The game of Morra: a fun alternative to rock-paper-scissors

People have long thought up ways to solve small disputes at random, from flipping a coin to playing a game of rock-paper-scissors.

The other day I came across another game that dates all the way back to the ancient Greeks and Romans.

The game is called Morra. While there are many variations on the game of Morra, there are typically a few common rules.

You don’t need any special equipment to play Morra: you just use your hands like. In the two-person game, each person’s move is to extend a certain number of fingers on their hand and simultaneously guess something about what the other player will show (either the number of fingers or the sum of all fingers). For instance, you might show two fingers and guess the other person will show three. Points or money are awarded for correct guesses.

Because of its simplicity, Morra is a fun alternative to rock-paper-scissors. The interesting part is the strategy is slightly more nuanced. Today I want to analyze and solve a specific version of the game of Morra.

The rules of this version of Morra

As I said before, Morra is a game with many variations. I read about this version of the game from the book The Compleat Strategyst, a Rand publication that is a good read about zero-sum games.

The rules of the game are this:

–There are two players
–Each player can extend either 1, 2, or 3 fingers
–Each player has to simultaneously guess what the other will show
–If exactly one person guesses correctly, the person wins the SUM of the numbers showing
–If neither guesses right, or both do, then the payoff is zero

An example of gameplay can help illustrate the rules.

Suppose you and I play Morra. Let’s say that I extend 2 fingers and guess 2, and you extend 2 fingers but guess 3. In that case, I was the only person who guessed correctly so I would win. I get a payoff that is the SUM of both of our hands, so I win 4 points (or winnings could be money).

As stated, Morra is zero sum game that can be solved using game theory analysis. Let us model the game and figure out the optimal strategy.

Analyzing the game

Each person has to do two things in the game: each has to extend 1, 2, or 3 fingers, and each has to guess 1, 2, or 3 for what the opponent is holding.

We can model each person’s strategy as an ordered pair (fingers to show, number to guess). For compactness, we can drop the parenthesis and commas and write 12 to mean “I will show 1 finger and I will guess 2″.

Since there are 3 choices for what to show, and 3 choices for what to guess, each player has a total of 9 strategies: 11, 12, 13, 21, 22, 23, 31, 32, 33

Now we can take the next step and write out a payoff matrix. If you and I both play 11, then each of us guess correctly, so the payoff is zero. If you play 12, and I play 11, then only I guess correctly, so I get the payoff of 2.

As each player has 9 strategies, there will be 81 payoff cells to calculate. I will spare you the tedious details, but here is the payoff matrix:

Because this is a zero-sum game, we only need to write the payoffs for player 1. Positive numbers represent times player 1 wins, and negative numbers correspond to times player 2 wins. The payoff matrix for player 2 is the exact same matrix, with each entry multiplied by -1.

Finding the solution

It would be very difficult to figure out the solution to this game analytically. Luckily we have numerical methods to solve this game that use something called linear programming.

I am going to omit the details fo the process and just get to the solution. (I was unable to find a suitable online solver for this 9×9 game–please let me know if you are aware of one).

The best strategy turns out to be a mix of just three choices. You only want to play the strategies 13, 22, and 31, and you play them with the ratio of 5:4:3.

In other words, the best strategy is:

–play 13 (show 1 finger and guess 3) with probability 5/12
–play 22 (show 2 fingers and guess 2) with probability 4/12
–play 31 (show 3 fingers and guess 1) with probability 3/12

The game turns out to be a bit of a rock-paper-scissors dynamic after all: you end up having three choices that are worth playing, and you want to mix amongst them.

Understanding the solution

If both players choose these strategies, then the expected value to each player is 0. This is not a surprise at all: Morra is a fair game just like rock-paper-scissors, and you can’t expect to win if your opponent is playing correctly.

Of course, if your opponent does not play correctly, then you are in a great position to profit.

If you play the equilibrium strategy, and the other person plays 11 by mistake, then you can expect a profit of 0.17 (if the game is played in dollars, then you can expect a profit of 17 cents). The same is true if your opponent mistakenly plays 33.

Similarly, if your opponent mistakenly plays 21 or 23, then you can expect to profit by 0.08.

Notice it is not obvious that playing 11, 33, 21, or 23 is a mistake unless you compute the optimal strategy and see it.

So if you want to be a bit devious, then you should teach your friends Morra and not tell them the best strategy. As described in the Compleat Strategyst,

“It is probably a good game to teach your friends, since the solution is easy to memorize, and yet difficult to intuit.”

Competitive Morra

Morra is a social game and other versions are played with three or even four people.

I came across a couple videos where people are playing Morra. They are speaking in Italian so I don’t know exactly how they are scoring or which version they are playing. But the game has an incredible rhythm to it, and I love how quickly they are playing.

If you’re going to be playing this fast, you better know the best strategy and play it appropriately or you will lose very quickly!

Youtube Video: Morra