Monty Hall Analogue

Attempt: The first problem I am getting the same answer as the original Monty Hall problem (2/3). however this doesn't make any sense, since the option to switch is what increases my odds .

It should make sense to get the same result, as it is the same scenario.   Two doors being selected for the host also selects one for the contestant, and then a goat is revealed by the host using knowledge of what's behind all the doors to select one from his two.

Thus there is a 1/3 probability that the prise is behind the contestant's door, and so a 2/3 probability for the prise being behind the host's unrevealed door.

As for the second problem, I do not have the slightest clue how to proceed since I do not know why my odds should decrease?

Since the goat was selected without knowledge (and thus randomly), the prise has an equal chance of being behind each door that is not that of the revealed goat.

How do you show this using conditional probability? i.e. how does one formulate precisely "with knowledge" vs "without knowledge" when writing out the conditionals for both probabilities? note that in each case you have P(car | goat shown)

We can treat whether the host is allowed to gain knowledge as a condition ($K$).   Now, independently of this, the host either has a car and one goat (event $C$) or two goats (event $C^\complement$).   If he is allowed to gain knowledge the event of revealing a goat ($G$) will be a certainty (and independent of how many goats), otherwise the probability of doing so depends on how many goats.

$$\begin{align}\mathsf P(C\mid G, K)&=\dfrac{\mathsf P(G\mid K, C)~\mathsf P(C)}{\mathsf P(G\mid K)}\\&=\dfrac{1\cdot\tfrac 23}{1}\\&=\dfrac 23\\[2ex]\mathsf P(C\mid G, K^\complement)&=\dfrac{\mathsf P(G\mid K^\complement, C)~\mathsf P(C)}{\mathsf P(G\mid K^\complement, C)~\mathsf P(C)+\mathsf P(G\mid K^\complement, C^\complement)~\mathsf P(C^\complement)}\\&=\dfrac{\tfrac 12\cdot\tfrac 23}{\tfrac 12\cdot\tfrac 23+1\cdot\tfrac 13}\\&=\dfrac 12\end{align}$$

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Combinatorics

Problem Solving

Monty Hall

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