# Problems with Monty Hall Problem Variation

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.

1. 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?

• You may have some missing brackets in your expression. But the real issue is whether you are trying to work out the probability of winning at the start of the game or the probability of winning given that one of the other doors has already been opened (i.e. not situation 1a) Jul 27, 2021 at 8:48
• Hi there! I just changed the expression I I think the question is asking for the probability of winning given the other door has already been opened. Thank you for your time! Jul 27, 2021 at 8:54
• Perhaps you should try the other interpretation? Jul 27, 2021 at 9:04

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