A dumbbell probability problem?

Question

This is a probability problem, not a physics problem, so dynamical behavior of solid bodies shall be disregarded. An ideal dumbbell of 2 identical spheres is connected by a solid rod, so that the centers of the spheres are 1 unit apart. There are 2 infinitely long adjacent rows of spherical indentations 1 unit apart, the indentations of both rows being aligned with each other such that the distances between them are also 1 unit apart. The dumbbell is thrown upon these rows at random, so that any one sphere of the dumbbell will come to a rest in any indentation with equal likelihood. What's the probability of the dumbbell landing with one sphere on row 1, and the other on row 2? Give reasoning, please?

I've tried to put in a lot of details in order that this problem would NOT be considered as a physics problem! This is a probability problem! Disregard all consideration such as machine spacing of indentations and bounce and friction, etc., as well as, "well, but what if one of the balls happened to land inbetwen the indentations?".

Yes, assume that the two spheres will always end up inside indentations, without "machine interference" getting in the way.

I could have also said, "There are two infinitely long adjaent rows of square boxes which are aligned side-by-side, so that each sphere of the dumbbell will always end up in some box", etc.

One sphere per box!

Answer 1

There are two approaches to this that I can think of, the trouble is one suggests that the probability is 1:3, the other that the probability is 1:2.

I'm going to argue 1/3

Given that one end of the dumbbell lands in any given slot, there are only three possible places that the other end could lie, and only one of these forms a bridge between the rows. Applying this logic to all of the slots individually, the possibilities add up to:

P = (1*∞) / (3*∞) = 1/3.

Answer 2

First, assume that the first dumbell sphere ends up center of the indentation.

Then, the second dumbbell is limited to the points on the circle radius 1.

First order, the probablity will be the arc of the indentation with respect to the center of the indentation where the sphere currently rests divided by 2 pi. This means that the center of the second sphere will be somewhere over the indentation and will eventually settle into the center of that indentation.

>>>Edit
The angle would be Theta = 4 * arcsin (r/2), where "r" is the radius of the indentation. So the solution would be

= 4*arcsin(r/2)/(2*pi)

*Note, the bar and the distance to the center of the second indentation will form two sides of an isosceles triangle joined together by r. Split r in half and Sin Beta = (r/2)/1. Theta will be 4* Beta.

End Edit<<<<<

To the second order, you have to take into account how deep the indentations are, because after the first dumbell sphere settles in place, it will shorten the effective length of the rod in the horizontal direction and thereby make a smaller angle and a smaller probability of the second sphere settling into the second indentation. But that is 3-D stuff.

......................................

And here I go solving the hard problem! ....

To solve this, you are going to have to make an assumption on how you throw the dumbbell. In other words, how wide past the center line between the rows can a dumbell be?

OK. First assumption.the sphere closest to an indentation will stick to that indentation. Second assumption, the second sphere can be oriented in any horizontal direction from the first sphere, i.e., 2*pi. Third assumption, the second sphere will end up sticking in the closest indentation. This would mean the answer would be 1/4 since the second sphere would only be closest if it were oriented within 45 degrees (pi/4, total angle would be 90 degrees or pi/2) of the perpendicular to the second row.

(This also assumes that the first sphere to stick is at or within the centerlines of the two rows and the second sphere is randomly oriented with respect to the first sphere. If the first sphere is allowed to be farther away than the centerline, more barbells would end up solely in one row.)

But if we put the constraint that the center of both spheres on the dumbbell have to be within the centerlines of the indentations, and that they could be oriented in any random direction, then the answer would be 1/2. Under this scenario, any time the bar is orientation within pi/4 of perpendicular to the second row, the second sphere would end up in that second row. Since you have limited the freedom of the barbell, the second sphere is limited to an angle of pi (180 degrees) with respect to the first sphere.

Answer 3

I know you put in a lot of details here, but I'm not sure if there are enough. Given no other information the answer can be zero. For example, say the spheres' radii are 10^(-10) inches and the unit you speak of is 1 mile. The likelihood of either of the spheres ending in any indentation is very close to 0, and even if one of them happens to end up in an indentation, the likelihood of the second one also ending in one is still practically 0.
Is it assumed that the two balls end up in indentations?

Answer 4

This is a version of Buffon's Needle

http://www.mste.uiuc.edu/reese/buffon/buffon.html