# Rock-Paper-Scissors but rock scores double

Suppose two perfect game theorists play rock paper scissors. Scoring is slightly different:

• 2 points when beating scissors with rock
• 1 point when beating rock with paper
• 1 point when beating paper with scissors

In case of a tie or loss, 0 points are awarded.

Assuming both players play an infinite number of games and both play optimally, what random distribution would give the highest expected score per game?

• Doesn't literally any distribution that has a non-zero probability for each of the throws have an infinite expected score? Dec 14, 2023 at 15:02
• Is a player trying to maximize her score or to maximize the difference between her score and her opponent's ? Dec 14, 2023 at 17:15
• @Evargalo Given two ideal players the expected difference is always going to be zero.
– fljx
Dec 14, 2023 at 22:39
• @fljx : but the definition of "ideal player" depends on the role they are assigned. If their goal is to maximize the difference, the expected difference is zero ; if their goal is to maximize each individual score, it might differ. Dec 15, 2023 at 8:12
• To make it a zero-sum game and avoid prisonner's dilemma to be a thing (which can make it tricky/impossible to solve), you may want to say that : "You lose 1 point when losing against paper or scissors and you lose 2 points when losing against rock". Only then the goal to maximize your points and the goal to maximize the difference with the opponent is the same. Dec 15, 2023 at 9:40

This probably isn't the answer you want but:

Given that the players are equally "perfect", they can both expect to gain exactly half the total points scored in the long run. This is true whether their strategies are matched or distinct.

So, the best strategy for both players to maximise their expected score per game, is to co-operatively maximise the total score.

This can be achieved by both players alternating between scissors and rock, one starting with rock, the other starting with scissors. That will gain them both an average score of 1 per game.

• This is called "Superrationality", a term coined by Douglas Hostadter. Dec 14, 2023 at 18:24
• @AlbertHendriks This answers the question as asked. The point here is that this isn't a zero-sum game. 1 win + 1 loss is better for both players than 2 draws, and co-operation is the optimal way to achieve that consistently over the long run. If you think the question should have been different, you're always free to ask your own.
– fljx
Dec 15, 2023 at 14:41
• @fljx - This probably isn't the comment you want, but the question asks "what random distribution", and this isn't a random distribution. Dec 16, 2023 at 2:13
• @fljx The term "zero-sum game" doesn't refer to number of points. Basketball is a zero-sum game, even though two teams could collude to get hundreds of points. Dec 16, 2023 at 23:29
• @Acccumulation But that's because basketball is ultimately scored in terms of games, not points, so it doesn't matter that points are not zero-sum. Here OP makes it clear that points are the final unit of scoring. Dec 17, 2023 at 13:49

The unique Nash equilibrium is where both players play Rock, Paper, Scissors with probabilities $$(r,p,s)=(\frac{2}{5}, \frac{2}{5}, \frac{1}{5})$$. In this equilibrium each player wins on average $$\frac{2}{5}$$ points per turn.

If you know your opponent is following some distribution $$(r,p,s)$$, then the expected values of your options Rock, Paper, Scissors are $$2s$$, $$p$$, and $$r$$ respectively. These are all equal if and only if $$(r,p,s)=(\frac{2}{5}, \frac{2}{5}, \frac{1}{5})$$.

For a Nash equilibrium, we should have that your distribution is an optimal response to your opponent's, and the opponent's distribution is an optimal response to yours. Suppose some Nash equilibrium where your distribution is not $$(\frac{2}{5}, \frac{2}{5}, \frac{1}{5})$$. Then your opponent's strategy would give weight 0 to at least one option, call it A. From this we deduce that your strategy gives weight $$0$$ to the option that beats A, call it B. In turn we get that the opponent's strategy gives weight $$0$$ to the third option, C. Continuing this line of argument one can quickly reach a contradiction.

• If both players play uniform distribution, the average score per turn is 4/9, which is larger than 2/5, so this strategy doesn't beat the uniform distribution, at least. Dec 15, 2023 at 5:14
• @justhalf Yes, if both players agree to play uniform distribution, then the average score per turn is 4/9. Even better, if both players agree to alternate between the outcomes (Scissors, Rock) and (Rock, Scissors) then the average score per turn is 1! So the players can profit by collaborating. But neither situation gives a Nash equilibrium. If you do play uniform distribution, then a rational opponent will play Rock against you every time (giving them 2/3 per turn and you 1/3 per turn). Dec 15, 2023 at 9:36
• @N.Virgo As the answer explains, this solution hits the Nash equilibrium. That means that the strategy is optimal in the sense that whichever player chooses to deviate will start losing more. So even in the long run, and with perfect memory, it is best for both players to stick with the given strategy.
– Bass
Dec 15, 2023 at 13:47
• The Nash equilibrium of the "stage" game is also an equilibrium of the repeated game, yes. But @N.Virgo is right that there are other equilibria of the repeated game, as in my answer!
– usul
Dec 15, 2023 at 17:17
• @N.Virgo yes, I think the Folk Theorem will apply there too, and the same reasoning as you say! Link for convenience: en.wikipedia.org/wiki/Folk_theorem_%28game_theory%29
– usul
Dec 16, 2023 at 13:54

There are a few things we could mean by "both play optimally". Here's an interesting one.

