# Rock, Paper, Scissors, Anti-Rock

Alice and Bob are both well-known perfect logicians who have the ability to internally randomize and often pass their time by playing rock-paper-scissors. One day, tired of the usual RPS game, Alice comes to Bob with a variation: Rock, Paper, Scissors, Anti-Rock. It works similarly to RPS, on the count of three each player shows one of the possible symbols, except this time there are 4 options:

• Rock, paper and scissors have the usual relationship: Rock beats scissors, scissors beats paper, paper beats rock, anything against itself is a tie.

• Anti-Rock, on the other hand, ties against rock (and itself), beats paper, and loses to scissors.

Upon being presented with this variation, Bob is skeptical: "It seems kind of dumb, and it feels like there'll be way too many ties, but I guess I'll give it a shot."

If they both play logically and they both play each of the four throws at least some of the time, how frequently will Bob and Alice tie in the new version of the game?

• So, how do you throw an anti-rock? Like a rock but fingers bent backwards? (Ouchy.)
– Bass
Commented Mar 30, 2020 at 21:18

Alice chooses columns, Bob chooses rows in this matrix: $$\pmatrix{ * & r & p & s & a \\ r&0&1&-1&0\\ p&-1&0&1&1\\ s&1&-1&0&-1\\ a&0&-1&1&0 }$$ $$+1$$ means Alice wins, $$0$$ is tie, $$-1$$ Bob wins.

There is no Nash equilibrium.

According to this solver/website https://cgi.csc.liv.ac.uk/~rahul/bimatrix_solver/ both players have two mixed Nash equilibrium strategies:

1) play rock/paper/scissors with probability 1/3

2) play rock/anti-rock with probability 1/2

The mixed rock/anti-rock stategy is against any strategy of the other player: rock/anti-rock per definition, but also against scissors/paper (in the mean).

The classic rock/paper/scissors is a tie against rock/paper/scissors.

Hope this makes sense.

• +1 for showing me the problem is badly posed, but there's a third equilibrium, where all four throws are played with positive probability! Commented Mar 31, 2020 at 0:44
• You could combine both mixed strategies (play the first with probability p and the other with probability 1-p) and get something like AR=1/4, R=5/12 S=1/6 P=1/6 (for p=1/2). Commented Mar 31, 2020 at 6:05
• yes, thats true. @HRogers can this third equilibrium obtained by Zilvarro's suggestions?
– daw
Commented Mar 31, 2020 at 7:46
• Yes, it's true! I made a mistake in my own calculations but Zilvarro is right, and the problem doesn't have a unique solution since the tie frequency depends on p. Sorry for the confusion! Commented Mar 31, 2020 at 12:41

If you count for each hand the number of hands they win (W), lose (L) or draw (D) against, you get the following:
R : 1W 1D 1L
AR : 1W 1D 1L
P : 1W 2L
S : 2W 1L

So, if I was "logical", I'd always play scissors, as if your opponent is playing randomly you get the best odds to win. If both think that way, then they'd end up tying all the time using only scissors.
But then again, I guess being a perfect logician can mean you know that your opponent is also going to chose that strategy ?

• Yes, as perfect logicians they are each able to anticipate/attempt to counter the other's strategy. So if one believed the other would always play scissors no matter what, they would always play rock and secure an advantage. Commented Mar 30, 2020 at 16:01
• We're looking for a mixed-strategy Nash equilibrium here, I think?
– Deusovi
Commented Mar 30, 2020 at 16:08
• That'll get you most of the way there! Commented Mar 30, 2020 at 16:11

This assumes that a win is worth 1 and both a loss and a tie are worth 0. That is, the players try to maximize their own number of victories instead of the victory difference.

Then:

Whenever I want to play Anti-Rock it would be better to just play scissors instead, because AR wins only against paper while scissors also wins against AR. If both players follow this logic, the game degenerates to regular RPS, in which the unique mixed Nash-Equilibrium (or reasonable strategy for worst-case optimization) is to pick every choice with an equal probability of 1/3. This then also implies that 1/3 of the games will be ties, on average.