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In the first episode of Squid Game Season 2, two characters play Rock-Paper-Scissors Minus One against each other. Here, each player chooses rock, paper or scissors with each of their left and right hands (possibly the same for each hand, possibly different). The players observe the other's choices and then simultaneously take one hand away, leaving one strategy for a standard Rock-Paper-Scissors outcome.

So as a game theorist I asked myself, what is the Nash equilibrium to this expanded R-P-S game? How does it compare with the standard single-shot version?

Spoiler Alert: Solution follows. If you want to puzzle it out for yourself, pause reading now....

There are six pure strategies for the initial choice here: RR, PP, SS, RP, PS, and RS (clearly, it doesn't matter which hand plays which strategy, so for example RP is the same as PR). Let's call the first three strategies set S since each hand chooses the same option, and the last three set D, for different strategies with each hand.

Then the strategies in S can be shown to be weakly dominated by D strategies: if your opponent plays a strategy $s\in S$ and you play $d \in D$, then the worst you can do is assure yourself of a tie, and in 2/3 of the cases you can assure yourself of a win. (If your opponent plays RR and you play RS, then you should choose R in the second round and tie. If you played either RP or PS, then choose P in the second round and win.)

In equilibrium, the players randomize in the first round over the strategies in D, playing each with probability 1/3. If by chance they play the same d, then they choose the dominant strategy among these two in the second round to assure themselves of a tie. For instance, if they both choose RS in round 1, then they both play R in round 2, since choosing S can only result in a loss.

If players choose different D strategies, then one will be "dominant" relative to the other in the sense that one player will always be able to defeat whichever choice their opponent makes. For example, Player A chooses RP in the first round and Player B chooses PS. Then Player B is dominant because if Player A chooses R, B can choose P and win. And if A chooses P, B can choose S and win. On the other hand, if B chooses P they can do no worse than tie.

The second period payoff matrix (from Player A's perspective) will look like this:

  |   P        S
===================
R |   0        1
  |
P |  1/2       0

Here neither player has a dominant pure strategy, so the equilibrium is in mixed strategies: A plays R with probability 1/3 and P with probability 2/3. B plays P with probability 2/3 and S with probability 1/3.

Given these mixed strategies, there is a 4/9 chance that both play P and the game is a tie, there is a 4/9 chance that B wins, and a 1/9 chance that A wins. So the player with the dominant first-round choice has a four times greater chance of winning than does the dominated player.

Overall, with optimal play there is a 17/27 chance of a tie in this two-stage variant, almost twice as large as the 1/3 chance of a tie in the traditional single-shot RPS game.

So that's the analysis; here's the question: how many pure strategies does each player have for the expanded 2-period game?

PS -- If someone wants to take the time, they can re-watch this part of the episode and determine if the players were optimal or not. It goes by pretty fast, but for sure at the end one of the players chooses a dominated strategy.

David Pepper
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    I guess you should add some spoiler alert though – Davide Masi Dec 29 '24 at 21:42
  • Good idea. Done. – David Pepper Dec 30 '24 at 02:00
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    What is the question here? If you want to verify your proof, then the [tag:solution-verification] tag should be used (but even then, you need to specify which part of your proof you doubt). Alternatively, you can keep the question in the question body, and add an answer yourself to the question – Soham Saha Dec 30 '24 at 04:10
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    Actually I meant "spoiler alert" for what happens in the first episode of the series – Davide Masi Dec 30 '24 at 09:23
  • Great analysis. I credited your post in this video https://youtu.be/psMgkbi84HQ and blog post: https://mindyourdecisions.com/blog/2025/01/06/how-to-win-squid-games-rock-paper-scissors-minus-one/ – Presh Jan 07 '25 at 00:45
  • Thanks Presh -- I'm a fan of your videos, and it's great to get the analysis out there. And the answer of 195 pure strategies is correct -- it's an unusual number, but it follows from the structure of the game. – David Pepper Jan 07 '25 at 23:24

1 Answers1

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As has already been stated in the question, there are $6$ pure strategies in the first round: RR, PP, SS, RP, PS, and RS. We are counting symmetric strategies like RP and PR just once.

The number of Player A's pure strategies in the second round depends on what both Players A and B played in the first round.

If Player A played RR, PP or SS in the first round, A will have only one possible move and one pure strategy in the second round (due to A's hands being the same).

If Player A played RP, PS or RS in the first round, A will have two possible moves in the second round, and it will depend on which of the $6$ moves B performed in the first round. Therefore the number of A's pure strategies in the second round for a fixed first-round strategy is equal to $2^6=64$.

The total number of pure strategies of Player A is therefore:

$$ 3 \cdot 1 + 3 \cdot 64 = 3 + 192 = 195. $$

I have also checked the Squid Game episode. Spoilers ahead.

Keep in mind that this game has not been invented by the creators of Squid Game. It is actually a popular game in South Korea.

The following ordered pairs of hands were played (with the second round in parentheses): nothing and RS, RS and RP (S and P), RP and PS (R and P), RS and RP (R and R), PS and RS (S and S), RS and RP (R and R), PS and RS (S and S), RP and PS (P and P), RR and PS (R and nothing). RR is a dominated strategy and would have lost had the opponent played optimally (the other strategies chosen were reasonable). The player's miscalculation was due to the fast-paced nature of the scene as well as the threat of imminent death. Another interesting observation is that two of the games are repeated; the series has possibly duplicated the footage.

Kocijan
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