Can You Race Around the Monopoly Board?

The game of Oligopoly is played on a square board of $40$ spaces numbered $0$ to $39$. Every player starts on space $0$ and on each turn rolls two dice and moves forward that many spaces. Unlike Monopoly, nothing else ever moves you: no cards, no jail. On their first pass around the board, which space from $1$ to $39$ are players most likely to land on at some point (not necessarily on the first roll)?

The Fiddler, Zach Wissner-Gross, December 1, 2023(original post)

The official solution appeared in the post of December 8, 2023 and gives space $7$, with probability about $18.2$ percent, for the main puzzle, and space $13$ (and space $17$ with three dice) for the Extra Credit. All agree with the answers below.

Solution

Write $r_k$ for the chance that two dice total $k$. Of the $36$ equally likely rolls, $6 - |k - 7|$ total $k$, so $$r_k = \frac{6 - |k - 7|}{36}, \qquad k = 2, 3, \dots, 12 .$$ Write $p_n$ for the chance of landing on space $n$ at some point during the first pass, with $p_0 = 1$ because everyone starts there.

The token only ever moves forward, so if it lands on $n$ there is exactly one space it stood on just before, some $n - k$, followed by a roll of exactly $k$. These possibilities, one for each $k$, cannot happen together, and the roll made from $n - k$ does not depend on how the token got there. So $$p_n \;=\; \sum_{k=2}^{12} r_k \, p_{n-k},$$ counting $p_m = 0$ for negative $m$. Each value comes from the eleven before it. The first few are $$p_1 = 0, \qquad p_2 = \tfrac1{36}, \qquad p_3 = \tfrac2{36},$$ $$p_4 = \tfrac3{36} + \tfrac1{36}\cdot\tfrac1{36} = \tfrac{109}{1296},$$ and continuing the same way:

Up to space $13$ the peak is space $7$, with $$p_7 = \tfrac{6}{36} + r_2 p_5 + r_3 p_4 + r_4 p_3 + r_5 p_2 = \frac{1417}{7776} .$$

What about later spaces? From $n = 12$ on, all eleven weights $r_2, \dots, r_{12}$ appear in the formula, and they add up to $1$. So $p_n$ is a weighted average of the eleven values before it, and an average can never exceed the largest thing it averages. For $n = 14$ those eleven values are $p_2, \dots, p_{12}$, all at most $p_7$, so $p_{14} < p_7$. The same holds for $p_{15}$, whose window has moved on by one, and so on along the board: no later space ever catches up with space $7$. In fact the values settle down to about $1/7$. The average roll is $7$, so in the long run the token touches one space in seven. $$\boxed{\ \text{space } 7, \text{ landed on with probability } \frac{1417}{7776} \approx 18.2\%. \ }$$

The computation

The program does not use the formula above. It follows the game roll by roll: it keeps the exact probability of every position after one roll, after two, and so on, until every position has passed $39$. A position is never visited twice, so adding up the chance of being on space $n$ after each number of rolls gives the chance of ever landing there.

from fractions import Fraction
from itertools import product
from collections import Counter

def landing(dice):
    roll = Counter(sum(t) for t in
                   product(range(1, 7), repeat=dice))
    total = 6 ** dice
    ever = Counter()
    now = {0: Fraction(1)}            # position -> chance
    while now:
        nxt = Counter()
        for pos, pr in now.items():
            for k, c in roll.items():
                if pos + k <= 39:     # still on the first pass
                    nxt[pos + k] += pr * Fraction(c, total)
        ever.update(nxt)
        now = nxt
    return ever

p = landing(2)
best = max(range(1, 40), key=lambda n: p[n])
print("most likely:", best, p[best], float(p[best]))

It prints

most likely: 7 1417/7776 0.18222736625514405

Extra Credit

The board has $10$ spaces per side, so the first side is spaces $0$ to $9$. Setting the first side aside, which space from $10$ to $39$ is least likely to be landed on during the first pass? What if you rolled three dice instead of two?

Solution

The table of $p_n$ from the main puzzle, carried on to space $39$, answers it. After the first side the values run $$p_{10} = .148,\quad p_{11} = .141,\quad p_{12} = .134,$$ $$p_{13} = .125,\quad p_{14} = .139,\quad p_{15} = .146, \quad \dots$$ and from space $14$ on they stay between $0.138$ and $0.149$, closing in on $1/7$. So the low point is space $13$.

The reason is visible in the chart. A single roll reaches at most $12$, so space $13$ is the first space that always needs two or more rolls. Two rolls favour totals near $14$, and $13$ sits on their lower flank. Spaces just before it still get a share from single rolls, and spaces just after it get the full benefit of two rolls. Space $13$ falls into the gap between the two humps. Its exact chance is $$\boxed{\ \text{space } 13, \text{ with } p_{13} = \frac{22{,}621{,}045}{3 \cdot 6^{10}} \approx 12.5\%. \ }$$

With three dice the same thing happens one hump later. A single roll of three dice totals at most $18$ and seldom more than $15$, and two rolls centre on $21$, so the gap falls in the high teens. The exact table, from the computation below, puts the low point at space $17$ with probability about $7.63$ percent, just under space $16$ at about $7.66$ percent. Their difference is small enough that only exact arithmetic settles it: $$\boxed{\ \text{with three dice: space } 17,\ \approx 7.63\%. \ }$$

The computation

The function from The computation above handles any number of dice. Ask it for the least likely space after the first side, with two dice and with three.

for dice in (2, 3):
    p = landing(dice)
    low = min(range(10, 40), key=lambda n: p[n])
    print(dice, "dice: least likely", low,
          f"{float(p[low]):.4f}")

It prints

2 dice: least likely 13 0.1247
3 dice: least likely 17 0.0763