1
$\begingroup$

Professional lottery machines like this one enter image description here

are accepted as producing genuinely random results. (Or random enough?).

However, it seems like the results are a largely a product of Newtonian mechanics, and are thus in some sense deterministic.

The question I have, what conditions would make them less random (short of actually rigging them) For example, how random would they be if they were operating in a vacuum?

$\endgroup$
  • 2
    $\begingroup$ While Newtonian mechanics is deterministic, the collisions in the machine cause extremely sensitive dynamic conditions which cause even the smallest differences in the initial conditions to blow up to a basically random position of the balls after a few collisions. The mathematical description of this sensitivity is called Lyapunov exponent and it describes how fast two trajectories of the system that are infinitesimally close in the beginning would diverge. One can, of course, rig the system by making some of the balls much heavier, but that is checked by officials. $\endgroup$ – CuriousOne Jun 3 '15 at 5:47
2
$\begingroup$

This is similar to the problem of the best way to shuffle a deck of cards. I think that, generally speaking, these types of mixing techniques (similar to just spreading out a deck of cards and making a mess pushing them around) are actually quite good at eliminating correlations if they are done for enough time. You are right in that this system is primarily governed by Newtonian mechanics, such that if you knew the initial conditions of the system well enough and had a powerful enough computer, you could probably predict the result, but for practical purposes this isn't so important. There are probably numerous ways of rigging a machine like this as well.

Some things that would make it less random include using fewer balls, having a smaller disk for them to move around in, and swirling the disk for a shorter period of time. Doing the experiment in vacuum would probably make the system a little easier to calculate, but again something like this would still be so difficult to calculate (you'd also have to very precisely know e.g. the location and structure of rough spots on the disk and the balls) that it isn't very important for anyone who actually cares about winning the lottery.

$\endgroup$
  • 1
    $\begingroup$ And even the slightest outside perturbations may influence the result (slight as in the gravitation of the person standing next to the rig). $\endgroup$ – Sebastian Riese Jun 3 '15 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.