Earlier quoted context omitted.
As if algebra or calc was common sense. I did learn about the Monty Hall Paradox but I'm still doubting it
Consider 3 doors: A, B, C. If you pick the right one you win the prize. Say you pick door A. Monty Hall opens B or C and reveals that there's nothing behind. Now he asks if you want to stay with A or switch to the closed door. What should you do to maximize your chances of winning? Right before any of the doors is opened we have no idea what to expect, so there are 3 equaly likely possibilities: 1. prize is behind A…
You are shown 100 doors 1 ... 100. Imagine you pick 1 door.
Of the other 99 doors Monty tells you which 98 of them don't have the prize.
He now asks if you want to switch to the 99th door that he didn't cancel out.
Intuitively that seems to make much more sense to me.