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I am not exactly sure if I understand this. But if you're right I am really wondering why the host of a newbie game would give an advantage about setup knowledge to experienced players.
Is it true that some setups such as cop+doctor are not possible?
Actually that would make things doable for town. If Moosy gets lynched, I will be shot at night. If I get lynched and flip, Moosy can still be lynched next day. Still tough game to win but doable.
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Also Superbia, I AM cc'ing moosy. If what you say is correct, my corresponding town power role should claim. It's a shame you're believing one claim but dismiss the other claim. There was no way I as doc would claim today, which is obviously why scum claimed cop. Scum can easily exchange one member vs cop or doc and you are currently playing exactly to scum likes.
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Meh I should've been patient and observed who backed who.
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The only one who can CC moosy is vigi or cop.
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The world you're selling is impossible with moosy being the town cop.
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On June 10 2016 16:35 Superbia wrote: The world you're selling is impossible with moosy being the town cop. You're not reading properly. I am doc. If what you say is right then Moosy is scum because I am doc. I hope the host will clarify the issue soon. Read this post:
On June 10 2016 16:22 beentheredonethat wrote: I am not exactly sure if I understand this. But if you're right I am really wondering why the host of a newbie game would give an advantage about setup knowledge to experienced players.
Is it true that some setups such as cop+doctor are not possible?
Actually that would make things doable for town. If Moosy gets lynched, I will be shot at night. If I get lynched and flip, Moosy can still be lynched next day. Still tough game to win but doable. It outlines my thought about the situation with Moosy being a scum fake claimer.
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On June 10 2016 16:18 Superbia wrote: I'm going back to work. IF someone does not understand the logic here I can explain it later. Don't be lazy though. Read my posts thoroughly and try to understand it first. I need help with your dice rolling interpretation. Between the two of us, either you presented the information incorrectly or I misunderstood it in our common language or I simply don't get the semantics. I don't know what dice rolling up is. I don't get your breakdown of tables and what each number value signifies for each of the six dice. Perhaps I misunderstood the OP, but in order for me to trust your absolute logic on the fact that BTDT is scum and Moosy is not, you will have to explain with more clarity, "Explain Like I'm 5," if you would.
As of now, it seems foolhardy to trust your word without understanding your rationale, because if you and Moosy are in cohorts as scum, then town is doomed. It is an all-or-nothing situation, and as such I feel that if you're town you owe the newbs and town an explanation that is more coherent.
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If Superbia's logic checks out, then BTDT fucked up real bad. If it doesn't, then Superbia fucked up real bad, trying to convince newbies. The former seems more likely, but I don't know enough to lean either way until I get a concrete answer on the matter.
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Claiming doc is no fuck up. Moosys claim is not genuine if Superbia is right. And it is a wrong call due to me being framed if Superbia is wrong. I hope Superbia is right because if he is not and Moosy really is cop, then town loses both power roles within one cycle and the game is lost.
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Wrong call -> wrong claim. Mobile again, autocorrection.
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On June 10 2016 09:02 MoosyDoosy wrote: Vote beentheredonethat for now to put the pressure on him and the scum team to respond in some way. You can always change your votes later anyway.
Why would you need to bother putting pressure on him if you redcheck'ed him?
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On June 10 2016 17:18 beentheredonethat wrote: Claiming doc is no fuck up. Moosys claim is not genuine if Superbia is right. And it is a wrong call due to me being framed if Superbia is wrong. I hope Superbia is right because if he is not and Moosy really is cop, then town loses both power roles within one cycle and the game is lost. I'm starting to think that I'm wholly retarded, because I don't follow your logic either. Can you explain it in greater detail, more inductive reasoning steps in between?
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On June 10 2016 15:53 beentheredonethat wrote: Just catching up. Posting from mobile since Moosy just claimed. Since there was no cc just yet, I think he is either genuine or no cop is in. I am not scum and I really want to point out that there is a framer in. I am the doc. I was roleblocked night 1, trying to save tumble, and I saved Jealous last night.
Given that no vig kill on Emperor happened, I think we do not have a vig in which makes Moosys cmaim stronger. Only thing is that his claim forced me to claim myself. We will be fucked if we lynch myself for Moosys framed check. Moosy will be killed at night then and if I count correctly, we lost. I posted extremely low volume all the time to not get on scum radar which so far worked.
