Solution to Monty Hall problem

The solution presented shows the three possible arrangements of one car and two goats behind three doors and the result of switching or staying after initially picking Door 1 in each case:

Door 1 Door 2 Door 3 result if switching result if staying
Car Goat Goat Goat Car
Goat Car Goat Car Goat
Goat Goat Car Car Goat

A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.

Increasing the number of doors

That switching has a probability of 2/3 runs counter to many people's intuition. If there are two doors left, then why isn't each door 1/2? The intuition may be aided by generalizing the problem to have a large number of doors so that the player's initial choice has a small chance of winning.

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three . In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. His initial probability of winning is 1 out of 1,000,000. The game host then opens 999,998 of the other doors revealing 999,998 goats. (Imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door.) The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. The chance that the player's door is correct hasn't changed. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.

To extend the above, it's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens the first 999,998 of them for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.

This example can also be used to illustrate the opposite situation in which the host does not know where the prize is and opens doors randomly. There is a 999,999/1,000,000 probability that the contestant selects wrong initially, and the prize is behind one of the other doors. If the host goes about randomly opening doors not knowing where the prize is, the probability is likely that the host will reveal the prize before two doors are left (the contestant's choice and one other) to switch between. If the host happens to not reveal the car, then both of the remaining doors have an equal probability of containing a car. This is analogous to the game play on another game show, Deal or No Deal; in that game, the contestant chooses a numbered briefcase and then randomly opens the other cases one at a time.