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) Commented 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! Commented Jul 27, 2021 at 8:54
• Perhaps you should try the other interpretation? Commented Jul 27, 2021 at 9:04

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

Write it out completely

• 1st guess is correct door (1/3)

• Monthy opens door (1/1)
• Monthy doesn't open door (0/1)
• 1st guess is wrong door (2/3)

• Monthy opens door (2/3)
• Monthy doesn't open door (1/3)

In $$2/3 \times 1/3 = 2/9$$ fraction of cases Monthy doesn't open an alternative door, doesn't give you an option to switch and you loose directly.

In $$2/3 \times 2/3 = 4/9$$ fraction of cases Monthly opens a door without prize and offers a switch while you have the wrong door.

In $$1/3 \times 1/1 = 3/9$$ fraction of cases Monthly opens a door without prize and offers a switch while you have the correct door.

Given that a second door opens, the odds of currently having the correct door are 3:4, so you better make the switch and that strategy will make you win in 4/9 cases.

Your 4/7 probability is the probability of winning, conditional on the option of a change. But in 2/9 cases there is no change possible and you loose directly. So the probability of winning is $$4/7 \times (1-2/9) = 4/9$$.

I am not sure what the 1/2 terms are doing in your calculation of 4/7.