Loading...
Project Euler 84: In the game, Monopoly, find the three most popular squares when using two 4-sided dice
Problem Description
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 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.
Analysis
Here’s our solution using simulation for 2,000,000 games. Sample output below is for six-sided dice.
Sample size: 2000000
Frequency for each space
|
CH3, 18479, 0.92395 CH1, 18496, 0.9248 CH2, 22400, 1.12 CC1, 37265, 1.86325 A1, 42550, 2.1275 A2, 43224, 2.1612 H1, 43282, 2.1641 T2, 43725, 2.18625 T1, 43918, 2.1959 CC3, 44532, 2.2266 B3, 45441, 2.27205 B1, 45509, 2.27545 B2, 46381, 2.31905 |
C2, 47331, 2.36655 R4, 48554, 2.4277 C3, 49521, 2.47605 G3, 50316, 2.5158 CC2, 51646, 2.5823 F3, 52056, 2.6028 U1, 52446, 2.6223 H2, 52472, 2.6236 G2, 52994, 2.6497 F2, 53158, 2.6579 G1, 53795, 2.68975 F1, 53856, 2.6928 C1, 54192, 2.7096 |
E2, 54428, 2.7214 U2, 56039, 2.80195 D1, 56451, 2.82255 E1, 56991, 2.84955 FP, 57917, 2.89585 D3, 58364, 2.9182 D2, 58802, 2.9401 R2, 59064, 2.9532 R1, 60164, 3.0082 R3, 61443, 3.07215 GO, 62558, 3.1279 E3, 64488, 3.2244 JAIL, 125752, 6.2876 |
Project Euler 84 Solution
Runs < 8 seconds in Perl.
Use this link to get the Project Euler 84 Solution Perl source.
Afterthoughts
-
No afterthoughts yet.
Interesting. I created a Markov chain with 120 states (40 squares x #doubles). I squared the 120×120 matrix repeatedly, and got to a steady state.
Posted by Frank Yellin | September 23, 2020, 10:12 PM