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Five votes are in.
Just talking out loud, here, but 2/6 players are sleeper agents. If chosen randomly, the odds of the first one being chosen is a sleeper agent is 1/3rd. The odds of the second one also being a sleeper agent, is 1/5. If the first player chosen isn't a sleeper agent (2/3rds chance), then the second player being a sleeper agent is 2/5. If the first player chosen is innocent at 2/3rds chance, there's a 3/5th chance that the second player chosen is also innocent.
There's surely an easier way of doing this, but it's been a few years since I took statistics and probability.
And means multiply, or means add. 2/3 and 3/5 = No sleeper agents = 40.0% 1/3 and 4/5 or 2/3 and 2/5= One sleeper agent = 53.3% 1/3 and 1/5 = Two sleeper agents = 6.67%
All adds up to 100%, so I probably did this right
@jacabyte, on 15 July 2012 - 11:00 AM, said in GTW 39:
Edit: I want SIB to tell us how to find the probability that 1 or more of the members of the first team were traitors like he did in GTW 36, because I can't do that even though I took college discrete math and we did probabilities in there. (I can tell you how many different ways there are of arranging people in a photograph, and many ways there were of selecting the first team (15), I just can't tell you how many ways there are of selecting a group such that 1 or more were traitors.)
That's a high school math red / blue marble problem. Just sayin'.
Chance of there being at least one sleeper agent in the first group of two is 60%, if names are chosen randomly. Things are a little more complicated than that, though, because the person choosing knows what their role is, and possibly what role somebody else is.
@adam_0, on 15 July 2012 - 11:19 AM, said in GTW 39:
They didn't teach us that in high school either.
Okay, so if a person picks themselves for the mission, they know their own role. If they choose randomly for the second person, the odds come down like this:
If the person choosing is a sleeper agent:
0 and 4/5 = No sleeper agents = 0% 1 and 4/5 = One sleeper agent = 80% 1 and 1/5 = Two sleeper agents = 20%
If the person choosing is innocent:
1 and 3/5 = No sleeper agents = 60% 2/5 = One sleeper agent = 40% 0 and 2/5 = Two sleeper agents = 0%
Only the person choosing (and possibly the sleeper agents), knows which case is true.
In the end though, I'm not sure how any of this is useful information. It comes down to trust, and seeing what happens on the actual mission to inform good decisions about future trust. I guess the lesson to learn here with these probabilities, is, you really should trust the person picking the names, or else vote to reject their proposals.
It really isn't useful, I just like math; we pretty much always have a 50% chance of putting a traitor on one of our committees.
Well, it's more like some of the time it's 40%, some of the time it's 60%, based on whether you can trust the person picking the names. It's a good idea keep the control over the names going on a mission with somebody that you trust because it makes all the difference. Trust has to be earned, but in the beginning, you have no choice but to fire a bit blind.
OK, here's the breakdown.
There are 15 ways to assign traitorships (2) to 6 people. Of those, 5 have a specific person being a traitor, that is to say the chance of someone being evil is 1 in 3. This is about what you'd expect (2 of 6, right?)
So, we have a 2/3 chance JacaByte was innocent, and a 1/3 chance he's not.
If he's innocent, he's picking randomly. The chance that retep998 is a traitor (and thus the first group contains traitors) is 2 in 5. If he's not innocent, there's no telling what he'd do, but the chance that the first group contains traitors is 100%.
The chance that the group contains traitors is thus (2/3) * (2/5) + (1/3) = 9/15 , or just over 50%.
This is why having more data is really a good thing.
@jacabyte, on 15 July 2012 - 12:10 PM, said in GTW 39:
That's only true for groups of 2 players. Assuming all innocents always pick randomly:
3 players: Chance of traitor pick (if player is innocent): 7/10 (2/3)*(7/10) + (1/3) = 4/5 , or MUCH better than chance.
4 players: Chance of traitor pick (if player is innocent): 9/10 (2/3)*(9/10) + (1/3) = 14/15. We better have our sh#! together by this point.
EDIT: Fixed math for 2-player example. Sorry about that!
This post has been edited by SoItBegins : 15 July 2012 - 03:46 PM
The problem is we won't be picking players randomly by that point. Or at least we shouldn't be.
Operation 2 Name Proposal 1 adam_0 Crow T. Robot SoItBegins
retep998 - Reject 10:28 AM Crow T. Robot - Approve 11:34 AM Mackilroy - Reject 11:38 AM adam_0 - Reject 11:52 AM JacaByte - Reject 12:00 PM SoItBegins - Reject 2:22 PM
Motion is rejected.
retep998, it is your turn to pick names for the second operation.
@soitbegins, on 15 July 2012 - 01:21 PM, said in GTW 39:
If he's innocent, he's picking randomly. The chance that retep998 is a traitor (and thus the first group contains traitors) is 1 in 5.
Picking randomly, if JacaByte is innocent, retep998 has a 2 in 5 chance of being a bad guy. There are two bad guys in the remaining five players, not one (if JacaByte is innocent). If you carry out the rest of your math with the correct odds, you get the 60% chance I said earlier, which should mean a win for the sleeper agents (lesson learned: don't let just anybody pick names). More data is good, but a lot of that comes from the outcome of the mission, and then looking back at voting records afterwards. By keeping the proposer as somebody trustworthy, based on all the information, you can push the odds closer to 40% if they pick randomly which will get you the win, and if the proposer is trustworthy and chooses names based on who is trustworthy, the odds go even more in your favor. People are going to have different opinions on who is trustworthy, though.
Choose wisely retep.
Well, s###. I'm not going to choose Jacabyte this round, so we can get some more information on other people. However, I will choose myself, because obviously I am innocent, and I wouldn't pick somebody else instead of me. So... using a random number generator... 2 so thats adam_0 1 so thats crow t. robot My proposal is as thus follows retep998 adam_0 Crow T. Robot
Please PM me, everyone, whether you APPROVE or REJECT the second proposed second operation: adam_0 Crow T. Robot retep998
Done. retep, your random number generator is horrible -- we already had a motion with Crow T. Robot and adam_0. I've rejected this motion too.
I have also rejected this motion since I don't see how this selection helps us determine who will be cooperative when it comes down to carrying out actual missions. I didn't notice that adam_0 and Crow was on the previous motion, as well as this one.
@jacabyte, on 15 July 2012 - 04:18 PM, said in GTW 39:
This is why you should keep track.
@mrxak, on 15 July 2012 - 10:32 AM, said in GTW 39:
Please PM me, everyone, whether you APPROVE or REJECT the first proposed second operation: adam_0 Crow T. Robot SoItBegins
@mrxak, on 15 July 2012 - 02:08 PM, said in GTW 39:
...I guess Crow and I are just popular? I dunno. I don't see how this will help us narrow things down though, so I will have to reject.
Operation 2 Name Proposal 2 adam_0 Crow T. Robot retep998
retep998 - Approve 3:03 PM Crow T. Robot - Approve 3:05 PM JacaByte - Reject 3:17 PM Mackilroy - Reject 3:23 PM SoItBegins - Reject 4:47 PM adam_0 - Reject 7:30 PM
adam_0, it is your turn to pick names for the second operation.
I choose myself , retep998 , who has been trustworthy in the past, and Mackilroy , who hasn't seen any action thus far.