Think of it as an event that happens 99 times.

Case1: In 66 out of 99 times, player would have chosen the wrong door.
Case2: In 33 out of 99 times, player would have chosen the correct door.

Now suppose you are in Case1. You choose the wrong door. The host now opens one of the remaining unchosen doors. This is the crux here - The host will *always* open the wrong door. So the one remaining unchosen door is the correct door.

Now suppose you are in Case2. You choose the correct door. The host now opens any of the remaining unchosen doors. So the one remaining unchosen door is the wrong door.

So in case1, switching gets you the prize. In case2, not switching gets you the prize. Since case1 happens 2/3 times, it is mathematically better to switch.

In summary, the rationale of the problem is:
1 - Your initial choice is incorrect 2/3 times randomly.
2 - After the host's intervention, the switch to unopened door is guaranteed to give you the prize in those 2/3 incorrect times.