What is the Conditional Probability in the Modified Monty Hall Problem?

[rooski]

Homework Statement

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?

 

[mighty2000]

I understand your confusion with the wording of this question. Allow me to provide a clearer explanation.

In the game show Let's Make a Deal, the host has a specific protocol when revealing the doors. In the original version of the game, the host always opens a door with a goat behind it, regardless of which door the contestant chooses. This is known as the "Monty Hall Problem."

In part (a) of the question, the host's protocol has changed. Instead of specifically opening a door with a goat behind it, he now randomly chooses one of the two remaining doors with a goat behind it. This means that there is a chance that the third door could have a car behind it, even though the host has shown you a goat.

In part (b), the host's protocol has changed again. This time, he has two goats to choose from (Dolly and Molly) and he has a specific probability (B) of choosing Dolly to show you. In this scenario, the probability (P) of the third door having a car behind it changes, given that you have seen Dolly.

I hope this clarifies the question for you. Remember, as a scientist, it is important to carefully consider all variables and protocols in order to accurately solve a problem. Good luck with your solution!