
Statistical paradoxes for sharping your Bayesian skills - gfrison1
https://gfrison.com/2019/statistical-puzzle-bayesian-conditioning
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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.

