Monopoly probability question?

Question

In Monopoly a player starts on the "Go" space and moves around the board according to a dice roll. There are a total of 40 spaces on the board. The "best" space to land on is Boardwalk, which is exactly 39 spaces from "Go".

Assuming fair dice and, for the sake of simplicity, no space sends you anywhere else on the board (even though this is, of course, untrue), how many times can a player expect to go around the board before landing on Boardwalk? Also, what's the probability of landing on Boardwalk the first time around the board?

The dice are two six-sided die

Answer 1

POSSIBLE MATHEMATICAL APPROACH

Let P(x) = probability of landing on position x in a given throw
Let D = sum of the dice in that throw
Then P(x) = ∑{t=2,12} * P(D=t) * P(x-t)
But P(D=t) = [6 - |t-7|] /36
So 36*P(x) = ∑{t=2,12} * (6 - |t-7|) * P(x-t)

It might be possible to get an analytical expression using some kind of transform. If we can get a value for P(x), then the expected no. of rounds will be the reciprocal.

I'll see if I can come up with anything before the question closes.


COMPUTATIONAL APPROACH

One issue you need to consider is that when you past Go, you do not always restart on position #1. You can incorporate that into your program by adding the scores until you get a total of 1 less than a multiple of 40.

sum = 0
while mod(sum + 1, 40) != 0
dice1 = random(1,6)
dice2 = random(1,6)
sum = sum + dice1 + dice2
end
n = (sum + 1)/40

Once it exits, sum will be 39, 79, 119 etc, then n gives the number of rounds necessary to accomplish that. Run this program many times and get the average. It should give something slightly less than 7.


ALTERNATIVE MATHEMATICAL APPROACH

Here is another approach which allows you to avoid the brute force method of calculating every possible scenario. Use the normal approximation for the distribution of the dice. It is true that n is not large enough, but it gives an answer very close to what others have been getting.

In one roll of a die, the mean is 7/2 and the variance is 35/12.
Let's suppose that a pair of dice is rolled n times (ie n turns in a round). The mean and variance of the dice total in those n turns are 7n and 35n/6 respectively.
Use the density function for the normal distribution:
P(a < X < b) = 1/2 * [erf{(b-μ)/σ√2} - erf{(a-μ)/σ√2} ]

Applying that to this problem, the probability pn of getting 39 in n turns is:
pn = 1/2 * [erf{(39.5-7n)/√(35n/3)} - erf{(38.5-7n)/√(35n/3)} ]

Now we need to sum this for all n. n = 1 to 100 is more than sufficient. It yields
P = ∑pn = 0.1428657889
E(rounds) = 1/P = 6.999576370

Answer 2

Average throw when throwing two dice: 7
(2, 12 same prob. 1/36, each; 3,11 same prob, 2/36 each; 4,10 prob, 3/36, each; 5, 9 prob 4/36, each; 6, 8, prob 5/36 and 7 prob 6/36).

Therefore, will land on Boardwalk, on average once every 7 trips.

The point at which the odds are > 50 % that the player will have landed on Boardwalk is the 5th trip around the board at which point the odds will be about ~ 54%. (1- (6/7)^5) that the player will have landed at least once. (Note, after 7 trips, which is the consensus, the odds are only 66% that the player will have landed on Boardwalk)

As for the probablity of landing on the first trip this is ~1/7 or 0.14286. By the time you have rolled the die 4-6 times, you are going to have a broad, flat distribution curve so that it is essentially random (i.e., 1/7) on whether the player will land on Boardwalk. (You could count up all the ways you could add up to 39 with 4, 5, 6 ... 19 rolls to figure the odds exactly.). This is confirmed by all of the computer simulations.

>>>>>>>>>>>>>>>>>>>>>

Edit: I really like Dr. D's "Alternate Mathmatical Approach." That is the best easy way to solve for landing on 39 the first time around. (Sure beats using a spreadsheet)

Answer 3

Using the following Maple procedure

f := proc(n )
description "probability of hitting nth Monopoly square beginning on square GO (square 0)";
if n<2 then
return 0;
fi;
if n=2 then
return 1/36;
fi;
if n=3 then
return 2/36;
fi;
if n=4 then
return 3/36;
fi;
if n=5 then
return 4/36;
fi;
if n=6 then
return 5/36;
fi;
if n=7 then
return 6/36;
fi;
if n=8 then
return 5/36;
fi;
if n=9 then
return 4/36;
fi;
if n=10 then
return 3/36;
fi;
if n=11 then
return 2/36;
fi;
if n=12 then
return 1/36;
fi;
return 1/36*f(n-2)+2/36*f(n-3)
+3/36*f(n-4)
+4/36*f(n-5)+5/36*f(n-6)
+6/36*f(n-7)
+5/36*f(n-8)+4/36*f(n-9)
+3/36*f(n-10)
+2/36*f(n-11)+1/36*f(n-12);
end proc;

I get that the probability of hitting Boardwalk on the first time around the board is exactly
5243284773098777232503/
55268479930183339474944
which is approximately
.09486935012.

This is the value of f(39). The probability of hitting Boardwalk on the k th time around is
f(40*k-1).

