# Monopoly probabilities

I recently had an interesting question posed to me regarding Monopoly. Suppose that one deal has one of the two other players paying you \$200 if they land on any of the 4 railroad spaces. A second deal has either player paying you \$100 if they land on 3 of the railroad spaces.

So to clarify, in deal 1 you own all 4 railroads, and only one of the two other players participates in the deal. In deal 2, you own 3 railroads (not clear which 3 are owned), but both players participate (but pay you lower money).

Which deal should you take, deal 1 or deal 2? In other words, what is your expected income generated from either deal?

I am assuming that there is an equal probability of landing on any space (there are 40, I believe). I know that this is incorrect, since the probability of landing on a particular space depends on your location on the board (as well as any chance/community chest cards in play). But for the purposes of this question, assume that each space has an equal probability of being landed on.

For Deal 1, the single player in the deal has a $4/40=0.10$ probability of landing on a railroad space. For Deal 2, each player has a probability of $3/40=0.075$ of landing on three of the railroad spaces (assume these are always the same 3 spaces). Since each player's position is independent, the probability of either player (or both) landing on a railroad space is

$$P(R_{1}\cup R_{2})=P(R_{1})+P(R_{2})-P(R_{1}\cap R_{2})=0.075+0.075-(0.075)^{2}=0.144.$$

Then it appears that, over many trips around the board, Deal 2 would be the best to take. Is this a correct way of going about this problem? Thanks for any help/advice.

• If in going round the board, the single player (option 1) lands on $2$ of your railroad spaces, does she pay you $(2)(200)$? – André Nicolas Aug 4 '14 at 19:13
• That wasn't made clear to me in the question. Let's assume that, yes, anytime they land on a railroad space, the player pays the specified amount. Thanks for asking – Kirk Fogg Aug 4 '14 at 19:16
• Something fun to note for a practical game, Reading Railroad is by and far the most landed on railroad in the game. So, from a practical point of view, that railroad has much greater value than Short Line, the least landed on railroad. And to continue the monopoly fun; ALWAYS buy Orange, Light Blue, and Railroads and you cannot go wrong. – Vincent Aug 4 '14 at 20:03

We assume that in going around the board once, you get a payment of $200$ dollars (first case) for each time the player lands on a railway space. So if she is really unlucky, you get $800$ dollars and begin to dream about a vacation.

We also make the probably unreasonable assumption that landing on any one of the spaces is equally likely, and that the events are independent. We also make the assumption that it is not possible to land twice on the same space in one tour of the board, something that is in fact not true, since sadly, in Monopoly as in life, one sometimes makes negative progress.

Four railways, one payer: Let $X_1,X_2,X_3,X_4$ be the incomes we get from each of the four railways. Then our total income is $X_1+\cdots+X_4$, which has expectation $E(X_1)+\cdots+E(X_4)$. This is $4\left(\frac{1}{40}\cdot 200\right)$.

Three railways, two payers: Let $X$ be the income we get from the first payer, and $Y$ the income we get from the second. We want $E(X+Y)$, which is $E(X)+E(Y)$. By an analysis like the one above, we have $E(X)=3\left(\frac{1}{40}\cdot 100\right)$. But $E(Y)$ is the same, so your expected income is $(2)(3)\left(\frac{1}{40}\cdot 100\right)$.

Remark: So deal $1$ is better, if we neglect the cost of acquiring that fourth railway.

It is more interesting to solve the problem under the assumption we can only get one payment per player round.

In the first case, the probability your opponent misses all your railways is $\left(\frac{39}{40}\right)^4$, so the probability she has to pay you $200$ is $1-\left(\frac{39}{40}\right)^4$. Multiply by $200$ for the expectation.

In the second case, a similar analysis shows that the expectation is $(2)(100)\left( 1-\left(\frac{39}{40}\right)^3 \right)$.