Left Shoot, Right Shoot: A Rock Paper Scissors Variant
In Hong Kong, there’s a variant of rock paper scissors called 左一拳右一拳, which can be roughly translated as “left shoot, right shoot”. In this post, I’ll walk through how it works and how to play optimally.
Rules
There are three steps in this game. First, both players play rock paper scissors with one hand. Second, both players play rock paper scissors with the other hand. Third, both players take back one hand, and the winner is determined by comparing the remaining hands using normal rock paper scissors rules.
Let’s run through a quick example. Say players A/B play ✊/🖐, then 🖐/✌. Now since A can only pick between ✊ and 🖐, if B keeps 🖐 and retracts ✌, B will never lose. If A anticipates B to play 🖐 and also plays 🖐, then they will tie.
But if B anticipates that, B can sometimes pick ✌ which beats 🖐. But if A anticipates that, maybe A will also sometimes play ✊… In true rock paper scissors spirit, there always seems to be a better strategy.
Strategy
But of course, all games have optimal strategies, when we allow strategies to incorporate randomness. The optimal play is:
- Pick anything for the first hand with equal probability;
- follow your opponent’s first hand for your second hand with 2/3 probability, unless that’s the same as your first hand, in which case pick the one that beats that;
- Keep the hand that both players have in common with 2/3 probability, unless both players have the same two choices, in which case pick the one that doesn’t lose.
Below, let’s go through an outline of the math involved. First, we have to define what both players are maximizing.
Objective
It might not be immediately obvious that it is nontrivial to define the goal of the game - of course you want to win instead of lose! But this is not always the case. For example, some people might really want to avoid losing, instead of trying too hard to win.
Rock paper scissors is commonly played as a way to fairly decide a binary outcome between two people (e.g. who gets to pick the restaurant). It is reasonable to apply the same to this game, meaning players will repeat the game until a winner is determined, allowing no ties. Playing optimally then means maximizing the probability of winning in a repeated game.
Since the game is fair, in the event of a tie, your chance of winning is still 1/2. So, we’re maximizing P(win)+P(tie)*1/2, which is the same as minimizing of P(lose)+P(tie)*1/2. Equivalently we can put those together and maximize P(win)-P(lose) for each game. In other words, we can pretend the loser pays $x to the winner, and maximize the expected value of winnings.
Analysis
First we can analyze step 3 - both players have to pick one out of two given choices.
Let’s run through the boring cases quickly. Boring case 1: either player has the same choice for both hands (dumb). Boring case 2: both players have the same two choices (the optimal pick is obvious).
Now to analyze the remaining case where both players have different choices. Let’s say A has ✊, 🖐 and B has 🖐, ✌ to pick from. Let’s say the winner of the game wins $3, the loser loses $3. We have four outcomes after both players take back one hand:
| A \ B | 🖐 | ✌ |
|---|---|---|
| ✊ | -3 \ 3 | 3 \ -3 |
| 🖐 | 0 \ 0 | -3 \ 3 |
Since this is a zero sum game, to maximize your winning, you want to make your opponent’s best option as bad as possible. This happens when both of their options are equally bad.
When A (B) plays 🖐 with 2/3 probability, B (A) gets $1 (-1) in expectation no matter which hand they take back. Hence, the optimal strategy for step 3 is to pick the option that can lead to a tie 2/3 of the time (🖐 in this case).
Now that we’ve established step 3’s strategy, let’s do the same to step 2. Say A picked ✊ and B picked 🖐, and both have to pick their second hand. Since we already worked out the expected value for all possibilities at step 3, we can again tabulate the expected value of all scenarios for A and B here.
| A \ B | 🖐 , ✊ | 🖐 , ✌ |
| ✊ , ✌ | -1 \ 1 | 1 \ -1 |
| ✊ , 🖐 | 0 \ 0 | -1 \ 1 |
If you paid attention, you’ll realize that this is just the previous table from step 3 but with winnings scaled down by 1/3. This means that at step 2, we’re playing the exact same game! So the optimal strategy must also be the same - we play the option that leads to a tie with probability 2/3.
Summary
As a bonus fact, we can calculate the probability of a tie. If both players end up with the same two choices after step 2, the game must end in a tie; otherwise there’s still a 2/3*2/3 chance of a tie. This yields 1/3 + 2/3 * (4/9 + 5/9 * 4/9) = 193/243 = 79.4% chance of a tie.
This concludes the analysis of the game. In Cantonese, both players would both say something like “Xxx, Xxx, xxxxXxx” (inscrutable Chinese) where the uppercase letters indicate when the steps happen. I wonder how this gameplay can be translated into English exactly.