Does the Monty Hall Paradox hold true if the Game Host can open either door?

Question

The Monty Hall Paradox is the answer to this question.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? (wikipedia)

If you are familiar with the paradox, you know that the answer is to always switch because the odds compress into the other door with 2/3 odds.

But if there are three different things, and the game host is allowed to open either non-picked door, and the revealed item is not the desired item, is it still advantageous to switch? Does the removal of the mechanic where the game host might be limited to only opening one door keep the odds at 1/3 for each door?

My initial instinct is that it stays the same, but this is already a mind puzzle anyway, so I'm not sure.

Answer 1

It depends on how the host chooses which door to show you. If it is that he will always show you a wrong door then this is the classic puzzle and you should switch.

If it is that the host shows you one of the two doors you did not pick and is chosen at random, then the odds are even if you stay or switch.

And if the host is using some other criteria, who the heck knows.

If it's a random door that you didn't choose there are 6 possibilities,ach equally likely. Imagine the back of the doors are labeled Wrong A, Wrong B, and Right but you can not see them.

1)You chose the right door and the host shows you the Wrong A. (STAY)

2) You chose the right door and the host shows you the Wrong B.(STAY)

3) You chose the Wrong A and the host shows you the right door.(NOT AN AVAILABLE OPTION)

4) You chose the Wrong A and the host shows you Wrong B. (SWITCH)

5) You chose Wrong B and the host you the right door. (NOT AN AVAILABLE OPTION)

6) You chose Wrong B and the host shows you Wrong A. (SWITCH).

Of the four available options that occurred (two of them simply did not happen--- this time) two have it that you should switch, and two that you should stay. Each of the four is equally likely.

BUT it depends on how the host choses. The host could always chose the left most door that you didn't pick. That would change your probabilities. Or as, the classic puzzle assumed, the host could always choose a wrong door.

Or, as per the real price is right, the rules whether the host is going to show you anything at all varies game to game would make the puzzle impossible to solve as the criteria are simply unknown.

Answer 2

If I understand this correctly you are talking about 3 objects A,B,C, let's say A is the desired one, then this is actually equivalent to the original game because B and C are both undesired they behave like being one thing:

A | B | C

either B or C is revealed, switch results in loosing.

A | B | C

C is revealed, switch results in winning.

A | B | C

B is revealed, switch results in winning.

Knowing where the least desired one is doesn't help you at all, because it would't change your strategy.

Answer 3

I never found this to be a mind puzzle, more so generally an unclear puzzle. The crucial aspect of the problem is that the presenter will always open a goat door, and in my experience people fail to be clear about this. Whatever lies behind your door, at least one of the other two doors contains a goat, and the presenter knows.

Now, if the host can open either of the two unpicked doors, the probability change depends on how he goes about choosing an unpicked door. Suppose he chooses either door with equal probability.

Strategy $(1):$ Don't change unless host opens a car door

If at first you chose the car door, with probability $1/3$, then whatever door the host opens will be a goat door and you won't change.

If at first you chose a goat door, with probability $2/3$, then with probability $1/2$ the host will open a car door and you will change to the car; otherwise, with probability $1/2$, host opens a goat door and you stick to your goat door.

Your overall chance of getting the car is $1/3+2/3\cdot 1/2=2/3$.

Strategy $(2):$ Always change

If at first you chose the car door, with probability $1/3$, then whatever door the host opens will be a goat door and you will change to the other goat door.

If at first you chose a goat door, with probability $2/3$, then with probability $1/2$ the host will open a car door and you will change to the car; otherwise, with probability $1/2$, host opens a goat door and you change to the other door, which contains the car.

Once again, your overall chance of getting the car is $1/3\cdot 0+2/3\cdot(1/2+1/2)=2/3$.

Conclusion: If the host chooses an unpicked door uniformly at random (and you're allowed to change to the car door if it is revealed), then both strategies have a $2/3$ probability of success.

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Weird, huh? To further drive the point home about 'how random' matters, let's generalize the problem. Suppose that when you pick a goat door, so that when choosing an unpicked door there is one goat door and one car door, the host picks the goat door with probability $0\leq p \leq 1$.

With Strategy $(1)$, we get a $1/3\cdot 1 + 2/3\cdot (1-p)=(3-2p)/3$ probability of success.

With Strategy $(2)$, we get a $1/3\cdot 0 + 2/3\cdot 1=2/3$ probability of success.

