I'm sure this analysis is not original.
The question to be answered is the following:
Which squares on a classic Monopoly game board are the most and least likely to be landed on?
I answered this question using a Markov Chain approach.
Markov Chains: Overview
A Markov Chain is a stochastic (i.e., random, or probabilistic) process wherein a system's "state" changes probabilistically over time. A key characteristic of a Markov Chain is that it is a "memoryless" process. That is, it doesn't matter which other states the system has been in before its current state in order to identify the probable state it will be in next; all that matters is the system's current state. The formal definition of a Markov Chain is any process that satisfies the following property:
P(Xn+1 = s | Xn, Xn-1, ... , X1) = P(Xn+1 = s | Xn)
where Xn is a random variable representing the system's state at time n, and s is a particular state. This property is called the Markov Property.
The game of Monopoly generally satisfies the Markov Property. If you are on a given square, it doesn't matter how you got there in order to determine the probable squares you could travel to next.
How a Markov Chain Model Can be Used for Monopoly
Markov Chains use the concept of "states" to describe a system. The collection of all states in a Markov Chain is known as the "state space". For any stochastic system, a state space can be defined such that the Markov Property holds. In Monopoly, it is fairly easy and intuitive to define the state space simply as the distinct squares of the board. Thus, you could define 41 possible states in a Monopoly Markov Chain. (There are 40 squares on the board, but "Jail (Just Visiting)" and "Jail" are considered two distinct states because they can function differently depending on the player's strategy for exiting Jail.)
A Markov Chain can be represented by a square matrix called the "transition matrix". The entries in such matrices represent the probability of transitioning to state j from state i. The originating states (state i) are the rows of the matrix, and the destination states (state j) are the columns. Because a system in state i must move to some other state j (even if j = i), the sum of each row in a transition matrix is 1. An example transition matrix for a three-state Markov Chain is provided below.
In the above example, if the system is in State 1 (row 1), there is a probability of 0.3 of remaining in State 1 at the next transition; a probability of 0.5 of transitioning to State 2; and a probability of 0.2 of transitioning to State 3. Notice that each row sums to 1.
In Monopoly, if a player is on a given square, the transition probabilities available to the player generally follow the distribution of rolling two six-sided dice to advance between 2 and 12 squares forward. This is not true for every square, such as Chance and Community Chest squares, or when doubles are rolled for the third time in a row, but we'll get back to that.
The Steady-State Distribution
If a transition matrix, denoted as P, has certain properties, then it leads to some interesting outcomes. Specifically, there is a theorem stating that if a Markov Chain is "irreducible" and "aperiodic", then there exists a unique vector 𝞹 such that 𝞹*P = 𝞹 and all the elements of 𝞹 sum to 1. This is called the "steady-state" distribution, in that it represents the convergence of the distribution of the system's state in the long-term. That is, 𝞹 is essentially a vector containing the probabilities (𝞹i) that the system will be in state i at any given point in time.
For the 3x3 example above, the vector 𝞹 = [0.4113475 0.2695035 0.3191489] (rounded) represents the steady-state distribution of the Markov Chain. The Markov Chain is irreducible and aperiodic (trust me for now; we'll discuss these in a bit), and the vector 𝞹 satisfies the equation 𝞹*P = 𝞹. We thus see that State 1 is the most frequently visited state in the long term and State 2 the least frequently visited.
In the case of Monopoly, 𝞹 represents the probability of a player being on any given square as the number of turns approaches infinity. In other words, this vector tells us the long-term likelihood of landing on all squares--precisely the question we sought to answer.
Before we move on, though, we need to be sure that our matrix P that we establish for a Monopoly board satisfies the necessary properties: irreducibility and aperiodicity.
"Irreducibility" means that the system can go from any given state to any other state after any number of steps. If you have ever played Monopoly, you know that this is true, as every square is landed on multiple times throughout a game. For example, you obviously could not reach Marvin Gardens from Baltic Avenue in one turn, but you could reach it in three turns by rolling 8, 11, 7; or four turns by rolling 6, 7, 9, 4; or seven turns by rolling 12, 12, 5, 10, 9, 11, 7, rounding the board in the process. (A similar exercise could be conducted for every possible pair of squares.) All that matters is that you can eventually reach any other square from a given starting square. Therefore, a Monopoly board is irreducible.
