Are you sure? I may have had some of my figures incorrect, but the logic is not:
The Monty Hall problem is a probability puzzle based on the American television game show Let's Make a Deal. The name comes from the show's host, Monty Hall. The problem is also called the Monty Hall paradox, as it is a veridical paradox in that the result appears absurd but is demonstrably true.
A well-known statement of the problem was published in Parade magazine:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? (Whitaker 1990)
As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car, from 1/3 to 2/3.
The Monty Hall problem, in one of its common formulations, is mathematically equivalent to the earlier Three Prisoners problem, and both bear some similarity to the much older Bertrand's box paradox. These and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly, and have led to numerous psychological studies that address how the problems are perceived. Even when given a completely unambiguous statement of the Monty Hall problem, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.
The player, having chosen a door, has a 1/3 chance of having the car behind the chosen door and a 2/3 chance that it's behind one of the other doors. As the host opening a door to reveal a goat gives the player no new information about what is behind the door he has chosen, the probability of there being a car remains 1/3. Hence the car is then with chance 2/3 behind the remaining unopened door (Wheeler 1991; Schwager 1994). Switching doors thus wins the car with a probability of 2/3, so the player should switch (Wheeler 1991; Mack 1992; Schwager 1994; vos Savant 1996:8; Martin 2002). In order to convert this popular story into a mathematically rigorous solution, one has to argue why the probability that the car is behind door 1 does not change on opening door 2 or 3. This can be answered by an appeal to symmetry: under the complete assumptions made above, nothing is changed in the problem if we renumber the doors arbitrarily, and in particular, if we switch numbers 2 and 3. Therefore, the conditional probability that the car is behind door 1, given the player chose 1 and Monty opened 2, is the same as the conditional probability that the car is behind door 1, given the player chose 1 and Monty opened 3. The average of these two (equal) probabilities is 1/3, hence each of them separately is 1/3, too.
The analysis can be illustrated in terms of the equally likely events that the player has initially chosen the car, goat A, or goat B (Economist 1999):
he above diagram shows that a player who switches always gets the opposite of their original choice, and since the probability of that choice being a goat is twice that of being a car, it is always advantageous to switch. In other words, the probability of originally choosing a goat is 2/3 and the probability of originally choosing the car is 1/3. Once Monty Hall has removed a "goat door," the contestant who chose the door with a goat behind it will necessarily win the car, and the contestant who originally chose the car will necessarily "win" the goat. Because the chances are 2/3 of being a contestant who originally chose a goat, probability will always favor switching choices.
