I was watching a TV show, and the setting is pretty simple:
There are 10 players in total, within them there are 2 "spies", players initially form 2 teams of 5 people, and there are 3 rounds of game, where at the end of each round, we can switch a player from team A with a player from team B (or no switches, team members remain unchanged), 2 "spies" win if they are in the same team when the game finishes, otherwise they lose and the rest of the players win
of cause this is a strategy game...spies know their identity and will try to manipulate others
TL;DR here is my question:
one of the player said it is better to maintain the original teams since it is unlikely that 2 spies are in the same team by initial arrangement (random)
well, I got that the initial probability of 2 spies in the same team is $4\over9$ which is indeed less that 50%, but randomly switch a player from team A with another player in team B after round 1 gives me the same probability...so there is no dominant strategy?
here is my calculation: (case 1) 2 spies start being in the same team, after round 1, we randomly switch two players between 2 teams, the probability of 2 spies still in the same team is:$${4\over9}*{3\over5}$$ 3/5 is the probability of non-spy player being selected from the spy team
(case 2) 2 spies start not being in the same team, after round 1, we randomly switch two players between 2 teams, the probability of 2 spies accidentally being in the same team is:$${5\over9}*({1\over5}*{4\over5}*2) $$
1/5*4/5 is the probability of a spy switched with a non-spy player, since 2 spies can end up in either team, (times 2 is required)
sum of these two probabilities gives me again $4\over9$. Is my calculation or the understanding of this game correct?
