Hi - Your edits to Monty Hall problem were reverted (your analysis is actually wrong - there are certainly 4 cases, but they aren't equally probable). I'm curious how much of the article you read before you made these edits. All of it? Or perhaps only the part including "Simple solutions"? The reason I'm asking is because we get very few fresh eyes willing to comment on the article. I've been arguing for quite some time that the existing text is not very convincing. If you're willing could you read the following and let me know whether you find it more or less convincing than however much of the article you already read? Thanks. -- Rick Block (talk) 00:22, 5 June 2011 (UTC)[reply]
Proposed text
Solution
Different sources present solutions to the problem using a variety of approaches.
Simplest approach
The player initially has a 1/3 chance of picking the car. The host always opens a door revealing a goat, so if the player doesn't switch the player has a 1/3 chance of winning the car. Similarly, the player has a 2/3 chance of initially picking a goat and if the player switches after the host has revealed the other goat the player has a 2/3 chance of winning the car. (some appropriate reference, perhaps Grinstead and Snell)
What this solution is saying is that if 900 contestants all switch, regardless of which door they initially pick and which door the host opens about 600 would win the car. Assuming each specific case is like any other, this means a player who initially picks Door 1 and sees the host open Door 3 wins the car with a 1/3 chance by not switching and with a 2/3 chance by switching.
Enumeration of all cases where the player picks Door 1
If the player has picked, say, Door 1, there are three equally likely cases.
Door 1
Door 2
Door 3
result if switching
Car
Goat
Goat
Goat
Goat
Car
Goat
Car
Goat
Goat
Car
Car
A player who switches ends up with a goat in only one of these cases but ends up with the car in two, so the probability of winning the car by switching is 2/3. (some appropriate reference, perhaps vos Savant)
What this solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1 and they all switch, of these 300 players about 200 would win the car. Assuming the cases where the host opens Door 2 or Door 3 when the player picks Door 1 are the same, this means a player who initially picks Door 1 and sees the host open Door 3 wins the car with a 1/3 chance by not switching and with a 2/3 chance by switching.
The probability of winning by switching given the player picks Door 1 and the host opens Door 3
Tree showing the probability of every possible outcome if the player initially picks Door 1
This is a more complicated type of solution involving conditional probability. The difference between this approach and the previous one can be expressed as whether the player must decide to switch before the host opens a door or is allowed to decide after seeing which door the host opens (Gillman 1992).
The probabilities in all cases where the player has initially picked Door 1 can be determined by referring to the figure below or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138 presents an expanded tree showing all initial player picks). Given the player has picked Door 1 but before the host opens a door, the player has a 1/3 chance of having selected the car. Referring to either the figure or the tree, in the cases the host then opens Door 3, switching wins with probability 1/3 if the car is behind Door 2 but loses only with probability 1/6 if the car is behind Door 1. The sum of these probabilities is 1/2, meaning the host opens Door 3 only 1/2 of the time. The conditional probability of winning by switching for players who pick Door 1 and see the host open Door 3 is computed by dividing the total probability of winning in the case the host opens Door 3 (1/3) by the probability of all cases where the host opens Door 3 (1/2), therefore this probability is (1/3)/(1/2)=2/3.
Although this is the same answer as the simpler solutions for the unambiguous problem statement as presented above, in some variations of the problem the conditional probability may differ from the average probability and the probability given only that the player initially picks Door 1, see Variants below. Some proponents of solutions using conditional probability consider the simpler solutions to be incomplete, since the simpler solutions do not explicitly use the constraint in the problem statement that the host must choose which door to open randomly if both hide goats (multiple references, e.g. Morgan et al., Gillman, ...).
What this type of solution is saying is that if 900 contestants are on the show and roughly 1/3 pick Door 1, of these 300 players about 150 will see the host open Door 3. If they all switch, about 100 would win the car.
A formal proof that the conditional probability of winning by switching is 2/3 is presented below, see Bayesian analysis.
Car hidden behind Door 3
Car hidden behind Door 1
Car hidden behind Door 2
Player initially picks Door 1
Host must open Door 2
Host randomly opens either goat door
Host must open Door 3
Probability 1/3
Probability 1/6
Probability 1/6
Probability 1/3
Switching wins
Switching loses
Switching loses
Switching wins
If the host has opened Door 3, these cases have not happened
If the host has opened Door 3, switching wins twice as often as staying
