The Monty Hall problem

[Angrboda]

I found the following to be an interesting tidbit. Indeed, the entire Wikipedia article is worth reading.

Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door (Adams 1990; Devlin 2003; Williams 2004; Stibel et al., 2008).

As Cecil Adams puts it (Adams 1990), "Monty is saying in effect: you can keep your one door or you can have the other two doors." The player therefore has the choice of either sticking with the original choice of door, or choosing the sum of the contents of the two other doors. The 2/3 chance of hiding the car has not been changed by the opening of one of these doors because Monty, knowing the location of the car, is equally likely to reveal a goat whether the player originally picked the car or the goat (in fact: in both cases, he is certain to do so). If the host did not know where the car is hidden, but opened a door revealing a goat by chance, the revelation would indeed change the probability of the player's original choice being correct to 1/2. The critical difference is that an ignorant host is more likely to successfully reveal a goat when the player's original choice is the car than he is when the original choice is a goat, so the revealing of a goat by chance does provide evidence in favor of the player's original choice winning the car.

In other words, offering the player to switch doors after revealing a goat is no different than if the host offered the player to switch from their original chosen door to both remaining doors, and only after the player chooses, the host shows that one of the two doors contains a goat (which at least one always will). The switch in this case clearly gives the player a 2/3 probability of chosing the car, and the revelation that a goat is behind one of the player's two doors, which is always the case, does not change this probability.

[Coffee Jesus]

I constructed a modified tree diagram.



Each square represents a 1/6 unit of probability.
The situation where I pick the right door is the only situation in which the host has a choice as to what door to open. Whichever he opens, he was more likely to open it if I picked wrong rather than right, which decreases the likelihood that I picked right. This decrease in probability cancels out the effect of the host eliminating a wrong door.