If we calculate Monty Hall problem from individual point of view, we obtain the following options
╔═══════════╦════════╦══════╦══════════════╗════════╗
║ Your Pick ║ Prize ║ Open ║ Don't switch ║ Switch ║
╠═══════════╬════════╬══════╬══════════════╣════════╬
║ 1 ║ 1 ║ 2 ║ win ║ lose ║
╠═══════════╬════════╬══════╬══════════════╬════════╣
║ 1 ║ 1 ║ 3 ║ win ║ lose ║
╠═══════════╬════════╬══════╬══════════════╣════════╣
║ 1 ║ 2 ║ 3 ║ lose ║ win ║
╠═══════════╬════════╬══════╬══════════════╣════════╣
║ 1 ║ 3 ║ 2 ║ lose ║ win ║
╠═══════════╬════════╬══════╬══════════════╬════════╣
║ 2 ║ 1 ║ 3 ║ lose ║ win ║
╠═══════════╬════════╬══════╬══════════════╣════════╣
║ 2 ║ 2 ║ 1 ║ win ║ lose ║
╠═══════════╬════════╬══════╬══════════════╣════════╣
║ 2 ║ 2 ║ 3 ║ win ║ lose ║
╠═══════════╬════════╬══════╬══════════════╬════════╣
║ 2 ║ 3 ║ 1 ║ lose ║ win ║
╠═══════════╬════════╬══════╬══════════════╣════════╣
║ 3 ║ 1 ║ 2 ║ lose ║ win ║
╠═══════════╬════════╬══════╬══════════════╣════════╣
║ 3 ║ 2 ║ 1 ║ lose ║ win ║
╠═══════════╬════════╬══════╬══════════════╬════════╣
║ 3 ║ 3 ║ 1 ║ win ║ lose ║
╠═══════════╬════════╬══════╬══════════════╣════════╣
║ 3 ║ 3 ║ 2 ║ win ║ lose ║
╚═══════════╩════════╩══════╩══════════════╩════════╝
According to this options, after the opening of the door which does not have the prize the probability of the win becomes %50. Since, the presenter has two options when the participant select the door that has the prize.
On the other hand, from the viewpoint of the game, when the participant selects the door that has the prize, there is no difference opening one of the door that is not selected by the participant. The options from the perspective of the game given below
╔═══════════╦════════╦══════════════╗════════╗
║ Your Pick ║ Prize ║ Don't switch ║ Switch ║
╠═══════════╬════════╬══════════════╣════════╬
║ 1 ║ 1 ║ win ║ lose ║
╠═══════════╬════════╬══════════════╣════════╣
║ 1 ║ 2 ║ lose ║ win ║
╠═══════════╬════════╬══════════════╣════════╣
║ 1 ║ 3 ║ lose ║ win ║
╠═══════════╬════════╬══════════════╬════════╣
║ 2 ║ 1 ║ lose ║ win ║
╠═══════════╬════════╬══════════════╣════════╣
║ 2 ║ 2 ║ win ║ lose ║
╠═══════════╬════════╬══════════════╬════════╣
║ 2 ║ 3 ║ lose ║ win ║
╠═══════════╬════════╬══════════════╣════════╣
║ 3 ║ 1 ║ lose ║ win ║
╠═══════════╬════════╬══════════════╣════════╣
║ 3 ║ 2 ║ lose ║ win ║
╠═══════════╬════════╬══════════════╬════════╣
║ 3 ║ 3 ║ win ║ lose ║
╚═══════════╩════════╩══════════════╩════════╝
Therefore, the nature of the game provide contestants %66 chance of winning the game. But from perspective of the participant, there is no difference between switching it or not.
I am not sure about my conclusions. I hope somebody make it clear in here.
The rows in your first table do not occur with equal frequency. You cannot just count them; you have to weigh them.
When the contestant picks a door and the prize is indeed there, then Monty can pick from two choices with equal probability (1/2).
When the contestant picks a door and the prize is elsewhere, then Monty must pick from the only choice with certainty (1). Thusly these rows each have twice the probability weight of each among the above mentioned rows.
Hence the probability for "win when you switch" weighs twice that for "loose when you switch". $$\mathsf P(\text{win when you switch})=2/3\\\mathsf P(\text{loose when you switch})=1/3~~$$