"Aperiodicity" means that the system doesn't get stuck in a fixed-timestep loop between a subset of states. That is, if you are in a given state, you don't know how many transitions you will have to make before you return to that state. An example of periodicity in Monopoly would be if a rule existed such that every time you landed on Park Place, your next six dice rolls were required to total 40, thus landing you on Park Place every sixth turn. Obviously there are no such rules in the game, as advancement around the board is defined by probabilistic dice rolls with no guarantee of the next time you will return to a square. Therefore, a Monopoly board is aperiodic.
So, we see that a Monopoly board Markov Chain satisfies the properties of irreducibility and aperiodicity. Consequently, a steady-state distribution 𝞹 must exist for such a Markov Chain.
For you linear-algebra-minded folk, the theorem alternatively states that for an irreducible and aperiodic Markov Chain with transition matrix P, one of the eigenvalues for the matrix P is 1 with a corresponding eigenvector of 𝞹 such that all the elements of 𝞹 are non-negative and sum to 1.
Defining the Monopoly Markov Chain Transition Matrix
Okay, underlying theory out of the way. Now for application.
A Markov Chain is defined by its state space and transition matrix. For a Monopoly board, we have already identified our state space as the 41 possible squares. Now we must define our transition matrix.
As mentioned before, for most squares on the board, the transition probabilities generally follow the symmetric distribution of the possible outcomes of rolling two dice. This distribution is as follows:
P(roll 3) = 2/36
P(roll 4) = 3/36
P(roll 5) = 4/36
P(roll 6) = 5/36
P(roll 7) = 6/36
P(roll 8) = 5/36
P(roll 9) = 4/36
P(roll 10) = 3/36
P(roll 11) = 2/36
As an example, consider what happens when a player is on Virginia Avenue--a typical square that does not have any special movement rules. This is State 16 in the state space. With a transition probability of 1/36, the next state will be State 18 (St. James Place), which represents what happens if a player rolls snake eyes. With a transition probability of 2/36, the next state will be State 19 (Community Chest), which represents what happens if a player rolls a 3 (and so on). For other squares on the board that are not reachable from Virginia Avenue with a single roll (such as North Carolina Avenue), the transition probability is 0.
Simple enough. But what about special squares from which movement can be irregular?
Wrinkle 1: Community Chest and Chance
There are three Community Chest squares and three Chance squares on a typical board. Upon landing on these squares, a player draws from the respective deck of cards and follows the instructions. Some of these cards instruct the player to move to other squares.
In the Community Chest deck, two of the 16 cards require the player to move to another square. One card sends the player to Go, and the other sends the player to Jail. If any of the other 14 cards is drawn, the player remains on the square and will continue the next turn with a standard roll of the dice.
The following tree illustrates what can happen on a Community Chest square and the corresponding probabilities of each outcome.
The following tree illustrates what can happen on a Chance square and the corresponding probabilities of each outcome.
In this process, we have derived what are known as conditional probabilities. Conditional probabilities answer questions such as: What is the likelihood of advancing to a given square, given that a player drew a certain card?
When a game of Monopoly begins, the Community Chest and Chance card decks are shuffled, and then the order of the cards usually remains fixed for the course of the game. But the way the Markov Chain model is set up assumes that the cards are randomly drawn without regard to a specific, fixed order. Thus, it would be theoretically possible to draw "Go to Jail" twice in a row from the same deck in this model. However, because this model is intended to be inclusive of all possible gameplays, and thus all possible orderings of the decks, this nuance may be insignificant in the limit.
Nevertheless, this is still a simplifying assumption that we have made. If we were to model it more accurately, our state space would need to be much bigger. Our state space would need to be defined to contain the information of the current square, all possible orderings of these card decks, and which cards had already been drawn. Our state space would thus consist of 41*(16!)2 states. In addition, defining the Markov Chain in this way (i.e., with a fixed card ordering) would destroy the system's irreducibility, as the system could never reach a state containing a different card ordering.
