“The Monty Hall problem has a 50-50 chance either way”

It's a common belief that in the popular game show scenario involving three doors, where one hides a car and two hide goats, your odds become 50-50 once the host reveals a goat behind one of the unchosen doors. This idea stems from a seemingly logical simplification: with two doors remaining, it feels like an equal choice. However, this intuition overlooks a crucial element of the game that fundamentally alters the probabilities.

The truth is that sticking with your initial choice leaves you with a 1/3 probability of winning the car. When you first pick a door, you have a 1/3 chance of selecting the car and a 2/3 chance of selecting a goat. The critical factor is that the host, who knows where the car is, will *always* open one of the unchosen doors to reveal a goat. This action isn't random; it concentrates the initial 2/3 probability of having picked a goat onto the *other* unopened door. Therefore, switching doors gives you a 2/3 probability of winning, effectively taking advantage of the host's informed decision.

People often struggle with this concept because their brains tend to reset the probabilities after the host's action, treating the remaining two doors as a fresh 50-50 choice. They fail to account for the information conveyed by the host's deliberate choice of which goat door to open. The host's knowledge and actions are what fundamentally change the game from a simple 50-50 coin flip into a situation where switching doors significantly increases your chances of driving away with the prize.