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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?

OmniOwl
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  • You'll need to clarify a little bit: there are 40 spaces, so you start on space 1 and you're asking for the probability of landing on space 40. That's the probability of moving 39 steps forward. I assume you roll both dice and then add the two results to decide how much to move forward? – Ben Millwood Jul 14 '16 at 14:02
  • In any case, the lazy answer is "on average you move 7 spaces, so the long-run probability of landing on any particular space is 1/7". Becomes more accurate the longer the game is, but is a bit off initially because e.g. it is impossible to land on square 2. I'm not sure how good the approximation is by the time you get to 39 moves, but it's probably pretty decent. – Ben Millwood Jul 14 '16 at 14:04
  • Yeah you are correct in your assumption that you throw both and then put them together. – OmniOwl Jul 14 '16 at 14:07
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    Also note that if your game has mechanisms that take you to specific spaces, that will make even the long term odds not closing in on $\frac17$. The many ways of arriving at the prison space in regular monopoly is the reason the orange properties are the ones that give you most in return for your investments. – Arthur Jul 14 '16 at 14:25
  • @Arthur Yes, there are cards which will take you to specific spots, even past or away from the last spot on the board. But for the sake of simplicity we settled on pointing out the dice factor alone first. – OmniOwl Jul 14 '16 at 17:22
  • @BenMillwood Thank you Sir, 1/7 sounds pretty intuitive given that on average two dice move 7 steps. Would you care to provide a little bit more rigor in this connection? Does this has anything to do with exponential distribution or long-run probability of markov chain? – Kenneth Chen Dec 25 '19 at 22:54
  • I'm a bit vague on what exactly I had in mind when making the previous comment but I feel duty-bound to point out that if you just had one die, and every time you rolled it you just doubled the roll, then the average is still 7 spaces per roll but the probability you land on square 39 is zero. So there's at least some subtlety to it. – Ben Millwood Mar 29 '20 at 08:51

1 Answers1

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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 $\approx \frac 17$.

Hagen von Eitzen
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  • Thank Science! That was a nice and much needed response. Thank you! We started doing this kind of math ourselves but we couldn't really get to a point that sounded right. – OmniOwl Jul 14 '16 at 17:21