12-10-2007, 05:24 PM | #71 | |
That One Dude, you know him, right?
Join Date: Jan 2006
Location: Florida
Posts: 20
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Quote:
Unvote Vote: Masked Jedi
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"Isn't sanity really just a one trick pony anyway? I mean all of you get is one trick, rational thinking, but when you're good and crazy, oooh oooh oooh, the sky is the limit!" - The Tick |
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12-10-2007, 05:26 PM | #72 |
Lakitu
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Vote: Sithdarth
Why? Because there's a 100% chance that he's playing in this game. |
12-10-2007, 05:27 PM | #73 |
Can't spell.... or count
Join Date: Mar 2007
Posts: 27
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7 pages before I see the game has started?!
Anyway I agree with Ole Dude here Vote: Masked Jedi. |
12-10-2007, 05:57 PM | #74 |
Finishing the fight
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Vote: BHS
'cos he's a syc... syc... syco.... fancy word !
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New signature coming soon. |
12-10-2007, 06:14 PM | #75 |
Behold the Manatee!
Join Date: Sep 2007
Posts: 1
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Eight pages? Wow I'm late. I will
FoS: Sithdarth because not only did he use math, but he used bad math. Assuming roles were randomly assigned and having no knowledge of any role but your own, ODWTF has a same probability of having a scum role as anyone else. No, I have nothing better to go on. |
12-10-2007, 06:23 PM | #76 |
Rocky Wrench
Join Date: Apr 2005
Posts: 1,351
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Vote: Lunaknight
Do I really need a reason? I mean, really? |
12-10-2007, 06:49 PM | #77 |
Sent to the cornfield
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Vote: Newb
He's wearing a barrel. I ask Newb, why do you need such a copious amount of leg storage space? Why are pockets not enough? You have a lot of something to store obviously. Evil perhaps? |
12-10-2007, 07:11 PM | #78 | ||
Friendly Neighborhood Quantum Hobo
Join Date: Mar 2004
Location: Outside the M-brane look'n in
Posts: 5,403
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Quote:
In this case we have P(x) = (15Cx * 35C(n-x))/ 50Cn (those are supposed to be the symbol for combination) x = number of successes n = number of trials If we take 3 trials then n=3 and set x=1 for only one success then we get P=.45535714285714 setting x=2 P=.1875 setting x=3 P=.02321428571486 setting x=0 P=.33392857142857 Add em all up and you get one. That means there is a roughly 66% chance of one or more mafiates being in a random sample of three. (P at x=1,2,3 added together) In this case I'm more interested in the case of only one mafiate in any random set of 3. This is because we have the possibility of information. That is if we can somehow eliminate 2 out of the 3 random choices as not being mafiates then the remaining person has an approximately 46% chance of being a mafiate. Now initially with just a straight random choice you have a 15/50 chance of hitting a mafiate or 30%. By choosing 3 at random and eliminating two as non-mafia you can apply conditional probability. That is: P(B|A) = P(B)*P(A|B)/(P(B)*P(A|B)+P(C)P(A|C)) A = 2 known townies B = last person is mafia C = last person isn't mafia P(B) = .5 P(C) = .5 P(A|B) = .455 P(A|C) = .334 P(B|A) = (.5*.455)/(.5*.455+.5*.334) = .577 Just as a note one random choice gives you about a 15/50 chance (30%) of hitting a mafiate. With a random sample of 3 and information about 2 I've increased my chances of having found a mafiate by 28%. Now if we only know the role of one of the three: A = 1 known townie B = at least one of last two is mafia C = none of the last two are mafia P(B) = 2/3 P(C) = 1/3 P(A|B) = .643 P(A|C) = .334 P(B|A) = (2/3*.643)/(2/3*.643+1/3*.334) = .794 That is there is a roughly 80% chance that one of those two is a mafiate. (This is a result I totally didn't expect myself. I love math sometimes.) Though I shouldn't be surprised as this is very close the classic three door scenario where when one door is eliminated you up your chances of winning by changing doors. Its a perfectly valid application of probability distributions that happens everyday as part of quality control in factories. Its just in this case I've replaced defective parts with mafia members. Just because things are random does not mean you can't use probability and information theory to draw conclusions. In fact, it is because things are random that you can apply these theories. (That is unless Fenris used some non-random method of assigning roles.) Quote:
In other words since I have to guess randomly I want to make sure any information gleaned from that guess is as useful as possible. |
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12-10-2007, 07:24 PM | #79 | |
That One Dude, you know him, right?
Join Date: Jan 2006
Location: Florida
Posts: 20
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Quote:
Of course I could throw a bunch of wrenches into your math like how do you know exactly how many mafiates there are in each family? After all, who except a mafiate knows exactly how many mafiates are in each family. With a game this size it isn't unreasonable to think there may not be the normal 5 mafiates per family. But then, there might be. Who knows? I don't. But, yeah what I'm getting to is that I think your math is wrong, and that you state by your own words that it isn't infallible.
__________________
"Isn't sanity really just a one trick pony anyway? I mean all of you get is one trick, rational thinking, but when you're good and crazy, oooh oooh oooh, the sky is the limit!" - The Tick |
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12-10-2007, 07:31 PM | #80 |
Cat
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Vote: Sithdarth
For the tl;dr |
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