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# A Monopoly Problem watch

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1. 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?
2. Yeah congratulations - You really deserve them.
3. Mayfair with a hotel is always a killer to land on.
4. (Original post by Usman938)
Mayfair with a hotel is always a killer to land on.
its £2000 in the old school version
(i think)
5. (Original post by sidewalkwhenshewalks)
its £2000 in the old school version
(i think)
Right you are my friend, tis a hefty sum of money.
6. The best bit is getting the community shield card with the beauty pagent win.

That's a barrel of laughs that
7. (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.
8. (Original post by n1r4v)
I bet it is Mr. Pence, I bet it is.
9. 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.
10. (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?
11. (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,

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].
12. (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.
13. (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
14. (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,

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.
15. 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 , where is the probability of getting j when throwing two dice).
16. 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.
17. So, using my method, I get: 9.2049922657910876 (which is probably exact; the next 3 digits are 000).
18. (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.
19. (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).
20. 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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