What's wrong with my attempt to find the probability that the strategy of always switching succeeds, given that Monty opens door 2?
Solution 1:
If $O_2$ happens, then either $D_1$ or $D_3$. $D_1$ and $D_3$ are equally likely but in $D_1$, Monty had a choice, he could have picked $O_2$ or $O_3$ but did pick $O_2$. In $D_3$ Monty didn't have a choice, he always has to pick $O_2$.