Monty's choice of door to open when the car is behind the door initially chosen by the contestant

Question

In the Monty Hall game, suppose that whenever the car is behind the door initially chosen by the contestant (so that Monty Hall may open either one of the remaining two doors), he chooses to open the door labeled with the smaller number. Is the probability of winning upon switching still 2/3 or is it 1/2? My considerations suggest that not only is it still 2/3, but also it is 2/3 no matter how Monty chooses a door to open when he does have a choice. Am I wrong?

If computer simulations (Monte Carlo) confirm or contradict the assertion that it is still 2/3, I would like to know. I am not in a position to carry out simulations myself.

Available applets demonstrate the 2/3 probability but they may be programmed to make Monty choose either door with probability 1/2.

Answer 1

Your Monty Hall variant is known as the 'Monty Crawl' or 'Lazy Monty' (the door with the lower number is closer to Monty) scenario. With that additional assumption it depends on what door was opened. For example, let's take the problem description from the Wikipedia page on the Monty Hall problem:

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?

In this case, it should be clear that door 2 contains the prize if Monty is indeed lazy (if door 2 would have had a goat, Lazy Monty would have opened up door 2). So in this scenario you should definitely switch, as you have a 100% chance of getting the car by picking door 2.

OK, but what if we had:

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. 2, which has a goat. He then says to you, "Do you want to pick door No. 3?" Is it to your advantage to switch your choice?

Well, now it doesn't matter whether you switch or not. It turns out to be 50-50

To see the math for the latter (simulations are fine, but math is better :P )

Let $Pi$ be the event of you picking door $i$, $Gi$ the event of Monty revealing a goat behind door $i$, and $Ci$ the event of the car being behind doors $i$

So what we want to know is: Given $P1$ and $G2$, what is the chance of the car being behind door $1$ ... and what is the chance of it being behind door $3$? That is, what are $P(C1|G2,P1)$ and $P(C3|G2,P1)$?

Well, let's first point out that:

$$P(G2|C3, P1) = 1$$ (Monty is forced to open door $2$ if you pick door $1$ and the car is behind door $3$)

$$P(G2|C1,P1)=1$$ (Monty could open door $2$ or $3$, but since Monty is lazy, Monty will definitely open up door $2$ ... this is of course where this Lazy Monty variant differs from the original Monty Hall problems, where we set $P(G2|C1,P1)=P(G3|C1,P1)=\frac{1}{2}$. With Lazy Monty, we have $P(G3|C1,P1) = 0$ )

Of course we also have $$P(G2|C2 , P1) = 0$$ (it is impossible for Monty to reveal a goat behind door $2$ if it has a car)

OK, so: $$P(G2|P1) = P(G2|P1,C1)*P(C1) + P(G2|P1,C2)*P(C2) + P(G2|P1,C3)*P(C3)$$

$$= 1 * \frac{1}{3} + 0 * \frac{1}{3} + 1 * \frac{1}{3} = \frac{2}{3}$$

So: $$P(C3 | G2 , P1) = \frac{P(G2|C3 , P1) * P(C3|P1)}{P(G2|P1)} = \frac{1 * \frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$

Sanity check: $$P(C1 | G2 , P1) = \frac{P(G2|C1 , P1) * P(C1|P1)}{P(G2|P1)} = \frac{1 * \frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$

So yeah, in that scenario it does not matter whether you switch or not.

You can use these same formulas to explore what happens when Monty prefers a certain door with a certain likelihood (e.g. $P(G2|C1,P1)=\frac{3}{4}$ and $P(G3|C1,P1)=\frac{1}{4}$), but I'll leave that up to you.

Answer 2

Short answer: chance for the contestant to win the prize, under optimal policy, is $2/3$ in this variant, the same as in the usual version of the game under the change doors optimal policy.

There are 3 possible positions for the prize vs goat (consider the doors are in order 1 to 3, left to right): $$P-G-G\ \ \ \ \ \ \ G-P-G \ \ \ \ \ \ G-G-P$$ Suppose door 1 was chosen by the contestant (marked with a $c$, e.g. $\overset{c}{P}$), and Monty chooses the first possible door left to right (marked with an $m$, e.g. $\overset{m}{G}$): $$\overset{c}{P}-\overset{m}{G}-G\ \ \ \ \ \ \ \overset{c}{G}-P-\overset{m}{G} \ \ \ \ \ \ \overset{c}{G}-\overset{m}{G}-P$$

1. If contestant picks door 1 and Monty picks door 3: best option is changing doors. $\text{Pr(Win)=1}$
2. If contestant picks door 1 and Monty picks door 2: keeping the door or changing door will both lead to $\text{Pr(Win)=1/2}$

Therefore, given that the 3 configurations Prize-Goat have probability $1/3$, the overall probability of winning for a contestant choosing door 1 and following an optimal policy is: $$\text{Pr(Win|Door 1 chosen and optimal policy)}=\dfrac{1}{3}\times 1 + \dfrac{2}{3}\times \dfrac{1}{2}=\dfrac{2}{3}$$ Same result will be obtained if door 2 or door 3 had been originally choosen by the contestant.

Notice that $2/3$ is also the probability of winning under the change door policy in the original version of this game.

Answer 3

You don't need a computer to simulate this if the simulation is for you and not for a course.

You'll need a six sided die. For the die, let a roll of $1$ or $2$ be a $1$, let a $3$ or $4$ be a $2$ and let a $5$ or a $6$ be a $3$. You'll use these outcomes to select the door that holds the prize.

Write down your guess and then roll the die. It's more fun if you have a friend guess. Then write down whether you win or lose with each of the two strategies.

You'll find that the switching strategy still gives you a $\frac{2}{3}$ chance of winning as in the online applets.