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Posted 04 November 2013 - 06:36 PM
Posted 05 November 2013 - 09:59 PM
I do not have an answer yet, but I did some calculations that might help get this started.
Edited by dgreening, 05 November 2013 - 09:59 PM.
Posted 06 November 2013 - 01:53 AM
Edited by mmiguel, 06 November 2013 - 01:55 AM.
Posted 06 November 2013 - 02:08 AM
i probably oversimplified this and don't have the right answer for the 3x case, but oh well - i'm tired
Edited by mmiguel, 06 November 2013 - 02:16 AM.
Posted 06 November 2013 - 06:37 PM
Spoiler for Equal speed case
They can follow other than mirror paths to meet at the same point.
If they meet, they meet at the diagonal. The probability (for each one) to get to a particular point on the diagonal is:
Pascal's triangle/(2 power n).
I hope someone finds a formula for the product.
Posted 06 November 2013 - 09:57 PM
Harey is absolutely correct.
I started playing with some small values of n to get some sense for the problem.
It turns out that the chances of hitting an particular square along that diagonal looks like "binary normal distribution" [I am sure there is a better description for this]
Of all the possible paths available. The only way to get to one of the corners of the diagonal row [for corners that are not A or B] and that is to choose all of one type of turn [up, up, up ..., up]. So if that route was chosen, the odds of colliding are very low.
Moving inward from the corners. the only ways to get to the next 2 spots is to use the same sequence as above, but replace an "up" with a "right".
and so on
Consider the 3x3 grid. There are 6 possible paths.
- 2 of them [1100 and 0011] will take you to the 2 diagonal corners
- 4 of the [1001, 1010, 0110, 0101] will take you to the center square
- so distribution [of ways to get to that spot] along the diagonal will be 1, 4, 1
- in this case there is a pretty good chance of colliding on the center spot
Now consider a 4x4. There are 20 combinations.
- The diagonal is now 4 squares long
- 2 of the  and  end up at the corners
- 9 sequences will end up at the "middle" square on the top left
- 9 sequences will end up at the "middle" square on the lower right
- Thus the distribution will be 1, 9, 9, 1
- In this case the 2 players could pass each other on the adjacent squares and the likelihood of collision would be much lower.
- I would guess that cases where N is an even number will have a lower probability than adjacent odd numbers.
Posted 07 November 2013 - 08:48 AM Best Answer
Past prime, actually.
Posted 12 November 2013 - 07:41 PM
Spoiler for I see the probability for equal speed as
I think you are on the right track. I like your analysis.
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