Monty Hall Problem is named after the host of an american game show with the name "Let's make a deal". Problem has many variants. One of them is:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?Should the player switch? That's the problem.
I tried this. There are 3 doors. 1 of them contains cars and hence probability of choosing a door with car is 1/3 and that with goat is 2/3. When one of the door with goat is opened then the question reduces to two doors. So probability should be 1/2 now and it should not matter whether you switch or not. But it does and that's why it has been most discussed paradox problem. From Wikipedia entry -
Because there is no way for the player to know which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.Why the probability is not 1/2?
This difference can be demonstrated by contrasting the original problem with a variation that appeared in vos Savant's column in November 2006. In this version, Monty Hall forgets which door hides the car. He opens one of the doors at random and is relieved when a goat is revealed. Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006).Increasing the number of doors
It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. 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.from coding horror blog -
When told of this (that switching was more lucrative choice), Paul Erdos, one of the leading mathematicians of the 20th century, said, "That's impossible." Then, when presented with a formal mathematical proof of the correct answer, he still didn't believe it and grew angry. Only after a colleague arranged for a computer simulation in which Erdos watched hundreds of trials that came out 2-to-1 in favor of switching did Erdos concede that he was wrong.Sunday evening went well with this problem!
The problem is so unintuitive one of the most notable mathematicians of the last century couldn't wrap his head around it. That's ... well, that's amazing.


