Think of the game as drawing a ball, blindfolded, from a box, which contains 1 white ball and 2 black balls, except that every time after you pick a ball, a black ball will be thrown away by the host, so only one ball remains in the box.
If you stick with the ball that you pick every time you play the game, regardless of what the host does afterwards, then in the long run you will end up having a white ball in your hand 1/3 of the time and a black one 2/3 of the time.
However, if you choose to switch every time regardless of the colour you picked, then the times when you should end up having a white ball become black and vice versa. So, in the long run, you will end up having a black ball in your hand 1/3 of the time and a white one 2/3 of the time.
Any constructive criticism is welcomed! Thank you!
Edit
Re the comment of aschepler, you are forced to assume that you were blindfolded before picking the ball, and therefore never knew the color of the ball that you had picked.
I agree that this does analogize to the Monty Hall problem, which I will explain below. My only gripe with the query, as it is presented, is that the last paragraph, while accurate and valid is somewhat convoluted:
In the normal Monty Hall scenario, (1/3) of the time, you correctly picked the right door, and (2/3) of the time you did not. Always switching :
In the (1/3) of the time that you initially picked correctly, you have turned victory into defeat.
In the (2/3) of the time that you initially picked incorrectly, you have turned defeat into victory.
Your statement that
is somewhat convoluted. I would have expressed it as (assuming that you always switch):
the (1/3) of the time that you picked a white ball, you have turned victory into defeat.
the (2/3) of the time that you picked a black ball, you have turned defeat into victory.
Beyond that, in the case where you pick Door #1, Monty Hall's showing that behind Door #2 is a goat does roughly analogize to the host selecting a black ball from the ones remaining and discarding it.
The key points here are that when Door #1 has the grand prize, Monty has a choice of which Door (#2 or #3) to show. Similarly, when your initial selection is the white ball, the Host has a choice of which black ball to discard.
Alternatively, if Door #1 does not have the grand prize then, (assuming that Door #3 does have the grand prize), Monty has no choice but to reveal Door #2. Similarly, if your initial selection is a black ball, the host has no choice but to discard the lone remaining black ball.