monopoly (idea)

The squares most landed on in an average game of Monopoly are:

British board:
1. Trafalgar Square (Red)
2. Go
3. Fenchurch Street Station
4. Free Parking
5. Marlborough Street (Orange)
6. Vine Street (Orange)
7. King's Cross Station
8. Bow Street (Orange)
9. Water Works
10. Marylebone Station

American board:
1. Illinois Avenue (Red)
2. Go
3. B. & O. Railroad
4. Free Parking
5. Tennessee Avenue (Orange)
6. New York Avenue (Orange)
7. Reading Railroad
8. St. James Place (Orange)
9. Water Works
10. Pennsylvania Railroad

I don't have my hands on the relative percentages yet, but I'd be interested to hear from anyone who does.

Many players new to Monopoly automatically assume that each square has an equal chance of being landed on. However, this is certainly not the case, due to a number of features of the rules. To outline the main ones:

The Jail

There are a large number of ways to end up in jail: landing on Go To Jail, getting a Go Directly To Jail card, or rolling three consecutive doubles. Once you go to jail, you are forced to go past the orange squares again, which greatly increases the likelihood of landing on them and thus partially accounts for the presence of all three orange squares in the top 10.

The Chance and Community Chest cards

As with the jail, these distort probabilities by sending you directly to certain squares. These two square types account for 6/40 = 15% of squares, and 30% of the cards require moving the piece to a new square, so these cards clearly have quite a large effect. That stat was based on my own set which is probably missing some cards.

Unequal probabilities of each dice score

One die has an equal chance of turning up any of the six scores, but with two dice, there is a 1/6 chance of getting a total of 7 but only a 1/36 chance of getting a 2. The reason for this is high-school probability theory: there are six ways of making 7 with two dice (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), but only one way of making 2 (1+1). Therefore, landing on squares that are multiples of seven from Go is more likely.

This is largely counter-acted, however, because 40mod7 = 5; in other words, once you complete a full circuit of the board, the typical starting square is different, and the uneven distribution of the dice rolls is less important. So this is not as important a reason as you might think.

How was this worked out?

It's a very simple computer simulation. You work out the average number of goes in a game, then leave the computer to play a large number of games, of that length, with itself. At the end, you should have a good idea of which squares are landed on most, by way of a simple tally. Fortunately, the element of choice in where you actually land on the board is very small, so the computer needs little in the way of AI.

Data source: The Top 10 of Everything 1997