Project Euler Problem 84 Solution
Question
In the game, Monopoly, the standard board is set up in the following way:
GO  A1  CC1  A2  T1  R1  B1  CH1  B2  B3  JAIL 
H2  C1  
T2  U1  
H1  C2  
CH3  C3  
R4  R2  
G3  D1  
CC3  CC2  
G2  D2  
G1  D3  
G2J  F3  U2  F2  F1  R3  E3  E2  CH2  E1  FP 
A player starts on the GO square and adds the scores on two 6sided 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 twodigit 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 sixdigit modal string: 102400.
If, instead of using two 6sided dice, two 4sided dice are used, find the sixdigit modal string.
Python
#!/usr/bin/env python
from random import randint, shuffle
from collections import deque, Counter
squares = ["GO", "A1", "CC1", "A2", "T1", "R1", "B1", "CH1", "B2", "B3", "JAIL", "C1", "U1", "C2", "C3", "R2", "D1", "CC2", "D2", "D3", "FP", "E1", "CH2", "E2", "E3", "R3", "F1", "F2", "U2", "F3", "G2J", "G1", "G2", "CC3", "G3", "R4", "CH3", "H1", "T2", "H2"]
square_numbers = {square: i for i, square in enumerate(squares)}
num_squares = len(squares)
def build_deck(choices):
numbers = [square_numbers[square] for square in choices]
shuffle(numbers)
return deque(numbers)
transitions = {
"G2J": build_deck(["JAIL"]),
"CH1": build_deck(["CH1", "CH1", "CH1", "CH1", "CH1", "CH1", "GO", "JAIL", "C1", "E3", "H2", "R1", "R2", "R2", "U1", "T1"]),
"CH2": build_deck(["CH2", "CH2", "CH2", "CH2", "CH2", "CH2", "GO", "JAIL", "C1", "E3", "H2", "R1", "R2", "R3", "U2", "D3"]),
"CH3": build_deck(["CH3", "CH3", "CH3", "CH3", "CH3", "CH3", "GO", "JAIL", "C1", "E3", "H2", "R1", "R1", "R1", "U1", "CC3"]),
"CC1": build_deck(["CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "CC1", "GO", "JAIL"]),
"CC2": build_deck(["CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "CC2", "GO", "JAIL"]),
"CC3": build_deck(["CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "CC3", "GO", "JAIL"])
}
transitions = {square_numbers[square]: deck for square, deck in list(transitions.items())}
def transition(square_number):
while square_number in transitions:
deck = transitions[square_number]
new_square_number = deck.pop()
deck.appendleft(new_square_number)
if square_number == new_square_number:
break
square_number = new_square_number
return square_number
def move(square_number, die_size=6):
roll = randint(1, die_size) + randint(1, die_size)
square_number += roll
square_number %= num_squares
return transition(square_number)
def simulate(iterations=2000000, die_size=6):
result = Counter()
square_number = 0
for i in range(iterations):
square_number = move(square_number, die_size)
result[square_number] += 1
return result
def main():
simulation = simulate(die_size=4)
print(("".join("{:02}".format(square_number) for square_number, count in simulation.most_common(3))))
if __name__ == "__main__": main()
$ time python3 monopoly.py
real 0m8.699s
user 0m8.640s
sys 0m0.000s
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