Feynman's box and tennis balls problem

[vhbelvadi]

Introduction: This isn't really a homework question in that I recently came across Feynman's original problem sets that had accompanied his famous lectures of 1960 and decided to solve them (and I've been doing rather badly). But at some point, I figured it would fall into this category best.

Q. A box has a square bottom 1m on a side, vertical side walls 0.33m tall and no top. The box is set so its bottom is accurately horizontal, and 7 tennis balls are thrown into it. One more such ball is dropped vertically into the box. What is the probability that the dropped ball hits the bottom of the box without encountering one of the balls already in the box?

source: http://www.feynmanlectures.info/FLP_Original_Course_Notes/pages/ON1-215.html2. A rather vague attempt at a solution: Well, right now, I haven't gone too far from the basic probability equation (P(G) being the probability of hitting the ground, r being the radius of a tennis ball in metre):

P(G) = \frac{1m^2-\{7*\pi r^2\}}{1m^2} = 1-7*\pi r^2

(What with the bottom being perfectly horizontal and all (normal distribution?), I thought of working along the lines of the coin-tile problem here: http://mathworld.wolfram.com/CleanTileProblem.html )

But I doubt this is correct: 1. wouldn't we need to know the diameter of a tennis ball? Or, if we can work it out in terms of r, 2. should the vertical wall length not have something to do with it? Or am I going in a completely wrong direction?

 

[mfb]

1. wouldn't we need to know the diameter of a tennis ball?

If you don't know it, you can look it up or make an estimate (after you worked it out in terms of r, ideally).

2. should the vertical wall length not have something to do with it?

Why should it?

Check your condition for "ball is touching other balls before reaching the ground". The new ball is not a point. Once you fix that, you can find another issue, this will give a smaller correction to the calculated probability.

 

[Khashishi]

First you need to decide how accurate a result you want. It's a good idea to come up with a rough estimate first, then check that your more accurate result is close to the rough estimate. You've already come up with a rough estimate.

Ok, what is wrong with your estimate? Firstly, you didn't account for the size of the final ball. The centers of two balls can't be closer than 2*r without colliding, so the exclusion radius should be approximately doubled around each ball.
$P(G) \approx 1-2*7*\pi*r^2$
That should be fairly accurate in the limit that A(balls) << A(floor).
But, when the total ball area is not small, it becomes a lot harder. Halls cannot pack together to cover the entire area of the floor, so there's some packing fraction effect. I'm not sure how to handle this except by simulation, but maybe you can come up with some other approximations.

 

[Spinnor]

The probability will depend on how the 7 balls came to rest. In the unlikely event that that the 7 balls came to rest in a close packed hexagon, see below, (and packed into one corner) you would minimize your probability. The greatest probability occurs when the center of each of the 7 balls is a distance greater then 3/2r from another ball's surface or wall, in other words when they are spread out and away from the walls.

http://math.ucr.edu/home/baez/mathematical/hexagonal_close-packed_structure.gif

 

[mfb]

First you need to decide how accurate a result you want. It's a good idea to come up with a rough estimate first, then check that your more accurate result is close to the rough estimate. You've already come up with a rough estimate.

Ok, what is wrong with your estimate? Firstly, you didn't account for the size of the final ball. The centers of two balls can't be closer than 2*r without colliding, so the exclusion radius should be approximately doubled around each ball.
$P(G) \approx 1-2*7*\pi*r^2$

The 2 has to get squared, too.

An estimate for the deviations from this formula is harder, but also more interesting (in my opinion) :).