Probability, Steady State Frequencies in Monopoly

Steady State Frequencies in Monopoly

When a system consists of a finite number of states, and transitions between those states, having certain probabilities, the system is called a markov chain, after Andrei Markov, a Russian mathematician of the late 19^th century. Unless the markov chain is somewhat pathological, one can derive a steady state frequency distribution, i.e. the odds of being in each state once the system reaches equilibrium. We will illustrate with the game of Monopoly, which is well known to most people. BTW, this is a beautiful application of eigen vectors. You need to be familiar with matrices and eigen vectors for this to make sense.

Place your token on GO, and roll the dice. A density function describes the probability of winding up on each of the 40 properties. Let's number them 0 to 39, with GO = 0 and Boardwalk = 39.

Consider Connecticut Ave, property number 9. You need to roll a 9 to land on Connecticut, and that happens with probability 4/36. If the density function is d(), then d(9) = 4/36.

What about d(39), the odds of landing on Boardwalk? You might think you can't get to Boardwalk in one roll, but there is a way. Roll a 7 and land on Chance, then pick up the card that says, "Advance Token to Boardwalk". So you need to roll a 7, then pick up one particular card out of 15. Therefore d(39) = 6/36 times 1/15 = 1/90. Other remote squares such as Illinois Ave, Pennsylvania Railroad, and even GO itself, are also accessible. You just have to run through all the cases.

The density function so derived is the density function associated with GO. It represents the odds of landing on various squares if you start from GO. I will call this function d₀(). If you start on Mediterranean Ave, you can derive a new density function, that we will call d₁(). This describes the odds of landing on all the squares, when starting at position 1. If your token is on Reading Railroad, you will be intrested in d₅(). For instance, d₅(15) gives the odds of jumping from Reading Railroad to Pennsylvania Railroad. This is pretty close to 3/36, the odds of rolling a 10, but again, you could roll snake eyes and land on Chance, and draw the card that tells you to "Advance Token to Nearest Railroad." So d₅(10) is actually 0.085185.

Note that d_x(30) = 0. There is no way to land on the policeman, since he immediately sends you over to jail. For convenience, let d₃₀(x) = 0, since you can't start on the policeman and roll the dice.

Construct a density function for each of the 40 properties, then arrange them in a 40 by 40 matrix, which we will call M. In other words, M_i,j gives the odds of starting on property i and winding up on property j after one turn. Remember that row 30 and column 30 are zero, thanks to the policeman. Now - let's see how we can use this matrix.

Let v be a row vector of length 40, with a one in the first position and zero elsewhere. This vector represents a token sitting on GO. Multiply v times M, written vM, to get a new row vector that shows the probability of landing on each of the 40 squares. Basically, multiplying by v extracts the first row of M, the first density function, or d₀. This is indeed the odds of landing on properties when you start at GO. If v has its "one" in position x, because your token is sitting on property x, then vM extracts row x from the matrix, and produces d_x.

Now take the next step. You aren't sure if your token is on Connecticut or New York. It's a toss up. The odds are 50% either way. So let the vector v contain the number 0.5 in positions 9 and 19; it is zero everywhere else. Multiply vM and obtain ½d₉ + ½d₁₉. This tells you (probabilistically) where you will wind up, after your next turn.

Generalize this to an arbitrary probability vector v. You aren't sure of your location on the board, but you have probabilities for where you might be, and you want to know where you are going to wind up after your next turn. Your current location is represented by the probability vector v, whose entries sum to 1. For instance, you are on GO with probability v₀, jail with probability v₁₀, Free Parking with probability v₂₀, and so on. The matrix product vM gives your new location. Extending this idea, vMM gives your location after two turns, vMMM gives your location after three turns, and so on.

At this point I'm going to gloss over a large chunk of math, and assert, without proof, that the system does reach equilibrium. After you've been around the board 100 times there is a certain frequency distribution, a certain probability of being on each of the 40 properties, and this does not change. Your fuzzy quantum location, all around the board, remains the same, turn after turn. This is our steady state distribution. If it is represented by a probability vector s, then sM = s. Aha, s is an eigen vector, with eigen value 1. The steady state frequencies can be derived by finding the eigen vector s of M.

I'm not fond of manipulating 40x40 matricies by hand; it's a task best left to a computer. Even formulating the matrix is a bit tedious, so I will simply trust the work of others. The table below lists the squares and their frequencies in order, as you travel around the board.

There is a big spike at Jail, but you can't buy the jail. The top 11 purchasable properties are all utilities, oranges, and reds. No wonder I always liked these sets. Funny, my brother preferred the yellows and greens, though they are expensive to build, so he had to acquire early, and build early, in order to win. In contrast, I could get my orange set (somewhat) late in the game, put up hotels quickly (for half the price of the greens), and pull ahead. Well, no matter the strategy, lady luck plays a large role.