Problem — Monty Fall variant

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Monty Fall variant: the host trips and opens a door at random (he does not know where the prize is). Show that, conditioning on «he has opened a door with a goat», going back to $50$–$50$ is in fact the correct answer. Hint: recompute $P(B\mid A_i)$ as uniform over $\{2,3\}$.

Solution

Set-up. The key difference from the classic Monty Hall is that here the host opens a door at random without avoiding the prize. Let $A_i$ be the event «the prize is behind door $i$» and $B$ the event «the host opens a door with a goat». In the classic Monty Hall the host deliberately chooses a goat, so $P(B\mid A_i)$ is not uniform and the $1/3$–$2/3$ asymmetry arises. In Monty Fall, opening at random, $P(B\mid A_i)$ is the same for all compatible configurations: applying Bayes, $P(A_i\mid B) = \frac{P(B\mid A_i)P(A_i)}{P(B)},$ the two still-closed doors turn out to be equiprobable. Conclude that, conditioning on having seen a goat, the probability is $1/2$–$1/2$: switching is no longer worthwhile. Carry out the computations comparing them with the classic case.

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Topics: Probabilita
Concepts: Paradosso di monty hall · Probabilita condizionata
Methods: Bayes · Prob condizionata
Skills: Calcolo probabilita · Ragionare per casi
Exercise type: Problema probabilita