By the

there are many different

Nash equilibria of the repeated game,

including where one player

has average utility almost $$2 - \frac{2}{5}$$, and the other just over $$\frac{2}{5}$$.

Here's how it works:

The first player will play rock four rounds in a row, then scissors, and repeat this pattern. The second player will play scissors four rounds in a row, then rock, and repeat. But on rounds 1,11,101,1001,..., they will also swap and the second player will play rock. The first player's average utility is almost $$\frac{8}{5} = 2 - \frac{2}{5}$$. The second player's average utility is just over $$\frac{2}{5}$$. The swap rounds, which get rarer and rarer, only have a vanishing impact on the utilities. Why would the second player agree to this? Because if they ever deviate from this plan (for example play paper on round they aren't supposed to), the first player has committed to switch to always playing the randomization $$(2/5, 2/5, 1/5)$$. That will give both players exactly $$\frac{2}{5}$$ expected utility for the rest of the games. So the second player would rather go along with the plan and get a tiny bit more than that than deviate.

• Huh, interesting. So even in this symmetric (albeit non-zero sum) game, the Nash equilibirium can involve different outcome for each player? I learn something new today. Thanks for the reference to the folk theorem Dec 16, 2023 at 15:26
• @justhalf The issue is that usul is treating "the game" as the entire sequence of plays. If we look at the Nash Equilibrium of a single round, then it's completely symmetric. It's only when we allow "Blackmail your opponents with threats regarding what you'll do on further rounds" as a strategy that non-symmetric Nash Equilibria appear. Dec 16, 2023 at 23:32
• @justhalf this is a peculiarity of how Nash equilibrium works. We can pretend someone else tells both players what they are supposed to play. In this case, the first player is supposed to get most of the utility and threaten to deviate, and the second player is supposed to go along with it. Given that, each player would rather follow the instructions than deviate. There are different Nash equilibrium, some where the second player gets a lot more. But how we negotiate which equilibrium to play ... is a whole other question :)
– usul
Dec 17, 2023 at 3:29
• @justhalf A Nash equilibrium is a pair of strategies such that each player can't do better, given the other player's strategy. If the first player is sure that the second player will give in to the threat, then the best the first player can do is to take advantage of that. If the second player can convince the first player that they aren't going to give in, then the first player will change their strategy. Note that Nash equilibria aren't necessarily unique. So the symmetry isn't really broken: there's another Nash equilibria where the second player threaten the first. Dec 17, 2023 at 4:13
• Here's another game to consider: there are ten dollars on a table. Both players write down how many they want. If what they write down is less than or equal to ten, then they both get what they asked for. If it's more, then both get nothing. One Nash equilibrium is both ask for \$5. Another is that one asks for \$9 and the other asks for \$1. If the first player always asks for \$9, then the second can't do any better than asking for \\$1. Dec 17, 2023 at 4:13

If this is 0-sum and points won by the winner are lost by the loser,

Throw paper with probability 0.5, rock with probability 0.25 and scissors with probability 0.25. Then regardless of your opponent's strategy your expected winnings are 0. Since the game is symmetric, this can't be beaten.

• "In case of a tie or loss, 0 points are awarded" implies it's not zero-sum. I don't know if this changes you analysis.
– R.M.
Dec 14, 2023 at 23:16
• @R.M. it does change the analysis, in that it unfortunately makes it wrong Dec 15, 2023 at 13:25

The expected value of each throw is EV=2P(R)Q(S) + P(P)Q(R) + P(S)Q(P) where P(x) is the probability that we choose x as our throw and Q(x) is the probability our opponent chooses x as their throw. I take our opponent playing optimally to mean that they will be able to exploit our strategy to maximize their EV, so what we'd really like is that the expected value of each of our throws to be the same regardless of what our opponent throws.

This is done by setting 2P(R) = P(P) = P(S), or a 20% chance of throwing rock, and a 40% chance of throwing each of paper and scissors. This simplifies our EV equation to EV = .4(Q(S)+Q(R)+Q(P)) = .4

• Shouldn’t we consider negative values for the case our opponent wins? Dec 14, 2023 at 15:58
• @Jujustum I don't think so, since we are looking for highest score and not highest score differential Dec 14, 2023 at 16:01
• The optimal response to this strategy is to play rock every time with an expected score of .8, so I don't think it is optimal. Dec 14, 2023 at 16:05
gain: $$2rs+pr+sp$$
$$rs$$ is the biggest source of points, it should be increased.
For $$r+s$$ constant, the profit is at maximum if $$r=s$$, so we can use that
gain: $$2rr + (1-2r)r + r(1-2r)$$
maximum $$2r-2rr$$ is at $$r =0.5$$
$$(r,p,s)= (1/2,0,1/2)\implies$$ expected profit is equal to $$0.5$$