The Monty Hall problem.

[Anymouse]

Think of a coin toss. Whether the person tossing the coin understands probability, they will, given enough trials, get about 50% heads and 50% tails.

Given a clueless audience, they will either choose the same door and win 33% of the time, or switch and get 66% of the time. It plays out whether the audience/guests understand or not. A stumped mathematician does not change the probability, nor the outcome. The outcomes, given enough samples, will always play out. Even if they all choose the same door. It still confirms the odds.

The full Monty Hall problem includes the following caveat, which is why the odds are better by switching. When the contestant picks a door, Hall then exposes a goat behind one of the doors not picked. (He has thus eliminated one of the bad choices).
If the contestant starts by picking the door with the prize, Hall can expose either door. The choice then becomes keep (win) or switch (lose).
If the contestant starts by picking a door with a goat, Hall can only expose the other goat.
Marilyn vos Savant (editor of Parade magazine) tried to get people to visualise the solution thus: Hall will never expose the car. If the player has chosen a goat, he can only expose the other goat.
Now imagine if there were a million doors, and behind door 777,777 is the car. Hall opens all the doors with goats. Which is the best choice: sit or switch?
Another way to look at it is intuitively: after one makes the initial pick, the odds of winning are 1 in three. The player can only lose he has the car to begin with (since the host will expose a goat after the pick, the odds of picking that door become zero). Thus, the odds of the prize being behind the remaining door are 2 in three.
For a practical demonstration you can conduct the experiment yourself. Have another person take two red playing cards (goats) and one black playing card (car), place face down. That person acts as Hall, and after you pick a card, exposes one of the red cards. Repeat the experiment many times both not switching, and switching, and keep a tally. You will find that switching will yield the black card 2 times of three, keeping your choice will yield the black card 1 time of three.
For a full explanation of the problem, and the furore that arose as mathematicians tried to prove Marilyn vos Savant wrong (who correctly surmised the odds) and savaged her in the press until they grudgingly admitted she was right, see Wikipedia.
https://en.wikipedia.org/wiki/Monty_hall_problem