0

Probability implies that "The Galton Board consists of a vertical board with interleaved rows of pegs. Beads are dropped from the top and, when the device is level, bounce either left or right as they hit the pegs." However, I ask if this is always true or can a perfectly symmetrical round bead dropped into a perfectly level Galton Board indefinitely balance on a peg?

stafusa
  • 12,435
James Goetz
  • 541
  • I think this is a question that sounds like physics, as it seems to refer to physically real objects like beads and pegs, and physically real forces like gravity. But the phrase "perfectly symmetrical round bead dropped into a perfectly level Galton Board" makes it really about imaginary objects with immeasurable properties, i.e. not physics. – Jahan Claes Jun 21 '19 at 19:42
  • Looks very similar to https://physics.stackexchange.com/questions/175985/can-we-theoretically-balance-a-perfectly-symmetrical-pencil-on-its-one-atom-tip – BowlOfRed Jun 21 '19 at 19:48

2 Answers2

1

No.

In an idealized situation where the (classical) system is perfectly isolated from the rest of the universe, in vacuum, etc., the probability of that happening by chance is zero.

The reason it's zero is that this equilibrium point is unstable, meaning that any deviation from it, no matter how small, drives the bead increasingly away from this point, down the board. Here, "by chance" is important, since there are starting positions from where a bead do end up balanced, but, and that's the key point: there are infinitely more starting positions from where the bead doesn't.

The situation is comparable to hitting a bull's eye by chance: if $A$ is the area of the whole target (let's assume you always hit the target) and $a$ is the area of bull's eye, then the chance of an ideal dart tip (a point) hitting the bull's eye is simply $P=a/A$; now, if the bull's eye is not a finite region, but a point, then it has zero area, $a=0$, and thus $P=0$.

Of course, if you don't have a usual Galton board (where the beads are dropped without care and usually simultaneously in great number), but a specially designed one where every bead is dropped separately from the precise position from which it ends up balanced on the first peg, then you'll end up with all the beads resting on top of each other, balanced on the peg.

A few minor extra points to note:

  1. you don't need perfect beads and board for that: you can find unstable equilibrium points for general shapes.
  2. we're not talking about a system that's perfect with respect to energy conservation: if the collisions were perfectly elastic, instead of reaching equilibrium positions, the beads would bounce around forever.
  3. you can see the board as a chaotic billiard.

And, last but not least, even if you balanced the bead by hand, in practice it would not remain there for long - due to thermal noise, the device being illuminated and the overall existence of the rest of the universe and even the quantum mechanical uncertainty principle, as this answer on balancing a pencil on its tip makes very clear.

stafusa
  • 12,435
  • Are you saying that idealized classical mechanics implies a zero chance? Could you please elaborate on that? I actually want to know if it is a minuscule probability greater than zero, it approaches zero, of if it is plain old zero. I know a little about classical probability and far less about Newtonian mechanics. – James Goetz Jun 22 '19 at 01:40
  • @JamesGoetz It's zero in the ideal situation; alternatively, it approaches zero as you approach the ideal situation. I've elaborated the answer, it should be clearer now. – stafusa Jun 22 '19 at 07:02
0

Yes.

In an idealized mathematical universe satisfying the laws of classical mechanics, if you set up the initial conditions just right - and it sounds like that’s essentially what you’re referring to when you talk about a “perfectly symmetrical” bead being placed on top of a “perfectly level Galton board” - the system can remain in an unstable equilibrium “indefinitely”. That is simply a mathematical statement about the solution to some differential equation.

Of course, in our actual physical universe, there are no perfectly symmetrical beads, so because of that (and many other reasons having to do with quantum mechanics, the uncertainty principle, the gravitational influence of a butterfly in the Amazon rainforest, etc) this mathematical phenomenon is something you will never observe in real life.

GenlyAi
  • 850