Oh

DANGIT DANGIT DANGIT DANGIT TURN OFF MATH BRAIN TURN OFF!

*is now being forced to try to figure out if there is a way to make 2 of the same dice, with any number of sides and any numbers on the sides, such that every number has an equal chance of being rolled*

WHY MUST YOU TORMENT ME SO!!!

*The longest night in wolves*

Edited 11/12/2014 08:05:53

Pulsey, Frank being a wolf proves my statement, you clearly didn't read my response thoroughly enough.

Two wolves in one faction is fishy enough, so if Frank is in fact a wolf, you can rest relatively assured that I'm not one.

Quoting yourself is always weird, but there it is. Lord V is being manipulated by a false seer. At least we know one of his is real though...

My theory is there was a wolf for each faction, this is Doms first experiment with factions so he would have made it as balanced as possible. With a muse copying a wolf we go from 4 to 3 back to 4. I know not everyone agrees with my numbers, but that's how I'm making my calls.

Frank could have been a lucky guess as well, Lord V pulling an Arun on us, who knows? We have to trust a former wolf who's good at throwing red herrings...

Oh and I definitely am not/was not the muse.

>Tfw false seer could be a wolf who threw Frank under the bus in order to gain trust

From what Dom said there must be two seers. If that is the amount of seers claiming to be seers, we don't have a problem and I will go vote for all seered wolves until one turns out to be a villager (which I don't expect).

No, Taisho, Dom's first post says that there can be as many wolves in a cell as he likes. When he makes a deliberate effect to state such a rule, it seems unlikely he'll put at most 1 wolf in every cell in that case, not to mention that you have absolutely no proof to know theres only 1 wolf in every cell, apart from
just gut feeling.

I don't think it quite matters. You've been called you by someone who has so far given the village no reason to suspect his credibility, and like i said, even if you are a failed lynched, we then get the name of a false seer. In the interest of winning, you should be lynched.

Same for Lawlz. Even if Veigar's seer ends up to be fake, worse he can do is cause the death of one innocent villager, but by doing so he's exposed himself and can be lynched immediately.

Edited 11/12/2014 11:08:44

Pulsey. Do you want to play the next game? If so refrain from posting in this one and prove you can be trusted to play in the spirit of the game. Thanks.

Im not sure Veigar even has a seer. I am betting he is one.

Pulsey - please stop trying to influence the game too much, as you did choose not to sign-up..


Sorry for slow update guys - I lost my father over the weekend so trying to play catch-up at home - I'll collect the info from everyone today and get an update done this afternoon.

Wo jeez dom. Forget the game.

Dude, forget this game. Take all the time you need...

Jesus, my condolences Dom. I'm sorry about that

Can we put this game on hiatus?

@ Dom365: my condolences in this sad and difficult moments.

@ Traintown

*is now being forced to try to figure out if there is a way to make 2 of the same dice, with any number of sides and any numbers on the sides, such that every number has an equal chance of being rolled*

For finite dice with real numbers, this is only possible if all numbers on the sides are equal (no matter how many sides there are).

Proof: suppose there are at least 2 sides with at least two different numbers. Then there must be a maximum, say "A". Define m as the number of sides that are "A". Then there are at least m*m possibilities to roll A+A and, since A is the maximum, no other possibilities exist to roll A+A, so there are exactly m*m. Similarly, there must be a minimum B with n sides that are "B". Following the same reasoning, there are exactly n*n possibilities to roll B+B. Thus, we must have m=n or the proposition can't be true.
Now we know that A+B, which is not equal to A+A, nor to B+B can be rolled at least 2*m*n times, which is larger than m*m=n*n=m*n.
Consequently, it is not possible for finite dice that have different numbers on the sides.

* For unfair dice, the same reasoning can be follewed as the "number of sides" could simply be replaced by the probability of rolling a certain number, which can always be defined.

* For non-real "numbers", this also holds when a complete order can be defined. When there is no complete order, this is possible. Counterexample: 2 fair sides "yes" & "no" where no+no=yes (double negative). P(sum=yes) = P(D1=yes)*P(D2=yes)+P(D1=no)*P(D2=no) = 0.5 = P(D1=yes)*P(D2=no)+P(D1=no)*P(D2=yes) = P(sum=no)

Wait with just a glance at that, what about complex numbers? That's what I'm looking at right now...

My deepest condolences to you and your family, DOM..

I proposed to all, that we stop this game for the while until our friend DOM is ready to come back. Our most and deepest respects goes to DOM.

In this difficult moment, DOM needs our support, even if we are located in the most remote corners of the planet. DOM has been a good friend and he deserves that we treat him equally fairly.

DO you agreed guys?

Edited 11/12/2014 14:43:05

Yes absolutely. We should put this game on hiatus.

My heart goes out to you Dom. Best of wishes to you and your family in this difficult time.

I agree too. All the best to you and your family, Dom!

What shall we do?