An analysis of the problem using the formalism of Bayesian probability theory (Gill 2002) makes explicit the role of the assumptions underlying the problem.
In Bayesian terms, a probability P(A | I)\, is a number in [0, 1]\, associated to a proposition A\,. The number expresses a degree of belief in the truth of A\,, subject to whatever background information I\, happens to be known.
For this problem the background is provided by the rules of the game, and the propositions of interest are:
C_i\, : The car is behind Door i, for i equal to 1, 2 or 3.
H_{ij}\, : The host opens Door j after the player has picked Door i, for i and j equal to 1, 2 or 3.
For example, C_1\, denotes the proposition the car is behind Door 1, and H_{12}\, denotes the proposition the host opens Door 2 after the player has picked Door 1.
The assumptions underlying the common interpretation of the Monty Hall puzzle are then formally stated as follows.
First, the car can be behind any door, and all doors are a priori equally likely to hide the car. In this context a priori means before the game is played, or before seeing the goat. Hence, the prior probability of a proposition C_i\, is:
P(C_i | I)\,= \frac{1}{3}.
Second, the host will always open a door that has no car behind it, chosen from among the two not picked by the player. If two such doors are available, each one is equally likely to be opened. This rule determines the conditional probability of a proposition H_{ij}\, subject to where the car is — i.e., conditioned on a proposition C_k\,. Specifically, it is:
P(H_{ij} | C_k,\, I)\,\, =\,\begin{cases} \, \\ \, \\ \, \\ \, \end{cases} \,0\, if i = j, (the host cannot open the door picked by the player)
\,0\, if j = k, (the host cannot open a door with a car behind it)
\,1/2\, if i = k, (the two doors with no car are equally likely to be opened)
\,1\, if i \nek and j \ne k, (there is only one door available to open)
The problem can now be solved by scoring each strategy with its associated posterior probability of winning, that is with its probability subject to the host's opening of one of the doors. Without loss of generality, assume, by re-numbering the doors if necessary, that the player picks Door 1, and that the host then opens Door 3, revealing a goat. In other words, the host makes proposition H_{13}\, true.
The posterior probability of winning by not switching doors, subject to the game rules and H_{13}\,, is then P(C_1 | H_{13},\,I). Using Bayes' theorem this is expressed as:
P(C_1|H_{13},\,I) = \frac{P(H_{13}| C_1,\,I) \, P(C_1 | I)}{P(H_{13} | I)}.
By the assumptions stated above, the numerator of the right-hand side is:
P(H_{13}| C_1,\,I) \, P(C_1 | I) = \frac12 \times \frac13 = \frac16.
The normalizing constant at the denominator can be evaluated by expanding it using the definitions of marginal probability and conditional probability:
\begin{array}{lcl} P(H_{13}|I) &{}= &P(H_{13},\,C_1 | I) + P(H_{13},\,C_2|I) + P(H_{13},\,C_3|I) \\ &{}= &P(H_{13}|C_1,\,I) \, P(C_1|I)\, + \\ &&P(H_{13}|C_2,\,I) \, P(C_2|I)\, + \\ &&P(H_{13}|C_3,\,I) \, P(C_3|I) \\ &{}= &{\displaystyle \frac12 \times \frac13 + 1 \times \frac13 + 0 \times \frac13 }\ = \ {\displaystyle\frac12\ .} \end{array}
Dividing the numerator by the normalizing constant yields:
P(C_1|H_{13},\,I) = \frac16\,/\,\frac12 = \frac13.
Note that this is equal to the prior probability of the car's being behind the initially chosen door, meaning that the host's action has not contributed any novel information with regard to this eventuality. In fact, the following argument shows that the effect of the host's action consists entirely of redistributing the probabilities for the car's being behind either of the other two doors.
The probability of winning by switching the selection to Door 2, P(C_2 | H_{13},\,I), can be evaluated by requiring that the posterior probabilities of all the C_i\, propositions add to 1. That is:
1 = P(C_1|H_{13},\,I) + P(C_2 | H_{13},\,I) + P(C_3|H_{13},\,I).
There is no car behind Door 3, since the host opened it, so the last term must be zero. This can be proven using Bayes' theorem and the previous results:
\begin{align} P(C_3|H_{13},\,I) &= \frac{P(H_{13}|C_3,\,I)\,P(C_3|I)}{P(H_{13}|I)} \\ &= \left(0\times\frac13\right) /\, \frac12 = 0\ . \end{align}
Hence:
P(C_2 | H_{13},\,I) = 1 - \frac13 - 0 = \frac23.
This shows that the winning strategy is to switch the selection to Door 2. It also makes clear that the host's showing of the goat behind Door 3 has the effect of transferring the 1/3 of winning probability a-priori associated with that door to the remaining unselected and unopened one, thus making it the most likely winning choice.
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