Wrinkle 2: Triple Doubles
Another rule in Monopoly is that if a player rolls doubles three times in a row, then the player is automatically sent to Jail. In order to fully and accurately include this in the Markov Chain model, the state space would have to be modified to contain the information of the current square and the previous two dice rolls. Similar to the random ordering of the Community Chest and Chance cards, we will make a simplifying assumption to account for this rule as well.
The probability of rolling doubles on any given roll is 1/6. Each doubles pair has a probability of 1/36, and there are six such pairs which count as doubles. Thus, the probability of rolling doubles three times in a row is 1/6 * 1/6 * 1/6, or 1/216. Similar to the conditioning performed for Community Chest and Chance cards, we will condition on whether the roll is a third double roll. The following tree illustrates how the conditional probabilities are calculated.
This addition slightly modifies the preliminary entries in the transition matrix by multiplying each of the dice roll probabilities by 215/216. It also makes Jail accessible from any other square with probability 1/216. The only square that is unaffected by this change is Jail itself. When a player exits Jail, the doubles counter is "reset", and thus it is not possible to be sent to Jail by a third doubles roll from Jail.
Wrinkle 3: Strategy for Exiting Jail
There are two ways that a player can choose to exit Jail: either by paying $50 (or using a "Get out of Jail free" card) and rolling the dice as usual, or waiting to roll doubles, after which the player exits for free and advances the number of spaces returned from the doubles roll. A player can change the strategy at any time. For example, a player may unsuccessfully attempt to get out for free via doubles for three rolls, but then decide to pay the $50 because the player doesn't want to spend any more turns in Jail.
To account for each of these strategies, two transition matrices must be developed. The first assumes that a player always pays $50 for immediate exit and thus may advance anywhere from 2 to 12 squares from Jail following the typical dice roll probability distribution. The second assumes that a player always waits for doubles to exit and thus may advance 2, 4, 6, 8, 10, or 12 squares from Jail, each with equal probability of 1/6. The steady-state distributions for each of these two matrices will be identified, and the effects of each of these strategies on square frequency can be identified.
Sample Entries from the Transition Matrix
All the information needed to create two 41x41 transition matrices is now available. These two matrices are identical except for the row that represents "Jail" as the current state. Examples of entries from these matrices are provided below and are given as probabilities.
Current square: Go
P(Community Chest 1 | Go) = (215/216) * (1/36) [third doubles was not rolled AND a 2 was rolled]
P(Baltic Avenue | Go) = (215/216) * (2/36) [same logic]
P(Income Tax | Go) = (215/216) * (3/36)
P(Reading Railroad | Go) = (215/216) * (4/36)
P(Oriental Avenue | Go) = (215/216) * (5/36)
P(Chance 1 | Go) = (215/216) * (6/36)
P(Vermont Avenue | Go) = (215/216) * (5/36)
P(Connecticut Avenue | Go) = (215/216) * (4/36)
P(Jail (Visiting) | Go) = (215/216) * (3/36)
P(St. Charles Place | Go) = (215/216) * (2/36)
P(Electric Company | Go) = (215/216) * (1/36)
P(all other squares | Go) = 0
Current square: Community Chest 2 (between St. James Place and Tennessee Avenue)
P(Go | Community Chest 2) = (2/16) * (1/2) ["Advance to Go" card drawn]
P(New York Avenue | Community Chest 2) = (215/216) * (14/16) * (1/36) [third doubles was not rolled AND non-movement card was drawn AND a 2 was rolled]
P(Free Parking | Community Chest 2) = (215/216) * (14/16) * (2/36) [same logic]
P(Kentucky Avenue | Community Chest 2) = (215/216) * (14/16) * (3/36)
P(Chance 2 | Community Chest 2) = (215/216) * (14/16) * (4/36)
