You pick a door (e.g. #1), and the host (who knows what's behind each door) opens another (e.g. #3) which has a goat. This means there are two doors left, one with a car and one with a goat. You are now given the opportunity to switch your choice (from #1 to #2). Is it to your advantage to do so?

Since you can't know which of the two remaining doors has the car, and since your initial pick had a chance of one-third, you might think that it does not matter. The chance is still

^{1}/

_{3}and you might as well stay with your original choice, right? Wrong! Switching actually

**doubles**your chances to

^{2}/

_{3}.

The Monty Hall problem is loosely based on the American television game show ‘Let's Make a Deal’. Monty Hall was the show's original host. So what's the deal here?

Let's make a list of the possible scenarios. There are three possible arrangements of the car and goats. The table below shows what happens when door 1 is initially picked — note that the result is the same when door 2 or 3 is the initial pick.

Door 1 | Door 2 | Door 3 | On switch | On stay |
---|---|---|---|---|

Car | Goat | Goat | Goat | Car |

Goat | Car | Goat | Car | Goat |

Goat | Goat | Car | Car | Goat |

The table doesn't lie: when staying with your initial choice you win only one out of three times, while switching will result in a win two times out of three. Remember that Monty will never open the door with the car behind it. Him opening a door is thus an additional piece of information that skews the odds in favor of switching.

The easiest way of looking at it is that switching is only a

*disadvantage*when you initially pick the car.

**That**chance is

^{1}/

_{3}, so that logically means that switching is an

*advantage*in the remaining

^{2}/

_{3}rd of the cases.

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