Monty Hall Change in Problem Suggestion

Question

So, Let's say we have three doors and three guests in our show. A car is behind one door. The other two doors have goats . Please note.. I am not asking about the original Monty Hall problem here. This is another problem (but eventually it might be equivalent to the original one).

Each guest picks a different door. and the host opens one losing door and says goodbye to one contestant.Then He asks each of the remaining guests if they would like to switch.

Q1 What is winning probability of each of the remaining guests if they decide to switch?

Q2 If both have to switch doesn't that mean that the computed probabilities are useless?

• NOTE I am aware that we have two events which aren't disjoint here

Event A : Car is behind the losing door or first remaining guest's choice

Event B : Car is behind the losing door or second remaining guest's choice

and each has 2/3 chance to win. So, the sum here isn't 1

So, eventually.. If the answer is to Switch or NOT , doesn't it mean that we have same probability for the remaining doors. Thanks.. please give time for discussion and don't down vote for the question if it's against your mathematical beliefs. I am suggesting a new problem here, so be open to other's ideas :)

Answer 1

By symmetry, both remaining doors have $\frac 12$, so there is no advantage to switching.

To compute the probabilities directly, consider the contestant who has chosen $\#1$. Initially, there were three states each with probability $\frac 13$, according to whichever door contained the prize. Let's at Monty opens $\#3$ and shows it is empty, removing one of the states.

Conditioned on the state in which the prize really is behind door $\#1$, the probability that Monty would have opened $\#3$ is $\frac 12$ (since nothing distinguishes doors $1,3$). It follows that the probability that the prize is being $\#1$ AND Monty opens $\#3$ is $\frac 13\times \frac 12=\frac 16$.Thus the conditional probability that the door is behind $\#1$, conditioned on Monty opening $\#3$ is $$\frac {1/6}{1/3}=\frac 12$$.

The same computation applies to the state in which the prize is behind door $\#2$ (just switch the rolls of $\#1$ and $\#2$ in the earlier computation).

Thus, as expected, there is no advantage gained from switching (though it doesn't hurt).

To stress: In this version of the game, nothing distinguishes the two unopened doors. That's what I meant in the, terse, initial statement that symmetry instantly implied the answer was $\frac 12$. In the standard version, the two unopened doors are very different, since Monty would never open the door the contestant had selected.

As is often the case in problems like these, it is very easy to simulate the situation. Worth doing as intuition often lets one down.

Answer 2

I do find your problem interesting! The main questions you ask seem to be:

1. Does it matter if they switch or not in your game?
2. How exactly is your game equivalent to or different from the original Monty Hall Problem?

So in effect, we have to strategies on the table to try out in either game:

• Switching
• Not switching

Breakdown of your game

We assume that placing the car and the host opening an (available) door is performed uniformly at random. Then we have:

Switching $$ P(win)=P(wrong\ door)\cdot P(host\ opens\ other)=\frac 23\cdot\frac 12=\frac 13\\ P(lose)=P(right\ door)\cdot P(host\ opens\ either)=\frac13\cdot \frac22=\frac13\\ P(thrown\ out)=P(wrong\ door)\cdot P(host\ opens\ yours)=\frac 23\cdot\frac 12=\frac 13 $$ Not switching $$ P(win)=P(right\ door)\cdot P(host\ opens\ either)=\frac 13\cdot\frac22=\frac13\\ P(lose)=P(wrong\ door)\cdot P(host\ opens\ other)=\frac 23\cdot\frac 12=\frac 13\\ P(thrown\ out)=P(wrong\ door)\cdot P(host\ opens\ yours)=\frac 23\cdot\frac 12=\frac 13 $$

Breakdown of Monty Hall

Switching $$ P(win)=P(wrong\ door)\cdot P(host\ opens\ other)=\frac 23\cdot1=\frac 23\\ P(lose)=P(right\ door)\cdot P(host\ opens\ either)=\frac13\cdot \frac22=\frac13 $$ Not switching $$ P(win)=P(right\ door)\cdot P(host\ opens\ either)=\frac13\cdot \frac22=\frac13\\ P(lose)=P(wrong\ door)\cdot P(host\ opens\ other)=\frac 23\cdot1=\frac 23 $$

So maybe you can see, that in your game two things can happen when you chose a wrong door to begin with:

• You stay in the game
• You are thrown out

The "You are thrown out" case in your version is converted to a "You are guaranteed to win if switching" in the original problem. Hence it shifts an extra $1/3$ onto the winning-by-switching case and also changes the denominator of cases where you stay in the game.

Answer 3

In that case it wouldn't be better to switch for any player, and if you think it should just because the answer in Monty Hall game, it's because you have not understood why the Monty Hall works, so let's start from it.

At first each option is equally likely to have the prize, as we don't have any information that can tell us that one is easier to have it than other. We can therefore represent the total probability $1$ as the total area of the rectangle below, which in turn is divided in three other sections of the same size each (three parts of $1/3$ probability), corresponding to the three possible locations of the car.

Now, according to the rules of the game the host knows the locations of the contents and must reveal a losing door from the two that the player did not pick. That means that when the player is incorrect, the host is forced to reveal specifically the only other incorrect one, while when the player is the winner the host can reveal any of the other two; we don't know which in advance, they are equally likely for us, because both would be losing ones in that case.

So, if for example the player starts selecting door 1, adding the corresponding revelations in the rectangle above we get:

As you see, we have got a disparity. The rectangle that represents when the host reveals door 2 and the car in in door 1 is only half as big as the rectangle in which the host reveals that same door 2 but the prize in in door 3. That means that winning by switching is twice as likely as winning by staying.

If you prefer, you can get the new probabilities applying rule of three. The scenarios when the host opens door 2 are the green rectangles. From them, the case when door 1 has the prize originally had $1/6$ chance and the other had $1/3$ chance. Since the total probabilities must sum $1$, with rule of three you get that which originally had $1/6$ has $1/3$ at this point, and which originally had $1/3$ has $2/3$ at this point. Similar reasoning is applied to when the host reveals door 3, and you can draw by yourself the rest of the diagrams representing when the contestant starts selecting each of the other doors.

In the long run the sample proportion tends to approach the actual probability, so if you repeated the game $900$ times, in about $300$ the car should tend to appear in each door. Now, if you always start choosing door 1, from the $300$ in which the correct is door 1 only in about $150$ the host would reveal door 2 and in the other $150$ he would reveal door 3. Instead, in all the $300$ games that the car is in door 2 the host would reveal door 3, and in all the $300$ that the car is in door 3 the host would reveal door 2.

That disparity does not occur in the game you proposed. No player was given the privilege of always advancing to the second part. With no further information, we can only assume that the host is equally likely to reveal any of the two losing doors in any case, so the diagram would be:

Here, once the host discards a specific door and so the corresponding player is eliminated, only two rectangles remain as possibilities, but this time they are of the same size, so at that point each represents $1/2$ chance to occur.