5
strange, they dont play on our game server?
thats to bad
that would have been cool
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Post by Avogadro on Mar 1, 2006 14:10:29 GMT -5
yes too bad indeed their ladder had like 5000+ members
On the home page they had simon and garfunkle playing and i couldnt make out any writting. From Yumpak's emails he states that gaming is extrememly popular in korea and the rest of asia. Serioud business. Apperently our ladder, even at its highest point did not compare in numbers.
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Post by nuckleralugard on Mar 1, 2006 14:53:38 GMT -5
y i know they got like olympic games in gaming, its kinda cool
and i think its gona spread, in fact i think the asian gaming industri is gona merge with the west
the future will hold a translotor lobby, so pepole all over the world can speak to each other with there native tung.
its sad that those that make mp dont think about this, and unite the players of the world we have alot to learn from them, and they from us,
like when i see counter strike servers up you see some hosts from the est with China letter or japanes, but surly they must still be a great number of players still playing civ3 over there if they had 5000 members.
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Post by hazard on Mar 26, 2006 18:02:38 GMT -5
There was a whole separate ladder just for people living in Asia? Whoa
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