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. 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.
Combining doors
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, and will not, choose the opened door (Adams 1990; Devlin 2003; Williams 2004; Stibel et al., 2008). The player therefore has the choice of either sticking with the original choice of door with a 1/3 chance of winning the car, or choosing the sum of the contents of the two other doors with a 2/3 chance as shown.
The game assumptions play a role here—switching is equivalent to taking the combined contents if and only if the game host knows what is behind the doors, must open a door with a goat, and chooses between two losing doors randomly with equal probabilities.
The only difference between trading for both doors and the trade that is actually offered is whether the host opens one of the two doors. Opening one shows which of these doors the car must be behind if it is behind either. At least one of the two unpicked doors contains a goat, and the host is equally likely to open either of these doors so opening one gives the player no additional information; opening one does not change the 2/3 probability that the car is behind one of them (Devlin 2003).
And of course the most Important:
Why the probability is not 1/2
The most commonly voiced objection to the solution is that the past can be ignored when assessing the probability—that it is irrelevant which doors the player initially picks and the host opens. However, in the problem as originally presented, the player's initial choice does influence the host's available choices subsequently.
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).
In this version of the puzzle, the player has an equal chance of winning whether switching or not. Assuming the player picks Door 1 there are six possible outcomes that can occur, each with probability 1/6:
| Player picks Door 1 | ||||||
|---|---|---|---|---|---|---|
| Car behind Door 1 | Car behind Door 2 | Car behind Door 3 | ||||
| Host opens: | Door 2 | Door 3 | Door 2 | Door 3 | Door 2 | Door 3 |
| Host reveals: | Goat | Goat | Car | Goat | Goat | Car |
| Switching: | loses | loses | ? | wins | wins | ? |
In two cases above, the host reveals the car. What might happen in these cases is unknown—perhaps the contestant immediately wins or immediately loses. However, in the problem as stated, the host has revealed a goat, so only four of the six cases remain possible, and they are equally likely. In two of these four cases, switching results in a win, and in the other two, switching results in a loss. Staying with the original pick gives the same odds: a loss in two cases and a win in two others.
The player's probability of winning by switching increases to 2/3 in the original problem because in the two cases above where the host would reveal the car, he is forced to reveal the remaining goat instead. In the table below, these two cases are highlighted:
| Player picks Door 1 | ||||||
|---|---|---|---|---|---|---|
| Car behind Door 1 | Car behind Door 2 | Car behind Door 3 | ||||
| Host opens: | Door 2 | Door 3 | Door 3 | Door 3 | Door 2 | Door 2 |
| Host reveals: | Goat | Goat | Goat | Goat | Goat | Goat |
| Switching: | loses | loses | wins | wins | wins | wins |
This change in the host's behavior causes the car to be twice as likely to be behind the "third door", and is what makes switching twice as likely to win in the "host knows" variation of the problem.
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Understand? I think "Why the probability is not 1/2", is the biggest explaination there is, it's quite clear up there, if you bother to read everything:P
Oh wells, good luck in future lectures:D:D:D
Disclaimer: I copied and pasted from Wiki
