Nash Equilibrium for Game that requires an "action" and a "guess" alternating between two players

Question

Hi I'm trying to write an CFR (Counterfactual Regret Minimization) script to calculate the Nash Equilibrium for a popular hand game in Taiwan where you can play with another person. The rules is as follows:

RULES:

1. Two players Rock Paper Scissors to see who goes first.
2. Players can either throw one of 3 moves with their HANDs:
   • 0 (both fists closed)
   • 5 (one hand open, one closed)
   • 10 (both hands open)
3. Simultaneously, the active player needs to GUESS one of 5 possible counts that are possible from adding the total count of your hand + the opponent's (16 total combos):
   • 0 (both players have their fist closed / 1 combo)
   • 5 (only one hand open / 4 combos)
   • 10 (two hands open / 6 combos (2 + 4) )
   • 15 (three hands open / 4 combos )
   • 20 (all hands open / 1 combo)
4. The player wins the match if the guess is correct. If it's incorrect, the opposing player gets to guess.
5. The game continues until there's a winner.

So where I'm confused is based on this rule, starting from the root game node, and your first to act you have actually:

3 (NUM_HANDS) * 5 (GUESSES) = 15 Actions branching out of the root. ( 9 if you disqualify invalids)

Actions:

Hand 0 : Guesses: 0, 5, 10, (15), (20)

Hand 5 : Guesses (0), 5, 10, 15, (20)

Hand 10 : Guesses (0), (5), 10, 15, 20

Where as for the opponent who's defending, there is only 3 possible actions:

Actions:

Hand 0, Hand 5, Hand 10

So they players actually take turns between selecting 15 actions or 3 actions. My question is how do you set things up to calculate the cumulative regret for the active player (hero) and the defending player (villain) for each rounds?

Do you need two sets of probability matrix to represent hero / villain ? If so how to arrive at a general strategy? Or is the only way is to have two sets of strategy for each situation?

Thanks in advance!

Answer 1

This is way more complicated than necessary. The guessing player is effectively merely guessing how many hands the other player will open; it doesn’t make a difference how many hands the guessing player opens, since she knows that and can just add it to her guess.

Thus, both players effectively have three pure strategies, and these are fully symmetric, so the Nash equilibrium is that they both choose each strategy with probability $\frac13$. That is, the non-guessing player opens $0$, $1$ or $2$ hands, each with probability $\frac13$, and the guessing player guesses $0$, $1$ or $2$ hands, each with probability $\frac13$, and opens an arbitrary number of hands and adds that to the guess.