So long story short: Moosy is the cop. There is a framer in. I am doctor.
Wow... hmmm I'll have to re-read both of your motives.
I'm pretty confident one of you are bullshitting - a frame landing has a rare chance of working.
It even has a rarer chance of being on a non-wagon.
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On June 10 2016 17:24 scott31337 wrote:Show nested quote +On June 10 2016 09:02 MoosyDoosy wrote: Vote beentheredonethat for now to put the pressure on him and the scum team to respond in some way. You can always change your votes later anyway. Why would you need to bother putting pressure on him if you redcheck'ed him? I will venture an answer here for the sake of continuing your involvement: if Moosy is in indeed cop, then his role is only absolutely valid to himself; putting pressure on BTDT could force a slip-up from scum?
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Okay.
3 tables: - Each start at 0. - Each time a die is rolled, one of the tables is incremented by 1 according to the die result. - When a table reaches 3 and that table is incremented, it goes back to 0.
So tables go 0->1->2->3->0. Or we can say EVEN->UNEVEN->EVEN->UNEVEN->EVEN->...
This is part 1.
Now we tie each table's value to their respective role: Table T1: 0: GF, 1 framer, 2 GF, ...etc. OR:
T1: Even: GF, Uneven: Framer T2: Even: Vigi, Uneven: Cop T3: Even: Vet, Uneven: Doc
This is part 2.
Now when we divide 6 increments over these tables, everything being uneven is impossible. Why? Two answers:
Answer 1 is expanded: In order to reach everything as uneven, we need to increment all 3 tables by 1, which leaves us with 3 increments (dice rolls) left. But how do we divide the remaining 3? If we increase a table by 1, it reaches even. If we increase a table by 2, we are left with 1. Finally, if we increase a table with 3 it becomes even.
As such, it is impossible to reach this result.
Answer 2 is reversed. Mathematically we know that an uneven number of elements of all uneven numbers results in an uneven number when added up (try it out). The same way that an even number of uneven numbers always results in an even number when added up. As such, The addition of 3 (number of tables) elements all being uneven can never be 6 (an even number).
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On June 10 2016 17:33 Superbia wrote: Okay.
3 tables: - Each start at 0. - Each time a die is rolled, one of the tables is incremented by 1 according to the die result. - When a table reaches 3 and that table is incremented, it goes back to 0.
So tables go 0->1->2->3->0. Or we can say EVEN->UNEVEN->EVEN->UNEVEN->EVEN->...
This is part 1.
Now we tie each table's value to their respective role: Table T1: 0: GF, 1 framer, 2 GF, ...etc. OR:
T1: Even: GF, Uneven: Framer T2: Even: Vigi, Uneven: Cop T3: Even: Vet, Uneven: Doc
This is part 2.
Now when we divide 6 increments over these tables, everything being uneven is impossible. Why? Two answers:
Answer 1 is expanded: In order to reach everything as uneven, we need to increment all 3 tables by 1, which leaves us with 3 increments (dice rolls) left. But how do we divide the remaining 3? If we increase a table by 1, it reaches even. If we increase a table by 2, we are left with 1. Finally, if we increase a table with 3 it becomes even.
As such, it is impossible to reach this result.
Answer 2 is reversed. Mathematically we know that an uneven number of elements of all uneven numbers results in an uneven number when added up (try it out). The same way that an even number of uneven numbers always results in an even number when added up. As such, The addition of 3 (number of tables) elements all being uneven can never be 6 (an even number). Correct me if I'm wrong, but if there are 6 dice rolls for 3 tables, could the results not be 1+6, 2+5, 3+4, all of which are uneven and equal 7?
Am I totally braindead about this system???
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Also the result of the dice merely affects what table gets incremented. The increment is always 1.
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Imagine having 6 poker chips and trying to divide them into 3 uneven stacks
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I don't see the even/odd number thing making sense - but i'm on a few brews so my thinking could be off.
from the OP - There are two blues and says ONLY two blues, so that would assume there are two blues ONLY - not one or zero.
only one can be cop or vig
only one can be doc or vet
Am I correct?
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