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# Interesting Chessboard Math Problem Watch

1. Two pawns on opposite sides of a 64 square, standard chessboard. Every second, the pawns travel from the square they are on, to another adjacent square, which is chosen at random.

So, here are the questions:-

How many second would it take until the pawns both sit on the same square on the chessboard?
As an average I mean. Obviously, it could be a minimum of 7 and a max of infinity...

How long until the pawns cross paths with one another, i.e., if pawn 1 is going from A-z and pawn 2 is going from Z-A?
2. (Original post by KangaroSarah)
Two pawns on opposite sides of a 64 square, standard chessboard. Every second, the pawns travel from the square they are on, to another adjacent square, which is chosen at random.

So, here are the questions:-

How many second would it take until the pawns both sit on the same square on the chessboard?

How long until the pawns cross paths with one another, i.e., if pawn 1 is going from A-z and pawn 2 is going from Z-A?
Is the first a minimum of five seconds and the second a minimum of four?
3. (Original post by BitWindy)
Is the first a minimum of five seconds and the second a minimum of four?

No minimum mentioned at all. Any ideas? On the surface the problem seems simple, but the more you think about it, the more complicated it becomes!
4. the answer can be quite odd because of course there is a probability they will never meet. may i ask where you found this question?
5. (Original post by kelefi)
the answer can be quite odd because of course there is a probability they will never meet. may i ask where you found this question?
Yeah, I was thinking maybe something along the lines of stochastic? My friend does Maths at uni, his lecturer gave it to him! Got any ideas?
6. i mean, the answer 15 seconds, 50 seconds or even 5x10^50 seconds are all legitimate answers. there is no real upper limit, as it could theoretically by infinity. which of course would mean they never meet. and then of course taking any averages with infinity involved would be very shady ground to be treading.
7. What's the initial set up exactly? can you give there starting squares in chess notation?
8. (Original post by james22)
What's the initial set up exactly? can you give there starting squares in chess notation?
I'm guessing a1 and h8.
9. (Original post by james22)
What's the initial set up exactly? can you give there starting squares in chess notation?
Well, I've found what seems to be a bug in Mathematica's treatment of DiscreteMarkovProcess while trying to investigate the problem. That's as far as I've got.
10. Just a thought here, haven't really put much into it and haven't really done much stochastics (or combinatorics?) before. Would it be simpler (or even the same problem?) to say fix one of the pawns and assume the other one moves 2 squares each second and ask the average time for it to end up in the opposite corner?

As it's hard to describe the movements on a board in words I will assume you are looking at a board like you were playing chess and the pawn is in the left corner on your side and you're wanting to get it to the right corner on the other side.

I'm going to call an "up step" one step away from you and down one towards you. Left and right steps are obvious.

The problem is then what would be the expected time for the number of up steps = number of down steps + 7 and number of right steps = number of left + 7.

Then I don't know
11. (Original post by TheIrrational)
Just a thought here, haven't really put much into it and haven't really done much stochastics (or combinatorics?) before. Would it be simpler (or even the same problem?) to say fix one of the pawns and assume the other one moves 2 squares each second and ask the average time for it to end up in the opposite corner?

As it's hard to describe the movements on a board in words I will assume you are looking at a board like you were playing chess and the pawn is in the left corner on your side and you're wanting to get it to the right corner on the other side.

I'm going to call an "up step" one step away from you and down one towards you. Left and right steps are obvious.

The problem is then what would be the expected time for the number of up steps = number of down steps + 7 and number of right steps = number of left + 7.

Then I don't know
This would work on an infinite chess board, but not a fixed one as it is not symmetric.
12. (Original post by Smaug123)
Well, I've found what seems to be a bug in Mathematica's treatment of DiscreteMarkovProcess while trying to investigate the problem. That's as far as I've got.
I'm going to have a go using MatLab, though I can't code well so it may take days to get a solution. I wonder if there is a nice proof though.
13. Just a thought, if you were to try to solve this problem instead for a general nxm board, couldn't you get a reccursive relation by conditioning on the first move?

Though a 2 variable recurrsive relation may not be much use for calculation purposes if it cannot be massively simplified.
14. (Original post by james22)
Just a thought, if you were to try to solve this problem instead for a general nxm board, couldn't you get a reccursive relation by conditioning on the first move?

Though a 2 variable recurrsive relation may not be much use for calculation purposes if it cannot be massively simplified.
Yes, you could, but it would be really quite disgusting. I briefly contemplated setting it up, but decided not to touch it with a barge-pole.
15. (Original post by Smaug123)
Yes, you could, but it would be really quite disgusting. I briefly contemplated setting it up, but decided not to touch it with a barge-pole.
I'm going to see if that approximate answer is anything nice before I even consider trying a mathematical argument.
16. I get 20.671 after 1000 iterations. This is fairly computationaly heavy (at least the way I did it), so I'm not sure if I can do better.

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