How the Monty Hall Problem Works
63The Monty Hall problem is one of the trickiest and most interesting logic problems I have ever seen. If you haven't heard of the problem, here is how it goes:
The Problem
You are on a game show and the host shows you three doors. He announces that behind one door lies a car (which you want) while behind the other two lie goats. You do not know which is behind each door. The host asks you to select a door and you do, but before opening the door the host announces that he will open a door with a goat before you open yours, then he opens another door to reveal a goat. Now there are two doors left: one with a car and another with a goat. The host now asks you, "Would you like to switch your door?" The question is, what should you do? Should you switch doors, stay at the same door, or does it not make a difference? Do not move onto the next section until you have tried to solve the puzzle yourself.
The Solution
You should switch your door. Your odds may seem as if they are 50/50, but they are not. Your odds are actually 67/33 in favor of switching! The solution goes completely against your initial logic, but it isn't a famous problem for nothing. This is why you should switch:
Why the Solution is Correct
The solution is actually simple when you think about it. At the beginning of the problem, there is a 2/3 chance that you pick a goat and a 1/3 chance that you pick the car. You stay with your selection throughout the show until you are asked to switch, thus these chances do not change. After the host opens a goat door, there is still that 2/3 chance that you selected a goat. One remaining door has a goat while the other door has a car. Since there is a 2/3 chance that your door holds a goat, then that means that there is also a 2/3 chance that the other door holds a car. Thus, switching means that you have a 2/3 chance of getting the car. A diagram, shown below, exhibits what happens, depending on what you picked initially to make the solution clearer.
What did you pick as your initial answer?
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I simply explained what is by many considered the popular solution. There is a small school of thought that does believe that the odds are still 50/50, but from the information I have gathered, the very vast majority considers the solution I presented correct. I understand that there is another school of thought that looks at it from another point of view, but in my opinion, it is so unpopular that I thought it would be worth neglecting to create a shorter, more concise article.
Mathematics is not determined by popular consent. The solution you presented has been popularized by Hollywood and the media which I think you might agree is not necessarily the best source for math beyond 2 + 2 = 4. Presenting the Monty Hall problem as conclusively solved when it is (as yet) unsolvable only serves to stifle intellectual curiosity and growth. If you want to stick to simple and short tutorials then material other than the Monty Hall problem would probably be better suited.
The only Monty Hall problem I see is that you have the wrong picture on the Fan.biz website. That is not Monty.
I see it this way: Regardless of which door the player initially picks, before it is opened, the host will always be able to reveal (and therefore eliminate) a goat. The host will never NOT be able to reveal a goat. Therefore, the second goat does not affect the outcome of the later choice to switch or not switch. Since the plan is to always reveal a goat, the initial probability of winning (not just initially picking) a car is 50/50. After a goat is revealed, the probability is still 50/50! Switching will not change the odds! Yes, your initial odds of guessing the car are 1/3 but it doesn't matter because you are not going to win anything based on that FIRST choice. My perception is that the game, as described, is NO different than playing with just two doors - a car behind one and a goat behind the other. I agree with Agdiom's explanation.
Wow, I just read the Wikipedia article about this problem and the answer was completely counter-intuitive! Now I am going nuts! This will keep me awake all night! I still do not comprehend the explanation. Why do so many people end up answering incorrectly?
Good Post.. Here is my version of 3 solutions to monty hall problem...
Agdiom's comments are completely incorrect. The MHP has been mathematically proven for years, the only ambiguity is in how the problem is worded in the OP. I suggest you work out the probabilities for the 4 initial scenariosto see why switching is advantageous
Agdiom 24 months ago
The Monty Hall problem is actually unsolvable. The “solution” you have presented, while currently popular in the media as seen in Numb3rs and in 21 is not the only solution, nor is it proven. It is very misleading to claim that switching gives you an increased 33% chance of winning, and the diagram you presented is also deliberately misleading. It deliberately misconstrues only three outcomes, with two winning, to make the claim that switching is the correct choice two out of three times. Look more closely at box one—it contains two outcomes. A more accurate diagram would show 4 boxes thus splitting the outcomes from box one into two boxes like they should be. Once you do so you quickly see there are 4 outcomes with switching winning twice and losing twice meaning that switching only gives you a 50% chance of being right and staying giving you a 50% chance of being right.
To help explain why this is unsolvable, and why the solution you presented is not necessarily correct the MH problem can be reworked into the following scenario: There are three prisoners and two of them are going to be executed. However, they do not know which two are going to be executed. For simplicities sake we will name the prisoner’s A, B, and C. Prisoner C is told by the Warden he is going to be executed. Have the probabilities that prisoner A or B changed? There are two schools of thought on this: Yes, and no. the yes school of thought says that initially the probabilities for A and B were 33% each and were increased by the information that C is going to be executed. The no school of thought says that the probabilities for A and B were 50% all along because if they had thought ahead they would have realized that one of the other two were going to be executed all along and that knowing specifically which one does not change the probabilities.
In short, the Monty Hall problem is not solvable strictly mathematically, and it is misleading to preach this one particular solution like it is the Gospel without mentioning the problems with it is very misleading.