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?

  • 4
    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, 2021 at 7:54
  • 2
    This question is literally answered in the series.
    – BCdotWEB
    Oct 12, 2021 at 8:12
  • 1
    @BCdotWEB What's the answer?
    – Sandun
    Oct 12, 2021 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, 2021 at 8:16
  • 4
    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, 2021 at 17:53

3 Answers 3


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.


My assumption would be that if the numbers remaining didn't suit tug of war, they would have played a different game. As seen when the organ harvesters can't tell the doctor what the next game will be, the staff don't always appear to know in advance which games will be played next. This would suggest the games played, or at least the sequence of them, is not fixed from the beginning.

I think that they would have prepared a number of options of what they could play so they could then choose what to play next.


Option 1 : as stated by another poster, they would simply go with another game. Anyone who played 'Fall guys' or any similar game will think about it. It was my first guess for any question related to 'what if that game'. But I'm not so confident anymore. Mostly because it would not be so entertaining as the other options.

Option 2 : "the weakest link" rule, as used in game 4. Whoever failed to make a full team get a free pass. Of course, using that rule 'early' would make it impossible to use for the next game.

Option 3 : The other weakest link. Whoever fails to complete a team will be eliminated.

Option 3b : The other weakest link, with a chance. Whatever team that fails to complete will compete against a normal one.
It's not necessarily so big a drawback, as a tug of war is generally decided when one of the participants fail. So 9 strong participants might still be a better option than 8 strong and 2 weak.
The decision between options 2/3 and 3b would probably be decided on how to get an even number of teams.

Option 4 : The riot result was fixed. After some reflection this will be my final guess. The entire point behind the riot was to adjust the number of candidates to a multiple of 8. The show clearly shows the riot was considered as a game and was certainly staged to happen (they would certainly have done something during the night to help it start if needed). From that point, all they needed to do was to break the riot at the right moment. If it was their first try, they would likely fail to achieve the wanted result. But it's not their first try.

And as a bonus round, it goes back to option one: They can go with another game.

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