Posted September 27, 2019

dedoporno
A bloody pirate!
Rep: 1520
Registered: Apr 2012
From Bulgaria

Microfish_1
I love gold!
Rep: 673
Registered: Dec 2017
From United States
Posted September 28, 2019
stuff changed, i can take part late sunday

GamezRanker
*Honor the fallen in kind words and deeds*
Rep: -2374
Registered: Sep 2010
From United States
Posted September 28, 2019

Also an aside: I love that game series....nice pic....gonna have to swipe a copy.

Addition: I also have the 3d glasses for the one game.....fun times.
=============
You sure? If you don't mind and i'm not imposing(and OP is okay with it) that seems ok.
Post edited September 28, 2019 by GameRager

dedoporno
A bloody pirate!
Rep: 1520
Registered: Apr 2012
From Bulgaria

GamezRanker
*Honor the fallen in kind words and deeds*
Rep: -2374
Registered: Sep 2010
From United States

Microfish_1
I love gold!
Rep: 673
Registered: Dec 2017
From United States
Posted September 28, 2019
One more thing. Before we start, I want a room with a view.

ZFR
I love gold!
Rep: 5069
Registered: Jan 2010
From Ireland
Posted September 28, 2019

The unbeatable duo want to determine whom to send to do the NK. Problem is, *someone* doesn't want to use random.org; he loves rolling dice instead. He says that if he rolls an even number then the prettier half of the duo does the NK, and if he rolls an odd number then the smarter half will do it. Problem is, the only dice he has is clearly unbalanced: some numbers are more likely to come than others. Fortunately, the smarter half of the unbeatable duo devices a clever method of using the unbalanced dice to randomly pick one of them with a 50% probability. How?
(This actually has a proper mathematical solution).
You roll twice. If the parity is same, discard and roll twice again. Otherwise use the parity of first roll to determine the outcome.
So
even, even - discard, reroll twice
odd, odd - discard, reroll twice
even, odd - ZFR
odd, even - Joe
Since P(even, odd) is equal to P(odd, even), regardless of P(even) and P(odd), both players are picked with an equal 50% chance.
This works for any situation where a dice/coin/roulette_wheel needen't be fair. You can pick any two events, even if their probability is radically disproportional e.g. dice rolls 6 vs dice rolls non-6, just as long as P(x,y) is greater than 0 and both rolls are independent. However the more disproportional they are, the more likely you'll need to discard and reroll.

JoeSapphire
Consultant Liar
Rep: 1290
Registered: Jun 2011
From United Kingdom
Posted September 28, 2019

The unbeatable duo want to determine whom to send to do the NK. Problem is, *someone* doesn't want to use random.org; he loves rolling dice instead. He says that if he rolls an even number then the prettier half of the duo does the NK, and if he rolls an odd number then the smarter half will do it. Problem is, the only dice he has is clearly unbalanced: some numbers are more likely to come than others. Fortunately, the smarter half of the unbeatable duo devices a clever method of using the unbalanced dice to randomly pick one of them with a 50% probability. How?
(This actually has a proper mathematical solution).


ZFR
I love gold!
Rep: 5069
Registered: Jan 2010
From Ireland
Posted September 28, 2019


P(b,w) = (b/n) * (w/(n-1)) = (bw) / n(n-1)
P(w,b) = (w/n) * (b/ (n-1)) = (bw) / n(n-1)

GamezRanker
*Honor the fallen in kind words and deeds*
Rep: -2374
Registered: Sep 2010
From United States
Posted September 29, 2019


P(b,w) = (b/n) * (w/(n-1)) = (bw) / n(n-1)
P(w,b) = (w/n) * (b/ (n-1)) = (bw) / n(n-1)

ZFR
I love gold!
Rep: 5069
Registered: Jan 2010
From Ireland
Posted September 29, 2019

There is an event with a probability p. We need to find on average how many times we need to roll to get p.
Assume that the average is x.
Now, there is a probability p of getting it on first try. And a probability of not getting it on first try is (1-p).
If we don't get it on the first try, then we used up one try but, since x is the answer, then on average it takes x further tries to get it.
Taking that into account, we got a simple equation to solve.
x = (p * 1) + ((1-p) * (1+x))
Which we can solve to get
x = 1/p
In Joe's case, such an average does exist because P(infinity) = 0 (the probability that Joe will keep rolling forever is equal to 0). Proof? That's for you to find out.
However consider this game: You roll a d6 dice. If you roll 1, you win. If not, you roll again, only this time the number of dice faces is doubled. And so on. You roll d12 a second time, d24 the third time, d48 the fourth time... etc. You keep rolling until you roll 1.
In this case P(infinity) > 0. For bonus points prove this and calculate P(infinity); it's actually not that difficult.
And because of that, the average number of rolls doesn't exist. Or you could say the average number of rolls is "infinite", but that's not strictly true, because there is no number called "infinity", so the average can't be equal to that.
Anyway, I'll stop now.
About that mafia game, Joe?

Lifthrasil
Bring the GOG-Downloader back!
Rep: 3505
Registered: Apr 2011
From Germany
Posted September 29, 2019
Yea. Let's start! I want to lynch someone!

my name is sadde catte
i touch your heart.
Rep: 3818
Registered: Mar 2010
From United Kingdom
Posted September 29, 2019
Vote someone

trentonlf
Easily amused
Rep: 3220
Registered: Apr 2014
From United States

Lifthrasil
Bring the GOG-Downloader back!
Rep: 3505
Registered: Apr 2011
From Germany