Problems with Monty Hall Problem Variation

Question

Here is a variation of the Monty Hall problem on a class I am taking on Coursera:

Imagine that now host have the following instructions. Put a prize behind a random door. Let the guest guess a door.

1. If the guest chooses an incorrect door (with no prize), roll a dice (in such a way that the guest does not see this and does not know whether this happened);

a) with probability 1/3 (outcomes 1 and 2) open the door (that has no prize behind); the game ends;

b) with probability 2/3 (outcomes 3,4,5,6) open the other door with no prize and ask the guest whether she wants to change the guess.

2. if the guest chooses a correct door (with a prize), open one of the two other doors (making a random choice) and ask the guest whether she wants to change the guess.

What is the probability for the guest to get a prize if she uses "change" strategy (i.e., changes the guess)? We consider the fraction of winning days among all days (when she was given a chance to change or when she was not).

My approach is to get the probability of winning and changing / probability of changing:

(2/3 * 2/3 * 1/2) / [(1/3 * 1/2)+(2/3 * 2/3 * 1/2)] = 4/7

But the answer is incorrect... Can anyone help me with the problem?

Answer 1

My approach would be the following:

Given that we always use the change strategy, in 1/3 of the days, we fail because we chose the correct door and we always change our choice.

Then, we have the remaining 2/3 of the days. For us to be able to win, we have to be given the opportunity to change the door. This happens in 2/3 of the cases.

So $P(W) = \frac{2}{3} \frac{2}{3} = \frac{4}{9} $