Regarding the generalized Monty Hall problem:

The blue frame around the doors should go from door 2 up to door k + 1 (instead of k), as the host opens k = (k + 1) - 2 + 1 doors. The red frame should go from door k + 2 (instead of k + 1) up to n. This way there are indeed n - k - 1 = n - (k + 2) + 1 remaining doors, as stated in the equations / below the graphic.

In addition, it is unclear which equality holds when k = n - 2. The given inequality, 1/n * ((n - 1) / (n - k - 1)) >= 1/n, turns to an equality only for k = 0, i.e. when the host does not open any door (because when you randomly switch to another door *without additional information*, the probability of getting the car is the same as when you stay on your initial door).