In the 2021 Netflix series Squid Game, round 3 was Tug of War, which made the remaining 80 players divide into 8 teams of 10 members each, for a deadly tug of war game where each 2 randomly selected teams played against one another.

Because of the fight that broke out in the midnight the day before, exactly 80 players remained for Round 3, which conveniently allowed the formation of an even number of teams (this is essential because each game is played in 1v1 knockout method). However, what if the total number of players was an odd number? Since the first two games (and count the midnight fight if you consider it was planned) can result in either an even or odd number of survivors, how could the organizers be so sure that they'd get a number that would enable an even number of teams? For example, what if the number of survivors was 37, which was an odd number?

  • 2
    It's called narrative imperative ;) It's the same plot device that makes one-in-a-million chances happen nine times out of ten in stories.
    – Tetsujin
    Oct 12 at 7:54
  • 2
    This question is literally answered in the series.
    – BCdotWEB
    Oct 12 at 8:12
  • @BCdotWEB What's the answer?
    – Sandun
    Oct 12 at 8:13
  • @Tetsujin Interesting. However, the one we see isn't the only edition of Squid Game, the record books imply it existed since at least 1990s. The possibility of getting an odd number is 50/50, which means there could've been instances where they got odd number of players.
    – Sandun
    Oct 12 at 8:16
  • 1
    I would assume that the odd player would be spared just like the girl in the later episode when she didn't get a partner.
    – Styxsksu
    Oct 12 at 17:53

Given that we see exactly what happens when there is an odd player out (game 4), I don't quite see why the same thing couldn't happen in game 3.

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