The Likeliest Monopoly Square

This week’s Fiddler is about rolling dice in the board game Monopoly.

We have a square board with 40 individual spaces around it, numbered from 0 to 39. All players begin on space 0 (akin to the “Go” square in Monopoly) and roll a pair of dice to determine how many spaces they advance each turn. However, unlike Monopoly, there is no way to otherwise advance around the board (i.e., there’s no “Chance,” “Community Chest,” going to jail, etc.). In their first pass around the board, which space from 1 to 39 are players most likely to land on at some point (i.e., not necessarily on their first or last roll, but after any number of rolls)?

Extra Credit
The square board has 10 spaces on each side. The first side has spaces 0 through 9, the second side has spaces 10 through 19, the third side has spaces 20 through 29, and the fourth side has spaces 30 through 39. Because you’re rolling two dice, it’s impossible to land on space 1 in your first pass around the board. Several other spaces on the first side of the board are similarly unlikely. Putting that first side of the board aside, which space from 10 to 39 are players least likely to land on at some point during their first pass around the board? (Another question: What if you rolled three dice at a time instead of two?)

My solution:
[Show Solution]

3 thoughts on “The Likeliest Monopoly Square”

  1. Interesting! One must wonder, what is the expression for lim_n q_n when using j dice and is there a nice compact way of getting to that answer?

    1. I was thinking the same thing but didn’t figure it out until just now. It’s actually just the residue of $p(x)$ at $x=1$! The result yields $\lim_{n\to\infty}q_n = \frac{2}{7k}$ if you’re rolling $k$ dice at once. I added an extra paragraph at the end of the write-up to explain this.

      1. “Now I’m wondering if there isn’t a way to arrive at this formula purely from first principles!”

        — Yes, this appears to be a consequence of the Blackwell renewal theorem ( Moving forward by the amount shown on k dice amounts to adding an I.I.D. random variable X to your position on the board, and that random variable has mean E[X] = 7k/2. Therefore, according to the theorem, in the limit as n goes to infinity, the probability of landing on space n approaches 1/E[X] = 2 / (7k), which is the result you found using generating functions.

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