The old TV game Let’s Make a Deal hosted by Monty Hall could be summarized as
follows. Suppose you are on a game show, and you are given the choice of three doors.
Behind one door is a car, behind the others, goats. You pick a door, say number 1,
and the host, who knows what is behind the doors, opens another door, say number
3, which has a goat.
(a) Assume that the host’s protocol is instead: he is determined to show you a goat and with a choice of two, he picks one at random. Let P be the conditional probability
that the third door conceals the car. Compute P.
(b) Assume that the host’s protocol is: he is determined to show you a goat and with
a choice of two goats (Dolly and Molly) he shows you Dolly with probability B. Compute P given you see Dolly.
The Attempt at a Solution
This question is badly worded, in my opinion. For Part A, it can be interpreted as not being conditional probability at all. I firstly pick a door at random. Then the host picks a door at random. So the chances of the last door having a car behind it are still 1 in 3. Unless the question means to infer that the host picks a random door that has a goat behind it and cannot pick a door with a car behind it...?
For part B i am also confused. Does the host randomly pick a goat? Does he automatically pick Dolly? What relevance does that have pertaining to whether the third door has a car behind it?