If Monty Hall doesn't know where the prize is, should the contestant still switch doors, after Monty opens one door and unveiss a goat?

The other day I was asked a variation of the Monty Hall Problem.

The answer of the original question is, of course, $ 66\% $ in favor of changing doors, but this is based on the fact that the game show host knows where the prize is.

Suppose Monty does not know where the prize is, and after you pick, he opens one of the other two doors and it happens to be a goat. Is it still better to change doors when he asks?

I believe it is. (After all it still leaves us with two chances instead of one.) But some of my friends think otherwise. We are not mathematicians, just a couple of riddle-likers, so we are not sure of the correct answer. So I thought to post it here.

Edit

What are your thoughts on the following on Wikipedia? This quotation seems to support my answer.

Morgan et al. (1991) and Gillman (1992) both show a more general solution where the car is (uniformly) randomly placed but the host is not constrained to pick uniformly randomly if the player has initially selected the car, which is how they both interpret the well known statement of the problem in Parade despite the author's disclaimers. Both changed the wording of the Parade version to emphasize that point when they restated the problem. They consider a scenario where the host chooses between revealing two goats with a preference expressed as a probability q, having a value between 0 and 1. If the host picks randomly q would be 1/2 and switching wins with probability 2/3 regardless of which door the host opens. If the player picks Door 1 and the host's preference for Door 3 is q, then in the case where the host opens Door 3 switching wins with probability 1/3 if the car is behind Door 2 and loses with probability (1/3)q if the car is behind Door 1. The conditional probability of winning by switching given the host opens Door 3 is therefore (1/3)/(1/3 + (1/3)q) which simplifies to 1/(1+q). Since q can vary between 0 and 1 this conditional probability can vary between 1/2 and 1. This means even without constraining the host to pick randomly if the player initially selects the car, the player is never worse off switching. However, it is important to note that neither source suggests the player knows what the value of q is, so the player cannot attribute a probability other than the 2/3 that vos Savant assumed was implicit.

Solution 1:

Suppose the prize is in A, for argument's sake. There are 6 equally likely options a priori:

• you pick A and the host picks B (no switch)
• you pick A and the host picks C (no switch)
• you pick B and the host picks A (this did not happen, as we then wouldn't see the goat)
• you pick B and the host picks C (switch)
• you pick C and the host picks A (ruled out, as before)
• you pick C and the host picks B. (switch)

By knowing that we saw a goat when the host picked (without information) we have that we have 4 situations we could be in (all same probability) and in 2 of them we need to switch. So we now have a 1/2 chance.

Solution 2:

I had prepared a great argument to this, typed it all up, and realized I was wrong at the last moment. This question has already been answered, but I thought I'd add my explanation in case it helps anyone later.

Bayes Theorem: $$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$$ For this case, I'm using this:

• Let Winning be the probability of winning if you switch
• Let Goat be the probability of Monty Hall revealing a goat

$$P(Winning|Goat)=\frac{P(Goat|Winning) \cdot P(Winning)}{P(Goat)} $$

For the original Monty Hall problem, because Monty knows where the goat is, there's a 100% chance he'll reveal a goat whether or not switching will make you win (P(Goat|Winning)). There's also a 100% chance he'll just pick a goat (P(Goat)). There are 4 out of 6 possible winning outcomes if you adopt a switching strategy. The formula works out like this: $$P(Winning|Goat)=\frac{1 \cdot .66}{1}=.66$$

If Monty Hall does not know where the goats are, the probabilities are a little different and is a little complicated (at least it is for me).

The probability of Monty Hall picking a goat ($P(Goat)$) is pretty easy; that's just 2 out of 3 or about 66%.

The other two variables depend on the rules of the game. For example, if Monty Hall will let you switch to the car if he reveals the car, then the probability of winning the game if you adopt a switching strategy ($P(Winning)$) is 66% because there are 4 winning outcomes out of 6; however, if you're not allowed to switch to an open door, then the two outcomes where Monty Hall picks the car at random become losing scenarios and $P(Winning) = \frac{2}{6} = 33\%$.

Finally, P(Goat|Winning) means "the probability of Monty Hall picking a goat given that you will win if you switch." If you are guaranteed to win if you switch doors, that means that you must have selected one of the two goats.

For the first rule there are four outcomes:

• You picked goat B, Monty reveals car A, you switch, you win
• You picked goat C, Monty reveals car A, you switch, you win
• You picked goat B, Monty reveals goat C, you switch, you win
• You picked goat C, Monty reveals goat B, you switch, you win

So, of the winning switching outcomes, Monty picks a goat half the time. Thus, $P(Goat|Winning)=50\%$ for the first rule.

In the second rule, there are two winning outcomes:

• You picked goat B, Monty reveals goat C, you switch, you win
• You picked goat C, Monty reveals goat B, you switch, you win

Remember, if Monty reveals a car for the second rule, you lose automatically. As a result, $P(Goat|Winning)=100\%$ for the second rule.

If you are allowed to switch to the door Monty reveals, then: $$P(Winning|Goat)=\frac{.5 \cdot .66}{.66} = .5$$

On the other hand, if Monty revealing a car means you lose, then: $$P(Winning|Goat)=\frac{1 \cdot .33}{.66} = .5$$

Solution 3:

Suppose Monty is biased towards selecting goats whenever there is a choice between goat and car, with probability $p$.

There are three possible outcomes of this game.

1. There is a prise behind the door you selected, so Monty shall certainly reveal a goat and you shall lose if you switch. This event occurs with probability of: $$\begin{align}\mathsf P[P\cap G\cap S^\complement]&=\mathsf P[P]\cdot\mathsf P[G\mid P]\cdot\mathsf P[S^\complement\mid P,G]\\&=\tfrac 13\cdot 1\cdot 1\\&=\tfrac 13\end{align}$$

2. There is no prise behind the door you selected and Monty reveals the remaining goat, so you shall win if you switch. This event occurs with probability of: $$\begin{align}\mathsf P[P^\complement\cap G\cap S]&=\mathsf P[P^\complement]\cdot\mathsf P[G\mid P^\complement]\cdot\mathsf P[S\mid P^\complement,G]\\&=\tfrac 23\cdot p\cdot 1\\&=\tfrac 23p\end{align}$$

3. There is no prise behind the door you selected and Monty doesn't reveal the remaining goat, so you shall lose whatever you do. $$\begin{align}\mathsf P[P^\complement\cap G^\complement\cap S^\complement]&=\mathsf P[P^\complement]\cdot\mathsf P[G^\complement\mid P^\complement]\cdot\mathsf P[S^\complement\mid P^\complement,G^\complement]\\&=\tfrac 23\cdot (1-p)\cdot 1\\&=\tfrac 23(1-p)\end{align}$$

[$\tfrac 13$ is the probability that the prise shall be behind your selected door; in that event Monty has only goats to reveal. On the event that the prise is not there, then Monty will either choose a goat with probability $p$, or the prise with probability $1-p$.]

Since we are interested in the condition that Monty reveals a goat, the focus is on the first two outcomes.

$$\begin{split}\mathsf P[S\mid G] &=\dfrac{\mathsf P[P^\complement\cap G\cap S]}{\mathsf P[P^\complement\cap G\cap S]+\mathsf P[P\cap G\cap S^\complement]}\\&=\dfrac{2p}{2p+1}\end{split}$$

So under the condition that Monty revealed a goat when making unbiased selections ($p=1/2$), the probability that switching wins is $1/2$.

When $p=1$, that is "Monty always selects goats", we get the traditional answer for the Monty Haul scenario: $2/3$