Originally Posted by
Catta
But doesn't the math say that the chance of being scum twice in a row is stastically smaller?
You usually always adhere to math.
Danner totes scum guys, we should kill Danner.
Your first conclusion might not be wrong, but if right you get to the right answer for all the wrong reasons. Like I said to Largehorn last game, that's the best kind of right. But I would have failed you if this was Statistics 101. Time to fix your critical statistics errors. You get remedial classes!
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Let's start with a condition.
Condition A: Graeham is scum in a given game.
We know that less than all the players in the game are scum. Chances of graeham being scum in a given game, P(A) is < 100%.
P(A) < 1
From this, you can prove that the chance of being scum twice in a row is smaller than being scum once:
P(A) * P(A) < P(A)
So far, so good. This is your argument, I believe.
However, we know for certain that Graeham was scum last game. This is at this point a historic fact.
Condition B: Graeham was scum in the previous game.
P(B) = 1
If we make a new formula with this knowledge, it should looks like this:
P(A) * P(B) = P(A) * 1 = P(A)
Meaning, the probability of Graeham being scum this game is completely independent of the previous game, and given by P(A) alone.
But that formula contains a major error.
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We really want to calculate P(A | B). That is, the chance of graeham being scum in a game (this game), given that he was scum in the previous game. You can read the | character as "given". P(A|B) is "the probability of A, given B".
Per the Kolmogorov definition, this mouthpiece can be calculated as follows:
P(A|B) = P(A ∩ B) / P(B)
So we just got a more complex defintion. Stupid mathematicians never make this easy. P(A ∩ B) is a fancy way of saying "the probability of both A and B happening". For statistics, that means how much these two conditions depend on each other. If these two conditions - A and B - are completely statistically independent, then
P(A ∩ B) = P(A)*P(B)
. Otherwise, things get complicated.
But, if A and B are unrelated, then the answer looks like this, which matches my formula above:
P(A|B) = P(A ∩ B) / P(B) = P(A)*P(B)/P(B) = P(A)
I could choose the quick way out here, and claim that rolecard assignment is completely random, so these probabilities need to be unrelated, and thus P(A | B) = P(A)*P(B) = P(A). However, what if that is not the case?! What if Kryllian is secretly playing favours (or grudges) against Graeham, giving him a specific role because because he was scum in the last game? It is not impossible! And since rolecard assignment is a zero-sum game, if Kryllian had a favour (or grudge) against any other player, that too would affect the statistical rolecard probability for Graeham. In fact, it seems to my subjective opinion very unlikely that A and B are statistically unrelated, whether intentionally or subconsciously.
Taking this argument to the extreme, we end up at this situation. It makes it very hard to calculate P(A|B) exactly.
But let's point out where you are wrong: Say hello to Bayesian Probability.
Allow me to introduce you to my friend, Thomas Bayes. His theorem formed the foundation of probability mathematics as we know it. It's used for a lot of things, but the most down-to-earth usecase is showing the gamblers fallacy.
What is the probability of getting heads on a standard coinflip?
Answer: 1/2. Duh.
What is the probability of getting heads on a standard coinflip, twice in a row?
Answer: 1/4
What is the probability of getting heads on a standard coinflip, given that your last coinflip came up head?
Answer: 1/2.
Basically: the gamblers fallacy claims things of the past doesn't matter for future probabilities.
The gamblers fallacy is actually derived from Bayes theorem. Which looks like this:
P(A|B) = P(B|A)*P(A) / P(B)
In our case, we know for a fact that P(B) = 1. And since P(B) =1, we also knows that P(B | A) = 1. In fact, P(B | anything) = 1, since nothing will change the fact that B happened. Meaning we can do this substitution into that theorem.
P(A|B) = P(B|A)*P(A) / P(B)
insert for P(B)=1 and P(B|A) = 1
P(A|B) = 1*P(A) / 1 = P(A)
This is basically the same answer as the gamblers fallacy; the probability of an event, given something that already happened, is the probability of that event alone.
However, I just cheated. I abused Bayes theorem in ways it should not be used when I applied this to Graeham. Bayes Theorem really only works with statistically independent operations. Like a coin toss. A coin toss is independent of previous cointoss; the coin doesn't get more likely to throw either heads nor tails after throwing heads. However, I earlier made a pretty good argument, that mafia rolecard assignments are not 100% independent of previous games. That means bayes theorem is useless!
Or rather, not useless, but not perfectly accurate!
(If it was accurate, I just would have proved that the two operations are not unrelated. And that's a pretty major leap!)
Nonetheless, if we assume that Kryllian made a reasonable random rolecard distribution, then Graeham's rolecard should be reasonably random. Meaning - Bayes Theorem isn't far off from giving the exact answer.
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Without knowing all parameters surrounding how Kryllian distributed the rolecards, consciously and subconsciously, we have no way to accurately describe just how much A and B are related. It literally depends on Kryllian. I find it subjectively more plausible if A is not completely independent of B. I do not know if this is in a positive or negative direction as to determining the probability of A. However, the critical answer is: if Kryllian did a rolecard distribution that mostly random, then A is mostly independent of B - and Bayes Theorem is mostly correct.
In your opening statement, I believe you fell in the gamblers fallacy. This is logical wrongthink, and you should feel bad.
Punishment: write 100 times on the school board that you will not commit the gamblers fallacy.
However, you aren't completely wrong in claiming that the probability is smaller. You have somewhat under 50% chance of being right, I reckon. It could easily also be larger. But let's be real here. We are subjectively likely WAY WAY closer to the answer being P(A), than the answer being P(A)*P(A). And I believe you just claimed that. Unless Kryllian is cheating, you're wrong. I just can't give you the accurate probability of you being wrong.
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