It is 2/3.

This is not correct. The host knows the locations of the contents and according to the rules of the game he cannot reveal the door that has the prize and neither which the contestant picked. Since the contestant only starts choosing the car door 1 out of 3 times on average, not 1 out of 2, but the car cannot be revealed anyway, then the host is who purposely leaves it in the other door that he does not open the remaining 2 out of 3 times. So, always two doors at the end, but the player's selection will be correct with a frequency of 1/3, and the other with a frequency of 2/3. And if the game started with many more doors, like 1000, almost never the player would start picking the correct, so almost always the correct would be the other that the host leaves.

The best way to show it is better to switch is honestly simulations, which show that you indeed get the car 2/3rds of the time if you switch.

Nope. Your 1/3 (and the host's 2/3) was locked at your pick of a door. That doesn't change when he shows you a goat, his remaining door is still 2/3. Your initial guess was probably wrong so it is better to switch. If the host offered you to switch from your one door to his two (both unopened) doors, would you do it?

@@Hank254 how is 1 door 2/3? Each is a 3rd. 1/3 was opened leaving two 1/3 opportunities left or 50/50 of the remaining unopened doors

@@notallowed337 The one door you picked has 1/3 and the host has 2/3 because he has two doors. The host opening one of his doors doesn't change the odds of the door you picked. You already know he has to have a goat (if he has two doors and there is one prize) so when he shows you the goat is isn't changing anything. All it does is show you which of his two doors probably has the car. It is the exact same thing as if he offered you to switch to his two unopened doors.

@@Hank254 but his remaining door isn't 2/3. You're using the 2/3 out of context. Yes he has 2 doors each carrying a 1/3 chance of having the car thus yes 2 doors equals 2/3 chances. But once one door is opened showing nothing then the 1/3 chance dissolves with it. Now there's 2 doors remaining. Each now has a 50% of having the car. Or 2 doors remaining each with their initial 1/3 chance of having the car.

@@notallowed337 It really doesn't. It might seem like that's what should happen but that is what most people get wrong. The reason it doesn't is because when you pick a door, it breaks the doors into two groups... Contestant's and Host's. When the host deliberately opens his goat door, he is only changing the probability of his own doors. You can test this with three playing cards very easily. When you do, you will notice that when you switch, you always get the opposite of your original pick. Since you have a better chance of picking a goat on your initial pick, you are better off switching.

except you've already chosen and now that there's 2 doors its a 50% chance but if you change its still a 50% chance there's no point in changing

@@mag-narwhal You are operating from the idea that all possibilities are equal when your initial choice always has a 2/3 chance of being wrong. once the host eliminates one of the other two options it is statistically advantageous to switch. this is because if you chose wrong initially (which as previously stated is most likely) the host is forced to pick the only empty door of the two you didn't pick.

@@MythBoy99 if you chose right initially the host is still gonna let you try and pick an empty door theres no guarantee the new one is right plus if you win you'll feel 100 times better cause you got it first try

@@mag-narwhal Of course there is a chance you were right the first time, a 1 in 3 chance. which is why switching gives you a 2 in 3 chance of success. And the satisfaction you claim you would get from being correct first try is exactly the psychological element that prevents people from solving this more often.

The fact that you are choosing between two does not imply that they must be equally likely to be correct. Suppose we put a random person from the street in a 100 meters race against the world champion; you don't see the race but later you have to bet who of those two won. Would you say they are each 50% likely to have done it, or do you think the champion had a clear advantage? In Monty Hall it occurs the same. There are two doors remaining but each was left by a different person: one was randomly chosen by the contestant but the other was deliberately left by the host, who knew the positions and according to the rules of the game couldn't discard the contestant's option and neither which has the prize. That means that the other he leaves closed is the best one that could be found from the rest, because if any of them has the car, that is preisely which he will avoid to reveal. In this way, you can see the switching door as if it were the selection of a second player that was cheating, because that person looks inside the two doors that the first one did not pick, looking if one of them has the car, and takes which prefer from them. The revealed door is like which that person would reject from the two that checked out, that would always be incorrect, of course. So, two doors remain at the end: the first player' selection and the cheater's selection, but that does not mean they are equally likely to have won. The first would only hit the correct 1 out of 3 times, but the cheater everytime the first fails: 2 out of 3. And if the game starts with many more doors, like 1000, more difficult for the first to randomly pick which hides the prize, so easier that the cheater is who finds it checking all the other 999 doors.