................................................................................. Door A B C Two contestants have to pick You pick A. Door C opens Switch or stay? Switch to Door B is the answer, right? Contestant two picks B, C opens Switch to A. Right? Both contestants knowing the Monty Hall problem both agree to switch knowing their opponent is a sucker because their door had a 2/3 chance of being the right door How do both A and B have 2/3 odds of being correct? .................................................................................. lmfao Is there some magic power stoping player two from picking door C? If player two pick door C, switching means he get the car. Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots. Let's say the game start with 1 door (1 car, 0 goat). After you make a pick, the host add a goat door and ask whether you want to switch door or not. What is your winning chance if you stay with your first pick? The answer is obviously 100%. Super easy, right? D: Another way to look at it. Family feud style. Player one pick door 1 and door 2 is added. Player one know that door 1 is 100% a car but is sent off stage and player two comes out and has to pick between the two remaining doors. Player two pick door 1 same door as player one. D: Now, does player two has a 100% winning chacne or a 50% winning chance? D: Well 1/2 there are two choices.

The Monty Hall problem does not claim that whenever a wrong choice is removed you should switch to the other remaining one. It only claims that you should do it in the specific scenario that it was removed by a person that already knew the locations and was purposely avoiding to remove the winner option and also which the player picked. That's because the player only manages to start picking the correct option 1/3 of the time, but as the host is not allowed to reveal the prize anyway, he is who is forced to use his knowledge to keep it hidden in the other door that deliberately avoids to open in the 2/3 of the time that the player starts failing. So, the knowledge of the host is what makes his option good twice as much as the player's. In that way, if you want to compare the switching door with the selection of a hypothetical second player, that person should need to be cheating, because he would already know the correct option and his only restriction would be to not choose the same door that the first already selected. In that way, it is obvious that the cheater's choice will be good as long as the car is in any of the two doors that the first did not pick. Because of that, when you are deciding between the two final doors, you are basically betting who managed to get the prize: 1) you, that chose randomly from three, or 2) the cheater, that had opportunity with two doors. Better to bet on him. But the cheater already knows if he has the car or not; for him it does not exists that uncertainty about whether it is better to switch. In contrast, with two normal players each would win 1/3 of the time, as both choose randomly. Neither of them has advantage over the other. Notice that not every started game will manage to advance to the second part, because if the prize is in the door that neither of them selected, the host will not have available options to remove, or one of them will need to de eliminated. That occurs 1/3 of the time. But in Monty Hall, the host wouldn't have failed his choice in those 1/3 cases.

And with respect of what you say about the contestant 2 that comes later, there is no contradiction with it because the probabilities are not absolute. They are a measure of the information that you have, and different people can have different information. Just think about the host. For him the probabilities for the car being behind a door are not 1/2 nor 1/3 nor 2/3, but already 100% for one and 0% for the others, because he already knows where it is. His information is complete. So, to understand this you can think about the probabilities as the frequency with which you would find the prize in each option in the attempts that you have the same information that you currently have. For example, the contestant 2 only knows that door C is opened, but he does not know which of the others is the switching door, so the two cases remain as possibilities. In the long run, when entering the room with that exact information, in about half of the attempts the switching option would be door A, and in the other half it would be door B. So he would end up accounting each one winning 1/2 of the time, but that's because he would be adding the wins for each of the two cases: when a door is the staying option and when it is the switching one: 1/2 * 1/3 + 1/2 * 2/3 = 1/2 * (1/3 + 2/3) = 1/2 But the person that already knows that door B is the switching one this time does not need to account the games in which door A would be the switching door. He can filter those games. So, to acquire more information is basically to filter games from the set that you use to calculate the proportion.

Ohhhhhhhhhhhh they're not equilateral