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Comment author: jlborges 07 March 2016 11:02:08AM *  0 points [-]

This problem is not so difficult to solve if we use a binomial tree to try to tackle it. Not only we will come to the same (correct) mathematical answer (which is brilliantly exposed in the first post in this thread) but logically is more palatable.

I will exposed the logically, semantically derived answer straight away and then I will jump in to the binomial tree for the proof-out.

The probability of the situation exposed here, which for the sake of being brief I’m going to put it as “contestant choose a door, a goat is revealed in a different door, then contestant switches from the original choice to win the car”, is the same probability as being wrong in an scenario where the contestant only needs to choose one door and that’s it. This is 66%. The probability of the path the contestant needs to go through to win a car IF he/she always switches is exactly the same probability as being wrong in the first place.

Why?

There are two world-states right at beginning of the situation. This is being right, which means having chosen the door with the car behind with a probability of 33% and being wrong, which means choosing a door with a goat behind with a probability of 66%. Being wrong is the only world-state that will put the contestant in the right path IF and only IF he/she then switches the door. Doing so guarantees (at a 100%) the contestant to choose the door with the car behind. That’s why is so counter-intuitive because being wrong in the first place will increase (double!) the probability of winning the car given the path that the host is offering. Therefore, if the path the contestant needs to take (being wrong) for then to switch (switching to only one door, the one left, there is no probability here as there is only one choice left) has a probability of 66% the answer to this problem is 66%!

The binomial tree will illustrate this better:

World-state #1 will never, ever, give the contestant a car if he/she switches his/her first choice. The fundamental thing here and where it’s so easy to get confused is that the problem is essentially defining paths, not states.

Binomial trees are excellent tools for real-life options such as this example. However they are mostly used to price options and other financial instruments with some optionality embedded in them. The forking paths can get really complex. In this case it worked very well because there are only two cycles and the second cycle of the second world-state has only one option for winning the car. Many cycles and probabilities within the cycles and world-states will compound the complexity of these structures. Regardless, I find them very powerful tools to visually represent these kind of problems .The key here is that this is a branching-type probabilistic problem and the world-states are mutually exclusive with their own probability distributions. Something that the classic probabilistic analysis fails to represent and make the analyst being aware to, as it’s so very easy to see the problem as one world, one situation. They are not.