Anyone familiar with the Monty Hall problem?

We have in the UK a game show called Deal or No Deal. It consists of a number of boxes, say 20, and the contest has to select one at the start of the game that remains unopened. The rest of the game is spent choosing various boxes at random to eliminate until (potentially) the final 2 are left. The eliminated boxes are opened to reveal what was inside. The boxes all contain different amounts of money, and I say potentially left because at various points in the game, the contestant can choose to gamble what he think is is in his box (based on the eliminated amounts/boxes) on what the 'banker' offers him. If the game were played to completion however, we would be left with 1 box in front of the contestant and one other box. On occasions, an offer of exchange is made, and the question is raised, as in the Monty Hall Problem, is it advantageous to change.

To make this question/problem easier, we can assume the the box numbers take the place of the amounts of money, and the numbers are displayed inside the box so they appear unmarked. 20 is obviously the highest number, and therefore the best box to end up with, 1 the worst. Simpler than trying to remember the denominations of money involved.

In the original problem, Monty Hall has knowledge of the contents of the boxes, and it is suggested by many answers I have read that this is what creates the skewed probability.

If by chance we end up with the box 1 and the box 20 in play at the end of the game, because my initial selection was from 20 boxes, it is most unlikely that I picked box 20, the chances are I picked any one of the other 19 available, and so it is overwhelmingly likely to be in the other box. If the odds are 19/20 that i'll win if I change, can they become 50/50 simply by using different logic?