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Thread: Dice probability question - Settlers of Catan

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    Dice probability question - Settlers of Catan




    Hello everyone,

    Let me begin by saying that this isn't a homework question. Rather, this is a question that relates to a personal inquiry into certain statistical aspects of my favorite board game: Settlers of Catan.

    Overview of the game:
    Rather than provide a full set of rules, I am only going to post the rules that are pertinent to the formula I am in search of.

    Players place settlements at the intersections of tiles. Each tile contains a single number between 2-12 (excluding 7). During each player's turn, the dice are rolled. For each number rolled, any player who has a settlement adjacent to a tile that matches that number gets to collect a resource. Whenever a 7 is rolled, players who have ≧ 8 cards in their hand have to discard half of their hand, rounded down. (ouch!)

    If you have never played the game, this explanation may be a bit hard to follow. So, in the next section I am going to distill this information down to what is essential for the formula.

    What I want to know:
    Given a certain number of starting cards "c" and the probability "p/36" of collecting an additional card during each turn "t," what is the likelihood that a 7 will be rolled after one's cards in hand is ≧ 8?

    What I've tried:
    I did some preliminary wikipedia research and tried applying the probability mass function to this problem. However, this only got me so far as to be able to determine the probability that one's hand size will reach some number "c" after turns "t."

    I have been unable to factor in the probability of a 7 being rolled.

    What I'm asking for:
    Help!, but not an answer.

    I would like to be pointed in the direction of theorems or functions that I can apply to this problem so as to be able to complete what I have started.

    There are many more Settlers of Catan probability problems that I would like to solve. But, I do not want to come to a forum for each new problem. I would like to develop a foundation of knowledge that will enable me to solve future problems.

    Any help would be greatly appreciated.


    ps. For those of you who have played Catan, you may have noticed that I left out some aspects of the game that could affect the final formula. For example, it is possible to have two settlements on a single number--thus creating a situation in which a subset of "p/36" would return 2 cards rather than 1. It is also possible to have a city, in which case the same thing occurs. I left this aspect of the game out of my request because I want to start small. Once I feel I have a working formula for some basic conditions, I will explore the game in a more thorough manner.

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    Re: Dice probability question - Settlers of Catan

    Are you asking P(D=7 | c\geq8) or P(D=7 \cap c\geq8) or something else?

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    Re: Dice probability question - Settlers of Catan

    Thank you very much for your help!

    Unfortunately, having only taken one very basic stats course two summers ago, my understanding of the notation is limited.

    That being said, I did some research on the notation used in your response, but I am having some difficulty understanding the difference between the two options you posted.

    My understanding of the first option you posted is the probability that 7 will be rolled given that C is already ≧8.

    My understanding of the second option is the probability that, over a certain number of turns, both a 7 will be rolled and C will be ≧8.

    If I am correct in this understanding, I believe I am asking for the second of your two options. Just to be sure, I will try to clarify what I am searching for with a concrete example.

    Given six cards in hand and a probability of "x" of collecting a new card during each new turn, what is the probability that over the next four turns one will have to discard half of one's hand? That is, what is the probability that two cards will be collected and then a seven will be rolled?

    There are many ways in which this can occur.

    Turn 1: Collect a card
    Turn 2: Collect a card
    Turn 3: Collect a card
    Turn 4: Seven is rolled, discard half.

    Turn 1: Collect a card
    Turn 2: Collect a card
    Turn 3: No result
    Turn 4: Seven is rolled, discard half.

    Turn 1: Collect a card
    Turn 2: Collect a card
    Turn 3: Seven is rolled, discard half.
    Turn 4: Does not matter.

    Turn 1: Collect a card
    Turn 2: No result
    Turn 3: Collect a card
    Turn 4: Seven is rolled, discard half.

    Turn 1: No result
    Turn 2: Collect a card
    Turn 3: Collect a card
    Turn 4: Seven is rolled, discard half.

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    Re: Dice probability question - Settlers of Catan


    Assuming each turn you roll 2 independent dice, and obtain the sum D as the final outcome.

    Every Catan's player know that D follows a discrete triangular distribution:

    \Pr\{D = d\} = \frac {6 - |7 - d|} {36}, d = 2, 3, \ldots, 12

    I think to answer your question, first you need to specify the player collect cards on which outcome - this should be a known information before each turn. To simplify the matter assume the player will not place any settlement in the period of consideration (otherwise need to consider the strategy dynamically)

    So in every turn, you have a distribution the cards obtain - 0, 1, 2, \ldots. Of course as you said you may first try to consider there is only 1 settlement here.

    Under those assumptions the problem now can be described as a Markov chain.

    Suppose you have a settlement adjacent to numbers 4, 6, 9. Therefore the probability of obtaining 1 resource card is

    \frac {3 + 5 + 4} {36} = \frac {1} {3}

    Since you consider discarding half of the cards as the only event of interest, the transition matrix can be simplified to something like this (as you assume currently you got 6 cards and only consider the next 4 turns)

    - 1 absorbing state indicating the event of discarding already happened.

    - other states indicating the number of cards held in hand without any discarding happened


    P = \begin{matrix} 6 \\ 7 \\ 8 \\ 9 \\ 10 \\ \text{Discarded} \end{matrix}\begin{bmatrix} \frac {2} {3} & \frac {1} {3} & 0 & 0 & 0 & 0 \\
0 & \frac {2} {3} & \frac {1} {3} & 0 & 0 & 0  \\
0 & 0 & \frac {1} {2} & \frac {1} {3} & 0 & \frac {1} {6} \\ 
0 & 0 & 0 & \frac {1} {2} & \frac {1} {3} & \frac {1} {6} \\ 
0 & 0 & 0 &  0 & \frac {5} {6} & \frac {1} {6} \\ 
0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}

    (the transition probability of state 10 is not important here as you only consider the first 4 turns).

    The transition probabilities after 4 turns will be the first row of entries inside P^4

    And by computer the entries are

    \begin{bmatrix} \displaystyle \frac {16} {81} & \displaystyle \frac {32} {81} & \displaystyle \frac {1} {4} & \displaystyle \frac {7} {81} & \displaystyle \frac {1} {81} & \displaystyle \frac {19} {324} \end{bmatrix}

    Therefore the probability of discarding will be \frac {19} {324} under the above assumption. And the other 5 entries will be the probability mass function of the number of cards hold without discarding after 4 turns.

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