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A Monopoly Problem watch

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    Take a square board which has a row of ten boxes on each side (i.e. total number of boxes is 36). You are given a random starting place for a token, two squares next to each other and then another square a blank step away from those two squares, and two standard dice rolls per turn. What is the average number of turns required such that a token will land on one of those squares?
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    Yeah congratulations - You really deserve them.
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    Mayfair with a hotel is always a killer to land on.
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    (Original post by Usman938)
    Mayfair with a hotel is always a killer to land on.
    its £2000 in the old school version
    (i think)
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    (Original post by sidewalkwhenshewalks)
    its £2000 in the old school version
    (i think)
    Right you are my friend, tis a hefty sum of money.
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    The best bit is getting the community shield card with the beauty pagent win.

    That's a barrel of laughs that
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    (Original post by Snobpence17)
    The best bit is getting the community shield card with the beauty pagent win.

    That's a barrel of laughs that
    I bet it is Mr. Pence, I bet it is.
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    (Original post by n1r4v)
    I bet it is Mr. Pence, I bet it is.
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    Move Back 3 kills me when it's moving back to Mayfair.

    I do love Old Kent Road and the other brown one, so cheap, it's just good to have to annoy everyone.
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    (Original post by Kolya)
    Take a square board which has a row of ten boxes on each side (i.e. total number of boxes is 36). You are given a random starting place for a token, two squares next to each other and then another square a blank step away from those two squares, and two standard dice rolls per turn. What is the average number of turns required such that a token will land on one of those squares?
    I don't get the question.

    So I pick a random square and place a token on it.

    Then I pick two random squares directly next to each other (but not necessarily next to the one with my token).

    What's this about "another square a blank step away from those two squares"? What in the world does "a blank step away" mean?

    Now when I roll the 2 dice, do I add the score up and move the token that many places or something? I'm going clockwise, right?
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    (Original post by Kolya)
    Take a square board which has a row of ten boxes on each side (i.e. total number of boxes is 36). You are given a random starting place for a token, two squares next to each other and then another square a blank step away from those two squares, and two standard dice rolls per turn. What is the average number of turns required such that a token will land on one of those squares?
    If you are happy to use a computer, here's an approach:

    Wlog, let's say the target squares have indices 0,1 and 3.

    Set up a vector E[0...35] where E[i] will be the expected number of turns to land on one of the target sqaures starting from square i.

    Then E[0]=E[1]=E[3] = 0
    for i != 0,1,3,

    E[i] = 1 + \frac{1}{36}\sum_{j=1}^6\sum_{k=  1}^6 E[i+j+k \mod 36]

    This gives you an iterative formula for the E[i] that I think should converge pretty efficiently.

    [There are similarities between this and one of the Project Euler problems].
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    (Original post by Swayum)
    I don't get the question.

    So I pick a random square and place a token on it.

    Then I pick two random squares directly next to each other (but not necessarily next to the one with my token).

    What's this about "another square a blank step away from those two squares"? What in the world does "a blank step away" mean?
    If we number the squares then it will be, say, squares 1,2,4; or squares 2,3,5; or squares 24,25,27, etc...

    Now when I roll the 2 dice, do I add the score up and move the token that many places or something?
    Yes.

    I'm going clockwise, right?
    Yes, although it doesn't matter.
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    (Original post by JakeCordell)
    Move Back 3 kills me when it's moving back to Mayfair.

    I do love Old Kent Road and the other brown one, so cheap, it's just good to have to annoy everyone.
    Oh Seconded, especially when they get snake eyes after getting £200 from the GO square.

    The ultimate move
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    (Original post by DFranklin)
    If you are happy to use a computer, here's an approach:

    Wlog, let's say the target squares have indices 0,1 and 3.

    Set up a vector E[0...35] where E[i] will be the expected number of turns to land on one of the target sqaures starting from square i.

    Then E[0]=E[1]=E[3] = 0
    for i != 0,1,3,

    E[i] = 1 + \frac{1}{36}\sum_{j=1}^6\sum_{k=  1}^6 E[i+j+k \mod 36]

    This gives you an iterative formula for the E[i] that I think should converge pretty efficiently.

    [There are similarities between this and one of the Project Euler problems].
    Could you explain how you came to the equation for E[i]?

    I was thinking about using a random no. generator to pick a square, and another one to pick a dice roll (based on the probability of each possible roll), and run enough trials to get convergence to the correct answer, but this seems far too rough.

    And I guess it does seem a little Project Euler-ish, although the problem is one I came up with myself.
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    The formula is because you "add one" for the current turn, and then you end up on square [i+j+k] with probability 1/36, and the expected number of turns from [i+j+k] is E[i+j+k].

    (You could also rearrange as a single sum E[i]=1+\sum_{j=2}^{12} p_j E[i+j \mod 36], where p_j is the probability of getting j when throwing two dice).
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    try playing this board game called 'hotels'. Its like monopoly, but better, and SO underrated. You will thank me so much believe me - i have spent YEARS of my life playing.
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    So, using my method, I get: 9.2049922657910876 (which is probably exact; the next 3 digits are 000).
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    (Original post by Usman938)
    Mayfair with a hotel is always a killer to land on.
    But people aren't as likely to land on it. You want the group of the orange ones, they're always the best to get.
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    (Original post by DFranklin)
    So, using my method, I get: 9.2049922657910876 (which is probably exact; the next 3 digits are 000).
    Yep, I get approximately the same answer simulating random trials, although your method converges to the correct answer many, many times faster (and I have the problem of using pseudorandoms).
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    If you simplify the problem by working out the average number of trials required to land on just one particular property on the board, rather than the aforementioned 3 [starting from a random place of the 36 squares etc.], would there be a way of directly computing it, without the use of numerical methods? I'm having a go right now, but it looks like I'm going to be here a while.
 
 
 
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