P(Indiana Avenue | Community Chest 2) = (215/216) * (14/16) * (5/36)
P(Illinois Avenue | Community Chest 2) = (215/216) * (14/16) * (6/36)
P(B&O Railroad | Community Chest 2) = (215/216) * (14/16) * (5/36)
P(Atlantic Avenue | Community Chest 2) = (215/216) * (14/16) * (4/36)
P(Ventnor Avenue | Community Chest 2) = (215/216) * (14/16) * (3/36)
P(Water Works | Community Chest 2) = (215/216) * (14/16) * (2/36)
P(Marvin Gardens | Community Chest 2) = (215/216) * (14/16) * (1/36)
P(all other squares | Community Chest 2) = 0
Current square: Atlantic Avenue
P(Jail | Atlantic Avenue) = 1/216 [third doubles rolled]
P(Water Works | Atlantic Avenue) = (215/216) * (1/36) [third doubles was not rolled AND a 2 was rolled]
P(Marvin Gardens | Atlantic Avenue) = (215/216) * (2/36) [same logic]
P(Go to Jail | Atlantic Avenue) = (215/216) * (3/36)
P(Pacific Avenue | Atlantic Avenue) = (215/216) * (4/36)
P(North Carolina Avenue | Atlantic Avenue) = (215/216) * (5/36)
P(Community Chest 3 | Atlantic Avenue) = (215/216) * (6/36)
P(Pennsylvania Avenue | Atlantic Avenue) = (215/216) * (5/36)
P(Short Line Railroad | Atlantic Avenue) = (215/216) * (4/36)
P(Chance 3 | Atlantic Avenue) = (215/216) * (3/36)
P(Park Place | Atlantic Avenue) = (215/216) * (2/36)
P(Luxury Tax | Atlantic Avenue) = (215/216) * (1/36)
P(all other squares | Atlantic Avenue) = 0
Current square: Chance 3 (between Short Line Railroad and Park Place)
P(Go | Chance 3) = (10/16) * (1/10) ["Advance to Go" card drawn] + (215/216) * (6/16) * (3/36) [third doubles was not rolled AND non-movement card was drawn AND a 4 was rolled]
P(Illinois Avenue | Chance 3) = (10/16) * (1/10) ["Advance to Illinois Avenue" card drawn]
P(St. Charles Place | Chance 3) = (10/16) * (1/10) ["Advance to St. Charles Place" card drawn]
P(Boardwalk | Chance 3) = (10/16) * (1/10) ["Advance to Boardwalk" card drawn] + (215/216) * (6/16) * (2/36) [third doubles was not rolled AND non-movement card was drawn AND a 3 was rolled]
P(Reading Railroad | Chance 3) = (10/16) * (1/10) ["Advance to Reading Railroad" card drawn] + (10/16) * (2/10) ["Advance to nearest railroad" card drawn] + (215/216) * (6/16) * (4/36) [third doubles was not rolled AND non-movement card was drawn AND a 9 was rolled]
P(Electric Company | Chance 3) = (10/16) * (1/10) ["Advance to nearest utility" card drawn]
P(Luxury Tax | Chance 3) = (215/216) * (6/16) * (1/36) [third doubles was not rolled AND a 2 was rolled]
P(Mediterranean Avenue | Chance 3) = (215/216) * (6/16) * (4/36) [same logic]
P(Community Chest 1 | Chance 3) = (215/216) * (6/16) * (5/36)
P(Baltic Avenue | Chance 3) = (215/216) * (6/16) * (6/36)
P(Income Tax | Chance 3) = (215/216) * (6/16) * (5/36)
P(Oriental Avenue | Chance 3) = (215/216) * (6/16) * (3/36)
P(Chance 1 | Chance 3) = (215/216) * (6/16) * (2/36)
P(Vermont Avenue | Chance 3) = (215/216) * (6/16) * (1/36)
P(all other squares | Chance 3) = 0
Current square: Go to Jail
P(all other squares | Go to Jail) = 0
Current square: Jail (pay $50 to exit)
P(Electric Company | Jail) = 1/36 [A 2 was rolled]
P(States Avenue | Jail) = 2/36 [same logic]
P(Virginia Avenue | Jail) = 3/36
P(Pennsylvania Railroad | Jail) = 4/36
P(St. James Place | Jail) = 5/36
P(Community Chest 2 | Jail) = 6/36
P(Tennessee Avenue | Jail) = 5/36
P(New York Avenue | Jail) = 4/36
P(Free Parking | Jail) = 3/36
P(Kentucky Avenue | Jail) = 2/36
P(Chance 2 | Jail) = 1/36
P(all other squares | Jail) = 0
Current square: Jail (roll doubles to exit)
P(Electric Company | Jail) = 1/6
P(Virginia Avenue | Jail) = 1/6
P(St. James Place | Jail) = 1/6
P(Tennessee Avenue | Jail) = 1/6
P(Free Parking | Jail) = 1/6
P(Chance 2 | Jail) = 1/6
P(all other squares | Jail) = 0
Computing the Steady-State Distribution
Now that the transition matrix (or matrices) P has been defined, and we know that the Markov Chain it represents is irreducible and aperiodic, we can find the steady-state distribution 𝞹 that will reveal which squares have the highest and lowest probabilities of being landed on in the long-term.
