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| *Didn't have the time nor the courage yet to tackle the needle/circle problem.*
Maybe starting with a square instead of a circle is simpler in a first step, then try to extend it to circle. The square case should be somehow related to the Buffon needle problem. | |

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Hi,

also I didn't have the time. I was only thinking about analytic solution on it one time when driving home. As mentioned many times before we have the coordinates (x, y) of the center of the needle and the angle (alpha for example). Basically try to translate Buffon-Laplace needle (square) to half-circle with one side to be closed (forbidden to be crossed):

- Take half of circle

- Start with Buffon-Laplace needle problem (with all sides closed)

- Change 'left, up and down sides' of the square to be a circle.

- Calculate 'safe region' inside of this object where needle for sure will not touch or cross lines (let's name it region R1, all coordinates that are safe in all angles)

- Calculate the region between outer circle (left) side and inner 'safe' circle 0 beware, Needle problem is not the same - left side of it can not be crossed (needles MUST be inside of circle) so we need to solve P(outer circle) = P(crossing - needle problem with curved lines) - P(crossing only left side).

We have now R2 with probabilities with respect to distance & angle of them.

- Calculate the regione between 'safe space' and inner line - this again could not be straight forward Buffon needle problem because upper and downer regions does not have the same probability as middle one (in the most upper part for example the needle might not be under 90 degrees angle in respect to middle line - it might 'fall out' of the circle.

This is region R3. Maybe this region can again be divided to a square in the middle (R3a), angle-dependent body in the upper and downer part (R3b, R3c).

In the end we would need to divide the full region of half-circle and the sum of the probabilities.

This is just thinking how to 'attack' the problem... I hope I will have some time during the weekend to put it into equations and maybe some nice picture (will I get free beer if I provide nice picture??

).

Regards,

Jure