Restricted Rock-Paper-Scissors Nash Equilibrium
You play rock-paper-scissors with a friend, but your friend is not allowed to play paper. The winner collects $\$1$ from the loser; ties result in no payment.
Both players know the restriction (your friend cannot play paper). Both play optimally.
Find the Nash equilibrium mixed strategies for both players and compute your expected payout per round.
Hints
- Since your friend cannot play paper, your scissors can never win. Eliminate scissors from your strategy -- you should only ever play rock or paper.
- With a 2x2 game (you: rock/paper; friend: rock/scissors), find the Nash equilibrium by making each player indifferent between their two pure strategies. Set $\partial E / \partial p = 0$ and $\partial E / \partial q = 0$.
- Write the expected payout as $E(p,q) = 2(p+q) - 3pq - 1$ where $p$ is your probability of rock and $q$ is friend's probability of rock. The partial derivatives give $q^{*} = 2/3$ and $p^{*} = 2/3$.
Worked Solution
How to Think About It: Your friend can only play rock or scissors. Given that, ask: should you ever play scissors? No -- scissors can only tie (against scissors) or lose (against rock), and it never beats anything your friend can play. So you play only rock or paper. Now you have a 2x2 game. In a 2x2 zero-sum game like this, neither player wants to play a pure strategy -- if you always played rock, your friend would always play scissors to tie (and never lose). So both players mix. Find the Nash equilibrium: the mixing probability where each player is indifferent between their two pure strategies.
Quick Estimate: The game is symmetric-ish but not quite. Your friend can play rock (beats your scissors, which you don't play anyway) or scissors (loses to your paper). You can play rock (beats friend's scissors, ties friend's rock) or paper (beats friend's rock, loses to friend's scissors). Intuitively you should weight paper heavily since it beats rock and only loses to scissors. Rough guess: you play paper about 2/3 of the time, friend plays rock about 2/3 of the time. Expected payout probably around $1/3$ per round in your favor.
Formal Solution:
You play rock with probability $p$ and paper with probability $1-p$. Friend plays rock with probability $q$ and scissors with probability $1-q$.
Payoff matrix (your payout) for the 2x2 game:
| | Friend: Rock | Friend: Scissors | |---|---|---| | You: Rock | 0 | +1 | | You: Paper | +1 | -1 |
Your expected payout: $$E(p,q) = p(1-q)(+1) + (1-p)q(+1) + (1-p)(1-q)(-1) + pq(0)$$ $$= p(1-q) + q(1-p) - (1-p)(1-q)$$
Expanding: $$= p - pq + q - pq - (1 - p - q + pq) = p - pq + q - pq - 1 + p + q - pq$$ $$= 2p + 2q - 3pq - 1$$
At a Nash equilibrium, each player is indifferent between their pure strategies.
**Find $p^{*}$ (friend's indifference):**
Your friend is indifferent between rock and scissors when $\partial E/\partial q = 0$: $$\frac{\partial E}{\partial q} = 2 - 3p = 0 \implies p^{*} = \frac{2}{3}$$
**Find $q^{*}$ (your indifference):**
You are indifferent between rock and paper when $\partial E/\partial p = 0$: $$\frac{\partial E}{\partial p} = 2 - 3q = 0 \implies q^{*} = \frac{2}{3}$$
Nash equilibrium strategies: - You: rock with probability $2/3$, paper with probability $1/3$ - Friend: rock with probability $2/3$, scissors with probability $1/3$
Expected payout: $$E(2/3, 2/3) = 2(2/3) + 2(2/3) - 3(2/3)^2 - 1 = \frac{4}{3} + \frac{4}{3} - \frac{4}{3} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$$
Answer: Your expected payout per round is $\boxed{+\$1/3}$. You have a structural edge because the restriction benefits you: paper -- which your friend cannot beat -- is available to you, while your friend's best response is to play scissors more to discourage it.
Intuition
The restriction on your friend creates a strict advantage for you: paper can only lose to paper, which your friend cannot play. So paper is now a 'dominant' strategy in the qualitative sense -- your friend has to play scissors to punish it, but scissors loses to rock, which creates a tension. The Nash equilibrium resolves this by having you play paper less than 50% of the time (1/3), keeping your friend from going all-in on scissors.
The expected payout of $+1/3$ means you have a real edge per round, not just a slight one. In game-theoretic terms, this comes from the asymmetry in the action spaces -- restricting one player's actions in a zero-sum game always weakly helps the opponent. This principle appears in market microstructure too: if a market maker knows one side of the market cannot hedge efficiently, they can shade their quotes to extract a structural edge, exactly as you extract 1/3 per round here.