I said we were done with theory, but there's actually one more piece that we need.
There is another theorem stating that if a Markov Chain is irreducible and aperiodic, and a steady-state distribution exists, then Pt(i,j) converges to 𝞹j for all i as t approaches infinity. In English, this means that as P is multiplied with itself over and over again, the entries in each column converge to the same value, and that the value in the jth column is the steady-state probability for state j.
To see this in action, consider the 3x3 matrix at the beginning of this post. If we multiply this matrix by itself, the result is as follows:
(All entries have been rounded to three decimals.)
If we continue this matrix multiplication, then eventually the entries in each column will converge to the steady-state distribution, which was 𝞹 = [0.4113475 0.2695035 0.3191489], which we already see beginning to occur as the power to which P is raised increases. This occurs because the Markov Chain represented by the matrix was irreducible and aperiodic, and a steady-state distribution was guaranteed to exist.In the case of our Monopoly transition matrix (or matrices), we know that a steady-state distribution exists. Therefore, we also know that one row of Pn for a very high value of n will be close to the steady-state distribution of the Markov Chain as we have defined it.
Results
Recall that there are two transition matrices for our Monopoly Markov Chain. Matrix 1 represents Case 1, where the player's strategy to exit Jail is to always pay $50 and take a regular roll. Matrix 2 represents Case 2, where the player's strategy to exit Jail is always to wait for doubles to be rolled. The transition matrices for each of these cases were multiplied by themselves dozens of times, and the first row of the final matrix was taken as the steady-state distribution.
The results for these two cases are provided below.
Case 1: Pay $50 to exit Jail
Most likely to least likely squares:
- Jail
- Illinois Avenue
- Go
- New York Avenue
- B & O Railroad
- Community Chest 2 (between St. James Place and Tennessee Avenue)
- Reading Railroad
- Tennessee Avenue
- Pennsylvania Railroad
- Free Parking
- Kentucky Avenue
- Water Works
- St. James Place
- Chance 2 (between Kentucky Avenue and Indiana Avenue)
- Indiana Avenue
- Atlantic Avenue
- Community Chest 3 (between North Carolina Avenue and Pennsylvania Avenue)
- St. Charles Place
- Ventnor Avenue
- Pacific Avenue
- Go to Jail
- Boardwalk
- North Carolina Avenue
- Electric Company
- Marvin Gardens
- Pennsylvania Avenue
- Virginia Avenue
- Short Line Railroad
- States Avenue
- Income Tax
- Vermont Avenue
- Chance 3 (between Short Line Railroad and Park Place)
- Chance 1 (between Oriental Avenue and Vermont Avenue)
- Connecticut Avenue
- Jail (Just Visiting)
- Oriental Avenue
- Park Place
- Luxury Tax
- Baltic Avenue
- Community Chest 1 (between Mediterranean Avenue and Baltic Avenue)
- Mediterranean Avenue
The most likely state (Jail) has a steady-state probability of 3.722%, and the least likely state (Mediterranean Avenue) has a steady-state probability of 1.968%.
