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} $

Answer 2

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.