You can better see it imagining you had a box with $100$ balls, from which $99$ are black and only $1$ is white, which is what you want. You grab one randomly and keep it in your hand without seeing it. Do you agree that you are $99/100$ likely to have picked a black ball, so in $99$ out of $100$ attempts (on average) the white ball would still be in the box?

In case you agree, let's keep that ball in your hand and now suppose that another person deliberately pulls out $98$ black balls from the box. With "deliberately" I mean that that person sees what he is pulling; there is no risk that he removes the white ball by accident. In this way, there are only two balls remaining, one in your hand and one in the box, and one of them is necessarily the white.

What do you think is the probability that the white ball is which is in the box? If you say $50$%, what happened with the $99$ out of $100$ attempts in which it was still in the box? The revelation of the $98$ black ones didn't move it from the box to your hand.

Before the revelation of the $98$ black balls, the cases are:

```
Hand || Box
=============================================
1) In 99 out of 100 attempts -> 1 black || 98 black ones and 1 white
2) In 1 out of 100 attempts -> 1 white || 99 black ones
```

So, when the other person removes the $98$ black balls from the box:

```
Hand || Box
=============================================
1) In 99 out of 100 attempts -> 1 black || 1 white
2) In 1 out of 100 attempts -> 1 white || 1 black
```

So, it is true that you always end with two balls, one white and one black, but the important thing is that they are in two different positions (hand or box), and those two positions depend on the first selection. Moreover, that first selection determines that the white ball will end more frequently in the "box" position than in the "hand" position.

The way you are thinking the Monty Hall problem is like since you are always going to end with two balls, it would be the same if you started with both in the box and you had to grab one. But it is not the same. One thing is the probability to get the correct one when you randomly pick from two, and another different thing is the probability that the correct is already set in one position or in the other.

Note that if you randomly decide if you will pick the ball in the box or the ball in your hand, like flipping a coin, then you will get the white $50$% of the time. But that does not mean that it is $50$% of the time in the hand and $50$% in the box. It is because the extra times that you guess right picking the one from the box are compensated with the extra times you guess wrong picking the one from the hand. The $50$% $= 1/2$ is the average of the two cases:

$$1/2 * 99/100 + 1/2 * 1/100$$
$$= 1/2 * (99/100 + 1/100)$$
$$= 1/2$$

But if you always pick the ball that is in the box, your chances are:

$$1 * 99/100 + 0 * 1/100$$
$$= 1 * 99/100$$
$$= 99/100$$

In Monty Hall it occurs the same. Since there are two incorrect doors and a correct one, it is like if the $3$ doors were $3$ balls in the box, $2$ blacks and $1$ white. The initial selection is like when you start grabbing one ball randomly, and after the revelation the switching door is like the other ball that was left in the box. In $2$ out of $3$ attempts you pick a wrong door (like you would pick a black ball $2$ out of $3$ times) so in $2$ out of $3$ attempts the correct one will be the other the host leaves closed.