Case 2: Must roll doubles to exit Jail
Most likely to least likely squares:
- Jail
- Chance 2 (between Kentucky Avenue and Indiana Avenue)
- Free Parking
- Electric Company
- Illinois Avenue
- Tennessee Avenue
- Go
- B & O Railroad
- Reading Railroad
- St. James Place
- Water Works
- Virginia Avenue
- New York Avenue
- St. Charles Place
- Community Chest 3 (between North Carolina Avenue and Pennsylvania Avenue)
- Indiana Avenue
- Pacific Avenue
- Atlantic Avenue
- Boardwalk
- Ventnor Avenue
- Go to Jail
- Kentucky Avenue
- North Carolina Avenue
- Marvin Gardens
- Pennsylvania Railroad
- Pennsylvania Avenue
- Short Line Railroad
- Community Chest 2 (between St. James Place and Tennessee Avenue)
- Income Tax
- Vermont Avenue
- Chance 1 (between Oriental Avenue and Vermont Avenue)
- Chance 3 (between Short Line Railroad and Park Place)
- Connecticut Avenue
- Jail (Just Visiting)
- Oriental Avenue
- Park Place
- Luxury Tax
- Baltic Avenue
- States Avenue
- Community Chest 1 (between Mediterranean Avenue and Baltic Avenue)
- Mediterranean Avenue
The most likely state (Jail) has a steady-state probability of 3.712%, and the least likely state (Mediterranean Avenue) has a steady-state probability of 1.959%.
Case 1 1/2: Average steady-state probabilities of Case 1 and Case 2
Most likely to least likely squares:
- Jail
- Illinois Avenue
- Go
- Free Parking
- B & O Railroad
- Chance 2 (between Kentucky Avenue and Indiana Avenue)
- Tennessee Avenue
- Reading Railroad
- New York Avenue
- Electric Company
- St. James Place
- Water Works
- Kentucky Avenue
- Pennsylvania Railroad
- St. Charles Place
- Indiana Avenue
- Community Chest 3 (between North Carolina Avenue and Pennsylvania Avenue)
- Atlantic Avenue
- Pacific Avenue
- Community Chest 2 (between St. James Place and Tennessee Avenue)
- Ventnor Avenue
- Virginia Avenue
- Boardwalk
- Go to Jail
- North Carolina Avenue
- Marvin Gardens
- Pennsylvania Avenue
- Short Line Railroad
- Income Tax
- Vermont Avenue
- Chance 3 (between Short Line Railroad and Park Place)
- Chance 1 (between Oriental Avenue and Vermont Avenue)
- Connecticut Avenue
- Jail (Just Visiting)
- States Avenue
- Oriental Avenue
- Park Place
- Luxury Tax
- Baltic Avenue
- Community Chest 1 (between Mediterranean Avenue and Baltic Avenue)
- Mediterranean Avenue
Findings
Considering that Jail is the most frequently landed on square, it makes you wonder what kind of people Parker Brothers thought landlords were.
Some of the differences between Cases 1 and 2 are quite stark. For instance, States Avenue is #29 in Case 1, but #39 in Case 2. Its low position in Case 2 (relying on doubles to exit Jail) is likely because States Avenue is 3 squares away from Jail, so it cannot be reached upon exit from Jail with one roll. Furthermore, it cannot even be reached within two rolls of exiting Jail in Case 2.
In Case 2, the player's exclusive reliance on doubles to exit Jail results in high rankings for those properties that are within the reach of a doubles roll from Jail: Electric Company (#4), Virginia Avenue (#12), St. James Place (#10), Tennessee Avenue (#6), Free Parking (#3), and Chance 2 (#2). Notice that for this list, as the distance from Jail increases, the ranking increases (except for the Electric Company, which a player can be sent to via a Chance card). This occurs because these squares can be reached by one roll or by two rolls, and the two rolls may have many possible combinations adding up to the square's distance from Jail.
In both Case 1 and Case 2, squares such as Illinois Avenue and B & O Railroad score highly because they are accessible from Jail with 2-3 rolls, and there are many combinations of rolls that can result in one landing on those squares. A player can also be sent to each of these squares via Chance cards.
