How about this. You've got some kind of pachinko type machine, where a steel ball drops down into one of two funnels. The left funnel is half the size of the right funnel, so the ball is twice as likely to land in the right funnel. The left funnel eventually drops the steel ball into box 1, while the right funnel eventually drops the steel ball into box 2.

So even though there are 2 boxes, and the ball could end up in either, it's more likely to be in box 2.

That's basically the Monty Hall problem, except instead of funnels you have a door elimination process. The door you initially choose is box 1, and the door you haven't chosen (and that Monty hasn't opened) is box 2.