Monty Hall Problem Essay

Far from a simple question posed to an excited game show contestant, the Monty Hall Problem continues to fascinate and puzzle mathematicians worldwide. The situation originated on the 1960’s game show “Let’s Make a Deal,” in which a contestant was asked to choose between three doors, one of which had a car hidden behind it. This sounds fairly trivial; the contestant would clearly have a 1 in 3 chance of picking the “correct” door. But, the situation soon becomes more complex. Let’s say the contestant chose door #1. Instead, Monty opens door #2 (not chosen by the contestant), and shows them that there is no car behind it. So, here’s the question: should the contestant stick with door #1, or switch to door #3 in hopes of winning a brand new

Marilyn Von Savant, thought to have the highest IQ in the world, replied that if the contestant switches doors, he or she has made the right choice, in order increase to a 2/3 chance of winning. This sparked outrage from magazine readers and mathematicians alike, who intuitively thought that it should make no difference whether the contestant chose to switch doors or not. In reality, switching actually doubles your likelihood of winning. Here is the logic behind it: first the contestant will chose a door, and has a 1 in 3 chance of picking the door with a car behind it. Without showing the contestant what is behind the door they picked, the host opens a different door. The host knows where the car is hidden, thus will 100% of the time chooses a door without a car behind it. Between the two remaining unopened doors, the odds have now shifted. The host could only identify the non-car door based on two options, instead of three. Therefore, the contestant has a 1/3 chance of winning if they stay with their original pick, and a 2/3 chance of winning if they

P(A) and P(B) represent the probabilities A and B separate from each other, while P(A|B) represents conditional probability; we observe A assuming that B is true. The theorem states that posterior odds equal prior odds multiplied by the likelihood ratio. In other words, the theorem takes into account the original odds, along with an evidence adjustment in order to reach the current odds of a given situation. This theorem is proven by symmetry as if you use this strategy in advance, the randomization of the initial door choice allows for a justifiable answer. Here is the scenario for Bayle’s theorem. Let's say the candidate chooses door #1 and then Monty shows him a goat behind door #2. A will represent the situation in which the car is placed behind door #1 (originally picked by candidate), and B is the event that the host opens door #2 to show a goat. If event A is true, then the host will half of the time show us a goat behind door #2 and the other half of the time will show us a goat behind door #3. Instead, if event B is true, and the car is behind door #2, Monty will NEVER open this door. In the final situation, if the car is behind door