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Puzzling

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    $\begingroup$ Wouldn’t your strategy extended to, say, 4 players give the probability of winning as 4/9 instead of 2/3? When you pair them up as you said, there are three possibilities. Both pairs lose with probability 1/9, that’s a team loss. One pair loses while the other wins with probability 4/9, but that’s still a team loss because in such a scenario, the number of wins is 1 and the number of losses is 2. Finally, both pairs win with probability 4/9, which is a team win. $\endgroup$ Commented Sep 14 at 21:51
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    $\begingroup$ @FirstNameLastName Here is my own simulation: ato.pxeger.com/… $\endgroup$ Commented Sep 15 at 0:32
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    $\begingroup$ I see. I just saw your code and there you say “…any even number of excess players can be…”. I added that word to your answer. That was really the source of my confusion. I see what you mean by “neutralize” now. $\endgroup$ Commented Sep 15 at 0:36
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    $\begingroup$ @FirstNameLastName Their strategy is to always pick what loses to what they see on their partner's hat. $\endgroup$ Commented Sep 15 at 0:44
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    $\begingroup$ OH! that is an interesting boolean difference ! I misread or misinterpreted, what wins :) $\endgroup$ Commented Sep 15 at 0:45