What is the expected number of times to pass Go before landing on Boardwalk

Question

This question just pertains to the monopoly board; that is, no actual functionalities of the game(chance cards, go to jail, etc.). The only mechanism to move forward is the rolling of the two dice. The expected number of passes of Go before landing on the 36th square boardwalk.

I interpret this question as the following: If we imagine the sequence of squares as the sequence of natural numbers, and counting the first square as a passing of Go, it is the following expectation: $$E(\frac{X}{36} | (X \bmod 36 = 0))$$ Where X is the current position on the board, or current number in the sequence. How do I compute this expectation?

Answer 1

Answer: On average, you will pass Go approximately seven times before hitting Boardwalk (where your very first move counts as passing Go).

Explanation: Each turn, you move forward seven squares on average. This means you land on about one seventh of the squares of board on average during each pass around the board. In particular, during each revolution, there is about a one seventh chance you will hit Boardwalk. Therefore, it will take approximately seven revolutions on average to land on boardwalk.

This is only an approximate answer, but computer simulations confirm it is very close to the truth. See for yourself.

Answer 2

For an exact calculation:

Number the fields from 0 (go) to 39 (boardwalk). Let $E_k$ be the expected number of times you reach or pass Go before reaching Boardwalk if you are located on field k. We ignore the “go to jail” field, which would move you from field 30 to field 10.

By definition, $E_{39} = 0$. For other fields, there are 11 possible moves, with different probabilities, of which some may reach or pass Go. For each k we write down the formula, for example $E_{35}$ = $E_{37} / 36$ + $2 * E_{38} / 36$ + $3 * E_{39} / 36$ + $4*(1+E_0)/36$ + $5*(1+E_1)/36$ + …

This is a linear system of 39 equations since $E_{39}=0$. We change the equations $E_{38}$, $E_{37}$ ,,, $E_{10}$ to be in terms of $E_0$ to $E_{10}$. We then have $E_{10}$ on both sides of the equation, which lets us express $E_{10}$ in terms of $E_0$ to $E_9$ only, we substitute this for $E_{10}$ in all the other equations.

The same is done for $E_9$ to $E_0$, which finally reduces all the numbers $E_0$ to $E_{38}$ to constants. They will all be different, for example $E_{38}$ will be about one larger than $E_0$ to $E_{10}$.