Given two dice, what's the probability that land on the last spot on the board? So me and my colleagues are discussing board games and we land on the subject of the Danish "Matador" (Monopoly) and on that board there are 40 spaces. You start on Space 1 and are given two dice to make it around the board. The dice are standard 6-side dice.
There is one colleague in particular who argues that the game is not really random, even though it is as your movement is decided by dice throw and thus what land you can buy and so forth.
What would be the probability of landing on the last field on the board?
 A: It is certainly random, but not necessarily uniform.
Also, the probability of ever landing on the last field (i.e., allowing enough rounds to complete) is $1$.
I assume you want to know: What is the probability of landing on space 40 before completing the round?
We can compute the probability $P_n$ of landing on space $n$ before going beyond $n$:
Clearly, $P_1=1$. We may also set $P_n=0$ for $n\le 0$.
Then for all $n>1$ we have 
$$P_n= \sum_{k=2}^{12}p_kP_{n-k}$$
where $p_k$ is the probability of rolling $k$ (so $p_2=\frac1{36}$, $p_3=\frac1{18}$, etc).
One finds 
$$P_1=1;P_2=0; P_2=\frac1{36}; P_3=\frac1{18}; P_4=\frac{109}{1296};\ \ldots\ ; P_{40}=0.142805773\ldots$$
In the limit as $n\to \infty$ we should find $P_n\to\frac17$ (why?), and we see that $P_{40}$ differs from this by only $\approx0.0008$.
If one makes a graph from the $P_n$ computed above, one notices: The probability increases from $P_2=0$ to $P_8\approx 0.182227$, then decreases to $P_{14}\approx0.1247$ and then quickly approaches $\approx \frac 17$.