Hence, when $3-2p<2\iff p>1/2$, Strategy $(2)$ is best. BUT, when $3-2p>2$ $\iff p<1/2$, the scenario changes, and Strategy $(1)$ is best.

Our initial uniform case, with $p=1/2$, is the critical case in which both strategies are equivalent. The classical problem has $p=1$.

Answer 4

Call your initial choice $D1.$ Suppose that when the prize is at $D1$ the host opens door $D2$ with probability $x$ , and opens $D3$ with prob. $1-x.$

The chance that $D2$ is opened is $x/3$ (when prize is at $D1)+0$ (when prize is at $D2)+1/3$ (when prize is at $D3)=(x+1)/3.$ So the chance that that $D3$ is opened is $1-(x+1)/3=(2-x)/3.$

Now $P(D3$ has prize GIVEN that $D2$ was opened)$=P(D3$ has prize AND $D2$ was opened)$/P(D2$ was opened). But $(D3$ has prize AND $D2$ was opened) is equivalemt to $(D3$ has prize) so $P(D3$ has prize AND $D2$ was opened$)=P(D3$ has prize$)=1/3.$

Therefore $P(D3$ has prize GiVEN that $D2$ was opened) is $(1/3)/((x+1)/3)=1/(x+1).$

Interchanging the subscripts $1,2$ and replacing x with $1-x,$ we see that $P(D2 $ has prize GIVEN that $D3$ was opened is $1/((1-x)+1)=1/(2-x).$

Now suppose you always switch. $D2$ gets open with odds of $1(x+1)$ and your chance of winning when this happens is $(x+1)/3.$ So the proportion of time you win by switching when $D2$ is opened is $(x+1)/3 \cdot 1/(x+1)=1/3$ of the TOTAL time.

Similarly the proportion of time you win when $D3$ is opened is $(2-x)/3\cdot 1/(2-x)=1/3$ of the TOTAL time. So you still win $1/3+1/3=2/3$ of the time by always switching, regardless of $x.$ (Even when $x=1$ or $x=0$).

I live in farm country. Everyone has a good car and usually at least one other old car. Some of the old ones would not be worth a few goats.

Answer 5

The Monty Hall problem is basically a shell game.   The dealer puts a ball under one of three cups and shuffles them around.   You pick one cup, the dealer lifts another, and you have a choice of whether to stay with your original choice, or move.

Basically there are three events.   $A$ your first choice is correct, $B$ the dealer lifts an empty cup, and $C$ changing cups is a good idea.

In the classic scenario, the dealer always chooses the empty cup and you are asked to evaluate the probability that changing cups is a good idea given that certain event. $$\begin{align}\mathsf P(C\mid B) ~&=~ \mathsf P(C\mid A,B)~\mathsf P(A\mid B)+\mathsf P(C\mid A^\complement,B)~\mathsf P(A^\complement\mid B) \\ &=~ 0\cdot \tfrac 13+1\cdot \tfrac 23 \\ &=~ \tfrac 23\end{align}$$

Since as it is certain that the dealer always chooses an empty cup, then the event that your first guess is right is independent of the dealer's revelation.

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So, how does uncertainty that the dealer could lift an empty cup change this? We go to Bayes' Rule.   Assuming the dealer's choice between the two untapped cups is biased to wards favouring the ball with probability $p$ when given your first guess was wrong.

$$\begin{align}\mathsf P(C\mid B) &=~ \dfrac{\mathsf P(C\mid A,B)~\mathsf P(B\mid A)~\mathsf P(A)~+~\mathsf P(C\mid A^\complement,B)~\mathsf P(B\mid A^\complement)~\mathsf P(A^\complement)}{\mathsf P(B\mid A)~\mathsf P(A)+\mathsf P(B\mid A^\complement)~\mathsf P(A^\complement)} \\[1ex] &=~ \dfrac{0\cdot 1\cdot \tfrac 13 ~+~1\cdot p\cdot \tfrac 23}{1\cdot \tfrac 13 ~+~ 0\cdot \tfrac 23}\\[1ex] &=~ \dfrac {2p}{1+2p}\end{align}$$

So when the choice is unbiased, then $\mathsf P(C\mid B)= \tfrac 12$.   This is what most peoples intuition about the classic Monty Hall game tells them, and its based in failing to realise the host cheats in favour of the player by never picking the prize.