Mediterranean Avenue and Baltic Avenue, the first properties after Go, are among the least-visited properties (with Mediterranean being the absolute least of all). These are also the lowest-value properties on the board. The combination of those two factors should persuade a player that these properties are not worthwhile investments.
An interesting version of Monopoly would be one based on the principle of "low risk/low reward, and high risk/high reward". In this version of the game, comparatively rarely landed on properties such as Mediterranean and Baltic would be the most expensive properties on the board and would have a high payoff, while the most frequently landed on properties such as Illinois and Tennessee would be the cheapest properties on the board and would have a lower payoff.
Monopoly is primarily about obtaining groups of properties--a monopoly on a certain area of town. So, which monopolies are most likely to be visited? For each Case, the sums of the constituent steady-state probabilities in each monopoly were obtained and ranked.
Case 1
- Railroads: 10.13% (Reading, Pennsylvania, B & O, Short Line)
- Orange: 8.13% (St. James, Tennessee, New York)
- Red: 7.80% (Kentucky, Indiana, Illinois)
- Yellow: 7.27% (Atlantic, Ventnor, Marvin Gardens)
- Green: 7.17% (Pacific, North Carolina, Pennsylvania)
- Magenta: 7.10% (St. Charles, States, Virginia)
- Light blue: 6.33% (Oriental, Vermont, Connecticut)
- Utilities: 5.52% (Electric Company, Water Works)
- Navy: 4.45% (Park Place, Boardwalk)
- Purple: 3.95% (Mediterranean, Baltic)
Case 2
- Railroads: 10.51% (Reading, Pennsylvania, B & O, Short Line)
- Orange: 8.15% (St. James, Tennessee, New York)
- Red: 8.09% (Kentucky, Indiana, Illinois)
- Yellow: 7.36% (Atlantic, Ventnor, Marvin Gardens)
- Green: 7.20% (Pacific, North Carolina, Pennsylvania)
- Magenta: 6.96% (St. Charles, States, Virginia)
- Light blue: 6.34% (Oriental, Vermont, Connecticut)
- Utilities: 4.99% (Electric Company, Water Works)
- Navy: 4.44% (Park Place, Boardwalk)
- Purple: 3.96% (Mediterranean, Baltic)
Case 1 1/2
- Railroads: 10.32% (Reading, Pennsylvania, B & O, Short Line)
- Orange: 8.14% (St. James, Tennessee, New York)
- Red: 7.94% (Kentucky, Indiana, Illinois)
- Yellow: 7.32% (Atlantic, Ventnor, Marvin Gardens)
- Green: 7.18% (Pacific, North Carolina, Pennsylvania)
- Magenta: 7.03% (St. Charles, States, Virginia)
- Light blue: 6.33% (Oriental, Vermont, Connecticut)
- Utilities: 5.26% (Electric Company, Water Works)
- Navy: 4.45% (Park Place, Boardwalk)
- Purple: 3.96% (Mediterranean, Baltic)
The totals do not sum to 100% because of the non-property squares on the board.
Perhaps not surprisingly, the single four-property monopoly ranks the highest in all cases, and the three two-property monopolies rank the lowest. Certainly, the more squares comprising a monopoly, the higher the chance that a square within the monopoly will be visited.
Now, these rankings do not mean that these are the monopolies that should necessarily be sought out first. The expected return on these properties must also be considered. For instance, although the railroads are the monopoly most frequently landed on, the maximum income that can be obtained from them is $200 ($400 if a player is sent to a railroad via a Chance card), which is comparatively low against the $2,000 that can be obtained if a player lands on Boardwalk with a hotel. In addition, the payoff from a colored property (i.e., not railroads or utilities) is fairly minimal unless and until a player has a monopoly involving that property.
Nevertheless, a key takeaway is that the orange and red properties (St. James Place, Tennessee Avenue, New York Avenue, Kentucky Avenue, Indiana Avenue, and Illinois Avenue) are among the most-frequently landed upon squares, a claim which is generally common knowledge.
