## In the game, Monopoly, the standard board is set up in the following way:

A player starts on the GO square and adds the scores on two 6-sided dice to determine the number of squares they advance in a clockwise direction. Without any further rules we would expect to visit each square with equal probability: 2.5%. However, landing on G2J (Go To Jail), CC (community chest), and CH (chance) changes this distribution.

In addition to G2J, and one card from each of CC and CH, that orders the player to go directly to jail, if a player rolls three consecutive doubles, they do not advance the result of their 3rd roll. Instead they proceed directly to jail.

At the beginning of the game, the CC and CH cards are shuffled. When a player lands on CC or CH they take a card from the top of the respective pile and, after following the instructions, it is returned to the bottom of the pile. There are sixteen cards in each pile, but for the purpose of this problem we are only concerned with cards that order a movement; any instruction not concerned with movement will be ignored and the player will remain on the CC/CH square.

Community Chest (2/16 cards):

Advance to GO

Go to JAIL

Chance (10/16 cards):

Advance to GO

Go to JAIL

Go to C1

Go to E3

Go to H2

Go to R1

Go to next R (railway company)

Go to next R

Go to next U (utility company)

Go back 3 squares.

The heart of this problem concerns the likelihood of visiting a particular square. That is, the probability of finishing at that square after a roll. For this reason it should be clear that, with the exception of G2J for which the probability of finishing on it is zero, the CH squares will have the lowest probabilities, as 5/8 request a movement to another square, and it is the final square that the player finishes at on each roll that we are interested in. We shall make no distinction between “Just Visiting” and being sent to JAIL, and we shall also ignore the rule about requiring a double to “get out of jail”, assuming that they pay to get out on their next turn.

By starting at GO and numbering the squares sequentially from 00 to 39 we can concatenate these two-digit numbers to produce strings that correspond with sets of squares.

Statistically it can be shown that the three most popular squares, in order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO (3.09%) = Square 00. So these three most popular squares can be listed with the six-digit modal string: 102400.

If, instead of using two 6-sided dice, two 4-sided dice are used, find the six-digit modal string.

### The rules are similar regardless of the sides on the die. Therefore, the mechanism to solve the problem doesn’t change – we just adjust the probabilities for the die rolls.

With two 4-sided dice, there are 16 possible outcomes of a roll, each equally likely. We need to adjust the probability of moving a given number of spaces based on these outcomes. In addition, we need to use a Markov Chain to keep track of the probabilities of landing on each square on the board, given the current state of the game.

The Community Chest and Chance cards will change the probabilities. For Community Chest, there is a 2 out of 16 chance of going to a designated square (Go or Jail). For Chance, there is a 10/16 chance of going to a designated square (different locations).

We need to update the probability distribution after each roll.

For example: If our current state is that we’re on the GO square, after one roll we have a 1/16 chance of being on each of squares 2 through 9, a slightly higher chance of being on square 10 (due to the chance of going to jail from the community chest or rolling a double three times), and a 7/16 chance of remaining on the GO square (due to the chance of rolling a total of 1, moving to a chance/community chest square and drawing a card to go back to GO, or rolling a double three times and going to jail).

We continue this update for each possible square we could be on (treating Jail and Just Visiting as the same square), until the distribution stabilises. The distribution becomes stable after enough updates – this is due to the fact that the Markov Chain associated with Monopoly is ergodic (both aperiodic and irreducible), which means that it tends towards a stationary distribution.

The six-digit modal string will be the squares with the highest probabilities in the stationary distribution, which is the limit of our updating process.

As this is a highly computation-dependent problem, we can’t provide an exact answer without programming and computing power. However, the process will be similar to the one described above.

##### More Answers:

Path Sum: Two WaysPath Sum: Three Ways

Path Sum: Four Ways