Two people Monty Hall paradox

Question

We're on a game show, and we have to select between three doors, one of which has a Lamborghini behind it while the others have goats. After we've decided, the host opens one of the other two doors, which contains a goat. Should we move to the other unchosen door or stick with our original choice?

This is the famous Monty Hall paradox problem, where we should move to the other unchosen door. The logic is straightforward: the likelihood that we made the incorrect decision at first is $\frac{2}{3}$, which is also the probability that the unchosen door contains the automobile, thus we should make the swap.

Take a look at a different scenario. Assume you and a buddy are at the show together. You two choose the door independently, unaware of the other's decision. After making your selection, you see that your companion chose a different door from yours. The third door, the unchosen one, is now open, and under it is a goat. Should you make the change in this situation?

If we follow the same reasoning as previously, you should make the switch since the likelihood of making the erroneous option initially is $\frac{2}{3}$. However, by symmery, your companion should likewise make the same changeover this time. Then you and your buddy effectively switched the positions, which is paradoxical. What's the problem?

My Attempt: I was thinking about this issue. And this is my explanation, and I wonder if this is correct. The primary difference between this example and the first is that the additional knowledge of opening the unchosen door impacts the original probability space in which I made the decision. In the original setup, no matter which door I selected, there was always an unchosen door under which the goat might be found. So the new knowledge has no effect on the likelihood that I made the erroneous decision in the first place. In the new example of two persons, however, the additional knowledge had an effect on the probability. If you and your friend have both picked the door with goats, it is impossible to open the third unchosen door with goats. As a result, the additional knowledge actually restricted the probability space to a smaller subspace, where the likelihood of choosing the erroneous decision becomes $\frac{1}{2}$. As a result, it makes no difference whether we make the transition in this scenario. Is this reasoning correct?

Answer 1

Your reasoning is correct. We can be more rigorous if we want to use Bayes' theorem. Let $D_1$ be the event that the car is behind door 1 (the door you chose). Let $G$ be the event that Monty is able to reveal a goat. Then \begin{align*} P(D_1|G) &= \frac{P(G|D_1)P(D_1)}{P(G|D_1)P(D_1)+P(G|D_1^c)P(D_1^c)}\\ &= \frac{(1)(1/3)}{(1)(1/3)+(1/2)(2/3)}\\ &= \frac{1}{2}. \end{align*} Here we have $P(G|D_1^c) = 1/2$ because there is a chance that Monty cannot reveal the other goat if it is behind your friend's door. (In the regular Monty Hall problem, this probability is $1$ since Monty can always reveal a goat.)

Answer 2

Interesting theory.. However, Monty Hall has its own pre-requisite before you zeroin on the winning or losing probability of the player(s). In other words, the winning or losing probability of the player(s) depends on which door Monty opens i.e., Monty always has a choice of picking the door with the goat. Considering your scenario where both players pick the 2 doors with the goats, its impossible for Monty to choose a door with goat. The winning odds for the player(s) will not come into picture unless Monty chooses a door. In this case, since Monty is unable to choose a door its not just restricting the probability of winning to a smaller subspace, instead its entirely eliminating it. Unless Monty gets to pick the door with the goat at all possible scenarios, its not Monty Hall. Practically speaking, if Monty is not able to pick a door with goat, its obvious that both players are gonna know that the unopened door has the car, and hence both players will switch.

A more pragmatic approach would be you and your friend play together and choose the same door everytime. Once Monty opens the door with the goat, one of you switches doors and the other doesn't; which will give you the expected results. The player who switches his/her choice will have a better chance of winning.

I spent few hours thinking about this and came up with a more hands-on way to approach this. Try the excel below to play around. The code is open, feel free to change.

PS: You would need Microsoft excel to run this

Monty Hall Problem Solution