I just did this recently:

You have 3 choices: A B or C each is equally likely for a prize.

You choose one a random, you have a 1/3 chance of being right.

Now Monty reveals that either C is not the prize by showing it. You still don't know what is behind A or B.

He asks do you want to keep A or take B, what do you do?

That's the problem. The solution is that you should switch the odds are better. When you started you had a 1/3 chance of being right, that means that you had a 2/3 chance of being wrong. Think of that as a binary choice of it's own each with odds of 1/3 and 2/3. Now he's eliminated one of the choices so you might think you only have a 50/50 chance which you do but not when you consider that he has to show you the 'not prize'.

If you reason that you had a 1/3 chance of winning or a 2/3 chance of winning you should switch to get into the 2/3 chance group.

If you don't the best you can do is still 50/50.

If you have ever heard of the gambler's fallacy I think this is kind of the reverse of it. That goes via the wrong idea that I flipped tails 10 times in a row so the next one must be heads "to even it out", or tails because "I seem to be on a streak". Well no, in that case the chance is still 50/50 on the new coin flip.

Monty Hall is like this in reverse, because information is being added to the equation.