EDIT After looking at the results of other posters, I wrote my own simulation and got approximately .142552 for the once around probability of hitting Boardwalk. Next I verified that Maple was correctly executing my recursive algorithm; a Java implementation of the algorithm produced the same result as did Maple, namely 0.09486935012003653. So my recursive algorithm must be wrong. Does anyone see the problem? Thanks.

Answer 4

Markovian analysis could be used, but as noted it would be very complex and something that could not be placed on in this forum easily. By considering each of the 40 spaces a state you can construct a 40 x 40 transition probability matrix.


--- using coding in R I found the experimental probability of landing on Boardwalk the first time around to be 14.2%

die <- seq(1,6,1);
trials <- 500000;
sumOfRolls <- 0;
successfulTrips <- 0;

for(i in 1:trials)
{
while(sumOfRolls < 40)
{
roll <- sample(die,2,TRUE);
sumOfRolls <- sum(roll) + sumOfRolls;
if(sumOfRolls == 39)
{
successfulTrips <- 1 + successfulTrips;
}
}

sumOfRolls <- 0;
}

cat("successful trips:", successfulTrips,"\ntrials:",trials,
"\nestimated probability of landing on Boardwalk on first trip:", successfulTrips/trials,"\n");

OUTPUT:

successful trips: 71140
trials: 5e+05
estimated probability of landing on Boardwalk on first trip: 0.14228

----

The best way to find the expected number of trips about the board to reach Boardwalk is to consider Xt = 39 an absorption state for the Markov Chain and use first step analysis to find the expectation of t. This is a lot of calculation and I'll work on this but in the mean time you can reference these two books for first step analysis of Markov Chains.

http://books.google.com/books?id=Ndx2HgAACAAJ&lr=
http://books.google.com/books?id=171UAgAACAAJ&lr=

while looking at some more code I have found that, as theoretically expected, the average proportion of times you are on Boardwalk is 0.025.

If you consider X to be the number of trials to get to Boardwalk then X has the Geometric distribution with success probability 0.025. This is a interesting approximation and would be valid in the long run. the mean of the Geometric is 1/p = 40. you expect to go around the board 40 times to hit Boardwalk.

----

also, by setting up the correct transitional probability matrix you can also easily account for the spaces that will send you to a different location on the board.

Answer 5

since the pobability changes every roll depending on the previous roll. some amazing intergration is involved to even get close. which after even the first roll which could be anything depending on the specific roll. the most probability roll for two dice is 7 but since there is multiple rolls involved the probability depends on previous situation. if you wanted something like the probability of hitting baltic avenue it would just be the probability of a 2 out of both dice (1 out of 36). and you could calculate the number of sevens it would take hit boardwalk. but again this is almost impossible. it couls be calculated with a large large amount of error

Answer 6

There are a few. If you look at just the dice roll, though, you'll miss it. You ahve to look at allth eother ways you can land on spaces. Like there are 5 ways to end up on jail (land there, land on Go to Jail, the Go to Jail Chance card, the Go to Jail Community Chest card, and rolling 3 doubles in a row). Also, since rolling doubles gets you out of jail quicker, the spaces 6 and 8 spaces after jail are hit regularly, making Orange a very good monopoly to own. Railroads and Illinois Avenue are also more common than others.

Answer 7

To solve this you would use something called Markov Chain analysis - look this up on Google. This will give you an answer in closed form (e.g. a discrete fraction) - but it will be horrendously complex.

Easier way is to simulate using a computer program. I wrote a quick program in Java to do this. I ran a million simulations and came up with:

odds of hitting boardwalk first time around = 0.142862
average # of times around board = 6.993856

The first time around the board is counted as 1 not 0.

In case you're interested I included the program I wrote below.

import java.io.*;
import java.util.*;

public class Monopoly {
public static void main(String[] args) {
int countFirstTimeAroundBoard = 0;
int totalTimesAroundBoard = 1;
for (int i=0; i<1000000; i++) {
int curSquare = 0;
int timesAroundBoard = 1;
while (true) {
int dice1 = 1+ (int)(6*Math.random());
int dice2 = 1+ (int)(6*Math.random());
curSquare += dice1 + dice2;
if (curSquare == 39) {
break;
} else if (curSquare >=40) {
timesAroundBoard++;
curSquare -=40;
}
}
if (timesAroundBoard == 1)
countFirstTimeAroundBoard++;
totalTimesAroundBoard += timesAroundBoard;
}
System.out.println("odds of hitting boardwalk first time around = " + countFirstTimeAroundBoard/1000000.0);
System.out.println("average # of times around board = " + totalTimesAroundBoard/1000000.0);

Answer 8

I think there is only one dice in this monopoly game.

A fair dice after 6 throws/rolls will total : 21

So it is gurateed that on the seventh time around the board it will land on boardwalk ie 21 x 13 = 273, and 39 x 7 = 273

On the first time time around the board, after 6 throws/rolls, we end at 21, 18 more steps to boardwalk at 39. So in the next four throws/rolls, it must be 3 4 5 6, ie cannot have 1 or two.

Answer 9

Probability Monopoly

Answer 10

3rd time