2
$\begingroup$

enter image description here

Based off of this classical monopoly board, my friend told me that it is better statistically to get 3 properties because you are more likely to land on the properties because they are close together. Because the properties are close together, it means that the probability of you landing on them is higher.

What I said is statistically it is better to get the railroads because there are 4 railroads on the board.

So the probability of landing on a railroad every time going around the board is 4 / 40 which is a 10% chance.

And the probability of landing on the 3 properties right next to eachother is 3 / 40 which is a 7.5% chance.

There are other factors involved like you have to roll 2 dice. This means 6 7 8 are the most common dice rolls, with 7 being the most popular.

What I want to know is, statistically through the course of a game where you might go around the board 20 - 30 times, is it more likely for you to land on the railroads or land on 3 properties (such as orange or red).

$\endgroup$
2
  • $\begingroup$ I Googled "monopoly strategy probability" and there were lots of hits. Didn't actually look, but maybe some of them will help with your question. $\endgroup$
    – David
    Commented Mar 5, 2014 at 23:21
  • $\begingroup$ How close your properties are together will not impact the average number of times people land on them, but it will impact the varience I think since if for example you have a whole side of the board it becomes very hard not to land on SOMETHING you have. $\endgroup$ Commented Mar 6, 2014 at 0:00

3 Answers 3

3
$\begingroup$

You are making the approximation that each square is equally likely to be hit. This is not correct, due to jail and the chance/community chest cards but let's accept that. In that case, the chance of hitting a specific square on a given circuit is $\frac 17$ as the average roll is $7$. By the linearity of expectation, you expect to hit a color group $\frac 37$ times the number of circuits and to hit a railroad $\frac 47$ times the number of circuits. In the long run, you will hit the railroads $\frac 43$ times as often. Whether this is better also depends on the value at each hit. In the US version of the game the railroads collect 200, while a hotel on the greens collects 1275/1400. The railroads are actually some of the most trafficked squares due to the cards.

$\endgroup$
2
  • 1
    $\begingroup$ "In that case, the chance of hitting a specific square on a given circuit is 1/7 as the average roll is 7" Where does that come from? $\endgroup$ Commented Mar 24, 2018 at 1:13
  • $\begingroup$ @RyanRosario Mean: (2 + 12) / 2 = 7 It's also the median as 2 and 12 each have a 1 in 36 chance (1 + 1 or 6 + 6) of being rolled while a 7 has a 6 in 36, or 1 in 6 chance of being rolled (1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, and 6 + 1). $\endgroup$
    – Pete
    Commented Jan 19, 2019 at 16:13
1
$\begingroup$

Your real mistake here is in assuming that your chance of landing on any particular property each time around the board is $1$ in $40$, which is where you got your 10% and 7.5% chances for landing on a railroad or 3 properties, respectively. You seem to be saying, for example, that in a game where you go around the board 20 to 30 times, you only expect to land on a railroad 2 or 3 times. I'm sure you'll agree, that's way too low!

$\endgroup$
1
$\begingroup$

In addition to what others have said, by creating a Markov chain, one can find the exact probabilities of each space on the board. Due to chance/community chest cards and, of course, the "Go To Jail" space, the probability of each space differs.

Since Jail is so commonly visited, the properties next to it -- orange and red, especially -- are visited a lot. In addition, spaces which cannot be landed on through chance/comm chest cards (such as chance and comm chest themselves) are visited significantly less. Chance 3 has around a $1\%$ chance to be landed on.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .