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You are standing on an infinite floor, made of an infinite number of infinitely long, mutually parallel wooden planks, each exactly 1 unit wide.

You have a thin needle (of negligible thickness) that is 1 unit long. You throw the needle, and it lands flat on the floor. What is the probability that the needle is contained entirely within one plank?

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Well if you look in the direction the planks are running, the width of the area the needle lands in will average out to .637 units, (average of half a sinewave). So the chance of a plank edge intersecting this is .637 in 1. So for the answer, the chance of it NOT intersecting is 0.363 in 1.

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I remember solving this problem a while ago:

There are two variables here:

The angle the needle lands at

The position of the needle in relation to edges of the wood.

I'm going to assume the planks are vertical.

Let T be the acute angle with the vertical that the needle lands at. The probability function of its angle is

f(T) = 1 / (pi/2) = 2/pi {0 <= T < pi/2)

The horizontal width of the needle = sin(T). Therefore the chance that the needle crosses a line at a given angle is sin(T)/1 (1 being the width of a plank) = sin(T)

Therefore the probability that the needle crosses an edge is:

integral(0,pi/2) {f(T)*sin(T) dT}

= 2/pi * integral(0,pi/2) sin(T) dT

= 2/pi * [-cos(pi/2) + cos(0)]

= 2/pi

The questions asks for the complementary probability which is:

(1 - 2/pi) or 0.363...

Edited by psychic_mind

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I used a much more complicated way than mick but my answer agrees with his (1-2/pi).

Imagine one plank running north/south and draw a line perpendicular to the plank edges across the plank. Take some random point p along this line a distance x from the left edge. If we place one end of a needle on p and rotate it, at some angles it will cross an edge and at others it won't. Start with the needle horizontal towards the left edge (man pictures would help here :P) and rotate it up an angle tL such that the other end of the needle is touching the left edge.

Then tL = arccos(x)

Between 0 and t the needle crossed an edge. Because of symmetry rotating down instead will yield the same angle. So on the left side, 2tL out of 2pi radians results in a crossover. Flipping the needle to the other side and doing the same thing gives:

tR = arccos(1-x)

so 2tR out of 2pi radians results in a crossover as well. In total [2arccos(x) + 2arccos(1-x)]/(2pi)% of the time the needle will cross over if one end lands a distance x from the left side.

Phew, just getting started. Now split up the line into a very large number of intervals of infinitesimal width dx. Add up the weighted probabilities of a crossover for one end of a needle in each interval all across the 1 unit length of the line:

Integrate between 0 and 1: [2arccos(x) + 2arccos(1-x)]/(2pi) * dx/1

The symmetry of arccos(x) and arccos(1-x) between 0 and 1 lets us simplify further to:

Integrate between 0 and 1: 4arccos(x)/(2pi)*dx

Anyways, when the dust settles we end up with 2/pi * [xarccos(x)-(1-x2)0.5] from 0 to 1

which gives, at last, 2/pi for the % chance of a crossover. So the chance of no crossover is 1-2/pi.

Phew...

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50%

If the needle falls parallel to the edges of the board then there is a 100% chance it will not be touching an edge.

If the needle falls perpendicular to the edges then there is a 0% chance of not touching the edges.

The average drop will result in the needle landing at a 45% angle compared to the edge, so the average distance between the two edges of the board will be 1/2 unit. Therefore, the odds are 50%.

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I'd say about 30%. It can land on one plank, or it can land on one plank and the plank to the right of it. Or it can land one one plank and the plank to the left of it.

Edited by moonthirsty

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I'd say about 30%. It can land on one plank, or it can land on one plank and the plank to the right of it. Or it can land one one plank and the plank to the left of it.

There are no planks to the left or right, Moonthirsty. Each plank is infinitely long. :)

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There are no planks to the left or right, Moonthirsty. Each plank is infinitely long. :)

It doesn't matter that each plank is infinitely long. Each plank WILL have two neighbours, either to the left & right or above & below. If there were no neighbouring planks then there would be only one plank. :)

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There are no planks to the left or right, Moonthirsty. Each plank is infinitely long. :)

0 DEG = 0 width = 100/100 probability

30 DEG = .5 Unit width = 50/50

45 DEG = .707 unit width = 29.3/100

60 DEG = .866 unit width = 13.4/100

90 DEG = 1 unit width = 0/100

AVG = approximately 38.54/100 probability

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This really reminds me of the exercise to calculate pi by throwing frozen hot dogs across the room.

Fun times.

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It doesn't matter that each plank is infinitely long. Each plank WILL have two neighbours, either to the left & right or above & below. If there were no neighbouring planks then there would be only one plank. :)

I misread his post. That's what happens when I try to do respond before my boss gets back. :)

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This really reminds me of the exercise to calculate pi by throwing frozen hot dogs across the room.

Fun times.

Yes indeed. Although a terribly inefficient way. :)

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This is Buffon's Needle Problem with l=d=1.

Buffon's needle problem asks to find the probability that a needle of length l will land on a line, given a floor with equally spaced parallel lines a distance d apart. The problem was first posed by the French naturalist Buffon in 1733 (Buffon 1733, pp. 43-45), and reproduced with solution by Buffon in 1777 (Buffon 1777, pp. 100-104). It was used in the 18th century to estimate the value of pi.

Buffon, G. Editor's note concerning a lecture given 1733 by Mr. Le Clerc de Buffon to the Royal Academy of Sciences in Paris. Histoire de l'Acad. Roy. des Sci., pp. 43-45, 1733.

Buffon, G. "Essai d'arithmétique morale." Histoire naturelle, générale er particulière, Supplément 4, 46-123, 1777.

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BUT!!! What if it falls in the seam between planks? Half in? Not in? BY the way I did stay in a Holiday Express once.

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Hmmm, I may be totally getting the wrong assumption here, but I would hazard a guess and say that it is a trick question, that it really depends on A). Real life-if the unit is a few thousand miles than pretty much no matter how thin the needle is you can't throw it so it would then depend on B). Where you are standing-if the direction you are facing is parallel or perpendicular to the tiles,and even C).What the tiles are made of-if it is glue than it would be easier to calculate, but if it is say marble it would roll(The needle)

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