Yes, if you were initially presented with only two doors and no additional information, choosing one of them would indeed give you a 50% chance of winning. However, the Monty Hall problem is different because it involves three doors and additional information provided by the host.

Let's break down the Monty Hall problem step by step:

1. You initially choose one of the three doors (let's say Door 1). At this point, you have a 1/3 chance of having picked the car and a 2/3 chance of having picked a goat.

2. The host, who knows what's behind each door, opens one of the remaining doors (let's say Door 3) to reveal a goat. The host will always reveal a goat, as they never open the door with the car behind it.

3. Now you're given the option to switch your choice to the other unopened door (Door 2). Here's where the intuition can be tricky. Since the host always reveals a goat, the initial 1/3 chance of having picked the car (Door 1) and the 2/3 chance of having picked a goat (Door 2 or Door 3) haven't changed.

4. By switching to Door 2, you're essentially betting that your initial choice (Door 1) was wrong, which had a 2/3 chance of being the case. So, by switching, you now have a 2/3 (or 66.67%) chance of winning the car.

In summary, the Monty Hall problem isn't a simple 50/50 choice between two doors. The host's action of revealing a goat and the fact that there were initially three doors are crucial factors that make the probability of winning by switching 2/3, not 50%.
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