[yqsk]

For those looking for the simple to understand solutions:

Bertrand’s box paradox:

The trick here is that you are not actually drawing a card, but a card side. There are 3 cards, but each card has two sides so there are 6 card sides and half of those sides are white.

So, now we have to pick one of the 3 white card sides at random. Of those 3, 2 have white on the other side and one has black. Therefore, if you pick a white side at random, there is a 2/3 chance that it is white on the other side.

Monty Hall Problem:

Say you're going to switch no matter what. Then the only way you can lose is if you pick the right door on your first try - a 1/3 chance. Conversely, as long as you pick either of two wrong doors first you win. Therefore, you have a 2/3 chance of winning if you switch vs a 1/3 chance if you stick with your original pick.

[avmich]

> Bertrand’s box paradox:

I suspect the problem isn't well stated, at least in the article. If one considers a random choosing a card, repeating until the card has a white side on it, and then putting that card a white side upwards, then probability doesn't seem to me be the one mentioned.