Monty Hall
Monty Hall
[A figure - Monty Hall - stands on stage, holding a microphone. There are three doors; two labelled "A" and "C", which are closed, and one that is being held open by Monty. There's a ramp to the right, down which a goat is being led by Beret Guy.] Beret Guy: ...and my yard has so much grass, and I'll teach you tricks, and... Goat: ♥

This comic is a reference to the US game show Let's Make a Deal , and more specifically the Monty Hall problem , a probability puzzle based on the show and named after its original host, Monty Hall . The premise of the show was that Hall would offer "deals" to contestants pulled from the audience in which they could win cash and prizes. Some deals involved games/tasks the contestant had to perform, while others simply involved the contestant making choices between a series of doors or boxes. In such games of choice, there were often several prizes and typically at least one "zonk", the show's name for an undesirable "gag" prize, which on the original Monty Hall version of the show were frequently animals such as goats.
In the classic version of the Monty Hall Problem , a contestant is offered a choice of three doors. Behind two of the doors are goats, and behind one of them is a car. First, the contestant chooses a door, which remains closed. The host then opens one of the two remaining doors and reveals a goat. The contestant is then offered a final choice of whether to switch their choice to the remaining closed door, or keep the door they originally chose. The problem involves an analysis of the probability of the contestant choosing the car given certain circumstances.
The problem assumes that a contestant would want to win a car, and would be disappointed to win a goat, which most contestants would have no ability to house, and no use for. The comic shows that Beret Guy , upon the host revealing that door B has a goat behind it, chooses to take the goat to keep as a pet, which makes them both very happy. This is much like, and may be an allusion to, the Simpsons episode Bart Gets an Elephant , in which Bart opts for the gag prize of an African Elephant rather than the $10,000 award. According to an interview with Monty Hall, several contestants actually decided to keep the animals; although rare, it was allowed since the animals were offered as prizes (and they were a lot more expensive than the consolation cash prize).
The title text references the car and the other goat, untouched behind the remaining doors.
For an in-depth analysis of the Monty Hall Problem, see its article at Wikipedia
The apparent "paradox" of the Monty Hall Problem is that many people's initial reaction once the host opens a door to reveal a goat, is that there are two remaining doors, one with a car and one with a goat; and therefore there is an equal probability the car is behind each door. Many people therefore believe that switching makes no difference to the odds of winning a car.

However, assuming that the host has knowledge of which doors contain goats, and that their choice of which door to open is always an unchosen door containing a goat, it is actually twice as likely that the contestant will win the car if they switch than if they keep their original choice. This is because the contestant initially had a one-in-three chance of choosing the car and a two-in-three chance of choosing a goat. Switching always wins the car in those two-thirds of cases where the contestant initially chose a goat. The probability of winning by switching is therefore the same as the probability that the contestant initially chose a goat.
The switch essentially gives the contestant both remaining doors instead of just the one door originally chosen. Because the host always has at least one goat available, the fact that the host reveals a goat does not provide the contestant any new information about their initially chosen door. The initial door still has a two-in-three chance of being a goat, and switching still has a two-in-three chance of winning. Opening a goat-door simply shifts all of the probability of the remaining two doors being a car to the remaining unchosen door.
Simple explanation :
Imagine there are 100 doors instead of just 2, and after you pick a door, the host opens all but one, revealing all goats. Do you switch to the remaining door or keep your initial pick?  Just as there is a 2/3 chance of picking the car when switching in the 3-door scenario, there is now a 99/100 chance of picking the car when switching in the 100 door scenario.  In this scenario, it becomes obvious that it is not a 50/50 chance when two doors remain.
Important Side Note :
There has been great debate about the precise wording of the problem, and what assumptions or rules might apply. These variants can greatly change the probabilities.
One variant has the host open one of the two remaining doors at random, which could result in the car being revealed, and the game ending. In that scenario, if a goat has been revealed, the probability that the first pick is correct is now 1/2 and switching is not advantageous.
In 1/3 of all possible games, the first pick is correct. The host cannot pick the car. In 1/3 of all possible games, the first pick is wrong but the host does not reveal the car. In 1/3 of all possible games, the first pick is wrong and host will reveal the car. We now know those cases are impossible.
With only 2/3rds of all possible games remaining, the chance that switching will win the car is now (1/3)/(2/3) = 1/2. Likewise, not switching also has a 1/2 chance of winning. Note that this variant requires that the host picks a door at random.
Another variant has the host only offering to switch if the first choice is correct.  In this case, switching always loses.
[A figure - Monty Hall - stands on stage, holding a microphone. There are three doors; two labelled "A" and "C", which are closed, and one that is being held open by Monty. There's a ramp to the right, down which a goat is being led by Beret Guy.] Beret Guy: ...and my yard has so much grass, and I'll teach you tricks, and... Goat: ♥
