OK, so my thoughts as of now.
Tamius has basically been called out as a spy by everyone and hasn't resisted that notion credibly. I'm going to mark that one down as definitely true.
Here is what I did to get the procentages I posted:
While doing this I found a mistake and corrected it.²
Also, I forgot to check the probability against the regular probability of M2 to have a spy if we don't restrict the allowed playerpool for it.³
(I'm rounding here for readability, caculating is (mostly) done with raw numbers)
With 8 players there are 56 possible teams of three. We have 5 safe players, there are 10 teams possible of them. So out of 56 possible teams 46 have spies on them, which brings the success rate of M1 to 1 - 56 / 46 = 17,9% (given the spies would always sabotage if they get on the mission)
Now if we look at M2 and would exclude the first three players, we have the following three scenarios, given we know at least one of the three is a spy:
(1 = spy, 0 = resistance)
excluded | left
A) 100 | 11000
B) 110 | 10000
C) 111 | 00000
At first let's look at the possibilities of the exclusions.
A excluded = 3/8 * 5/7 * 4/6 = 17,9%
Multiply this by three, since 010 and 001 are also one-spy combos with the same probability
A excluded = 53,6%
B excluded = 3/8 * 2/7 * 5/6 = 8,9%
Also multiplied by three (101, 011)
B excluded = 26,8%
C excluded = 3/8 * 2/7 * 1/6 = 1,8%
No multiplication, there is only one case
Note that the sum of these is 82,1% the probability of M1 going wrong. Since we know there was at least one spy we need to interpolate these percentages to 100%. ²
82,1 * 1,22 = 100%
Therefore:
A excluded = 53,6 * 1,22 = 65,2%
B excluded = 26,8 * 1,22 = 32,6%
C excluded = 1,8 * 1,22 = 2,2%
Now that we know how likely the scenarios of excluded players are, we need to find out how likely we are to get a clean team in M2 given these restrictions of the player pool.
Pick 4 clean players out of the following left players:
A) 11000
B) 10000
C) 00000
A clean = 3/5 * 2/4 * 1/3 * 0/2 = 0%
B clean = 4/5 * 3/4 * 2/3 * 1/2 = 20%
C clean = 5/5 * 4/4 * 3/3 * 2/2 = 100%
The overall probability of a clean M2 team after excluding the M1 team from proposals will be the sum of the products of the three scenarios:
M2 clean = 0% * 65,2% + 20% * 32,6% + 100% * 2,2% = 8,7%
Regular M2 probability of a clean mission: ³
70 possible teams of 4 out of 8.
5 possible clean teams of 4 out of 5.
70 / 5 = 7,1%
Conclusion:
By excluding all three M1 players the possiblity of a clean M2 is higher by 22,5%.
Also, I should make a spreadsheet. This is way to hard to follow.
€dit: This is different from the point of view of a Resistance member that was part of M1. He can exclude the C scenario, so the odds of a clean M2 from his point of view fall down to 5,4%, which is lower by 23,9% than the regular probability of a clean M2.
Huh. So from the perspective of someone off the failed mission, the probability of there having been 2 spies on the failed mission is high enough that it actually justifies leaving everyone who was on it off the next mission? Because that feels really counterintuitive, though I see nothing blatantly wrong with your math.
Of course it's making a flawed assumption of probabilities being the driving factor, without accounting for player designs. Spies aren't likely to want a team with two spies on it, so it is even less likely than probability suggests that mission 1 would have multiple spies.Huh. So from the perspective of someone off the failed mission, the probability of there having been 2 spies on the failed mission is high enough that it actually justifies leaving everyone who was on it off the next mission? Because that feels really counterintuitive, though I see nothing blatantly wrong with your math.
That's assuming that every leader picks randomly - which is true enough for Resistance leaders, but spies are more likely to pick Resistance on their teams (or at least, I think they do), which would push up the probability of one spy on the mission.
Of course it's making a flawed assumption of probabilities being the driving factor, without accounting for player designs. Spies aren't likely to want a team with two spies on it, so it is even less likely than probability suggests that mission 1 would have multiple spies.
I don't have much of a counter. If I was a spy, I'd have done the same, put a well known first mission sabotager on the team and figure Tamius was sure to use his KACEOY. Obviously I'm not a spy, but nothing to prove or support that at this point, though the same is true of anyone else.I know I always used to assume that 1 spy on a sabotaged team was by far the norm. But with the spy coordination being as developed as it is now, I'm not quite ready to completely discount that you two aren't both spies.
I don't have much of a counter. If I was a spy, I'd have done the same, put a well known first mission sabotager on the team and figure Tamius was sure to use his KACEOY. Obviously I'm not a spy, but nothing to prove or support that at this point, though the same is true of anyone else.
You proposed a sabotaged mission where the guy with KACEOY didn't use his card. I'm not saying your as implicated by that as Tamius, but I don't consider you as good a choice as Madchemist. Other than that, choosing only people off the mission seems stupid, as if there was only one spy on Mission 1, then there's guaranteed to be a spy on Mission 2.I may distrust Tamius, but that doesn't mean I trust Madchemist, so not off to a great start (giving a reason why you distrust me would be useful Rovsea). Even if Mission 1 only had a single spy and said spy is Tamius, this team would only be clean if Falc and hazbot are spies, and I am far from convinced of that at this point.
Reject
Tamius not using KACEOY in no way implicates me. I proposed the team sure, but that doesn't implicate me either. Assuming Tamius is the spy, everyone last round proposed a spy on their team. Overall being the proposer of a sabotaged team is not particularly indicative of a player being a spy.You proposed a sabotaged mission where the guy with KACEOY didn't use his card. I'm not saying your as implicated by that as Tamius, but I don't consider you as good a choice as Madchemist. Other than that, choosing only people off the mission seems stupid, as if there was only one spy on Mission 1, then there's guaranteed to be a spy on Mission 2.
OK, so my thoughts as of now.
Tamius has basically been called out as a spy by everyone and hasn't resisted that notion credibly. I'm going to mark that one down as definitely true.
Given that, I'm inclined to distrust Cliges. Yes, he's been (correctly) vocal against tamius for his failure to use the card, but don't forget that tamius got the card in the first place from Cliges, and that tamius returned the favor with another card at the start of this round.
Huh. So from the perspective of someone off the failed mission, the probability of there having been 2 spies on the failed mission is high enough that it actually justifies leaving everyone who was on it off the next mission? Because that feels really counterintuitive, though I see nothing blatantly wrong with your math.
I'm still a bit wary of MC
Tamius does not need to be on the team to use his card. It was theoretically beneficial to have him on the team mission 1, since him making an accusation or not would greatly narrow down the spy teams if he used the card and the team was sabotaged. His failure to use the card thus reflects quite poorly upon him.Math. This game would be pretty impossible to win mathematically, so we should probably not go down that road further. I just made the calculations cause they seemed easy to do, but mostly I just wanted to prove tamius wrong / my gut feeling right
As the others said we need to take all the other information we have into consideration.
If there is anything I learned in my short time in this forum, that would be: always be wary of MC ^^
KACEOY can be used even if the holder of the card in not on the mission, right? Or do we have to put Tamius on it to make use of the card?