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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.

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    $\begingroup$ The paragraph you quoted is about the case where the host knows the location of the prize, and if you have initially selected the prize, he picks among the two goat-doors with some distribution (not necessarily uniformly). It is irrelevant to the earlier part of your question. $\endgroup$ May 28, 2011 at 20:29

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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.

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    $\begingroup$ what makes this any different then when the host knows? he will not pick A just as picking A will not happen here. If we picked A and the car was in B or C then switching will always give as a win. The same as when the host knows where it is. Or am I missing somthing? $\endgroup$
    – Jason
    May 28, 2011 at 12:01
  • $\begingroup$ @Jason: When the host knows and avoids opening the prize door, the equally probably events are: you choose A; you choose B; and you choose C. Of these 3 equally probably events, switching is better in 2 out of 3. $\endgroup$
    – Henry
    May 28, 2011 at 13:09
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    $\begingroup$ @Jason: No. Instead of the two options that were here assigned equal a priori probability and ruled out a posteriori (namely: you pick B host picks A, and you pick C host picks A), in the Monty Hall case when the host knows, they are ruled out (have 0 probability) even a priori because a host who knows will never open them. This is the whole point of the Monty Hall problem. $\endgroup$ May 28, 2011 at 19:02
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    $\begingroup$ @ShreevatsaR actually I dont think I do see my edit $\endgroup$
    – Jason
    May 28, 2011 at 19:34
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    $\begingroup$ I think what confuses people about this answer is your comment about ruling out the host picking A. Intuitively we could say that if the host has no information, it's effectively the same as you making a second guess (and there was no host). The only reason the probabilities change in the original form of the problem is because the host has to restrict his choices based on what he knows. $\endgroup$
    – Peter
    Dec 10, 2014 at 19:56
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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: $$P(Winning|Goat)=\frac{1 \cdot .33}{.66} = .5$$

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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: $2/3$

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  • $\begingroup$ Can you please expatiate how you computed $\tfrac 13, \tfrac 23p, \tfrac 23(1-p)$? Where do these probabilities hail from? $\endgroup$
    – NNOX Apps
    Dec 31, 2021 at 5:36
  • $\begingroup$ $\tfrac 13$ is the probability that the prise shall be behind your selected door. If the prise is there Monty has only goats to reveal. If the prise is not there, then Monty will choose a goat with probability $p$, or the prise with probability $1-p$. $\endgroup$ Dec 31, 2021 at 9:39
  • $\begingroup$ many thanks Prof. Kemp! 1. Do you mind using another letter rather than P, because you already use P for Pr? 2. Do you mind defining P (or your new letter), G, S? 3. can you please expound how $$\begin{split}\mathsf P[S\mid G] &=\dfrac{\mathsf P[P^\complement\cap G\cap S]}{\mathsf P[P^\complement\cap G\cap S]+\mathsf P[P\cap G\cap S^\complement]}?\end{split}$$ This DOESN'T look like Bayes Rule, which is merely $P(S|G) = \dfrac{P(S \cap G)}{P(G)}$. $\endgroup$
    – NNOX Apps
    Jan 5 at 5:06
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A third interpretation: If Monty reveals the car, then game is neither won nor lost so we assume that he shows goat, as otherwise there is no choice and in practice no game of chance to play at all.

  1. There is 1/3 for the player to pick right from the start, then goat is always shown and we lose 100% if switching after Monty shows us the goat.

  2. There is 2/3 for the player to pick wrong from the start, assuming Monty shows goat, we win if switching 100% of the time.

So it is still 2/3 chance. See my answer to closed question for better explanation.

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  • $\begingroup$ This is incorrect - of your 2/3 chance in point, half of the time, "the game is neither won nor lost." Therefore, if you chose wrong from the start, half the time the game will end early. This leaves you with 1/3 win, 1/3 lose, 1/3 neither, therefore 1/2 win if neither doesn't happen. $\endgroup$
    – Stephen S
    Aug 7, 2017 at 17:51
  • $\begingroup$ As I explain, I didn't count the neither cases but removed them entirely from the event space. The question did not mention how they were to be treated. $\endgroup$ Aug 7, 2017 at 18:32
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Here's a variation, which also includes answering the classic formulation, and whether Monty knows where the prize is or not, what door you picked, or both.

You’ve made a choice of door. Monty opens a door at random.

If the door he opens does have the car behind it, what are the rules? If that means you win it, then your chances going into the game of winning the car are now better than 2/3, which is the probability you’ll win the car in the classic formulation.

The probability he will open the door the car is behind is 1/3. You win. Could have been the door you in fact chose.

Now consider if the door he opens doesn’t have the car behind it (probability 2/3).

If he opens one of the remaining 2 doors you didn’t choose (at random or with knowledge aforehand, doesn’t matter actually) and it doesn’t have the car behind it, then it’s basically the same as the classic Monty Hall problem. The chance the car is behind the door you picked is 1/3. Remains so, doesn’t change. Consider opening that door 300 times now, either before or after Monty has opened a different door revealing a goat. About how many times will the car be behind it do you think?... Answer: about 100. (There’s no hope if you don’t agree with that.) The door Monty has opened has been eliminated (probability is 0 that the prize is behind it). Your chance of winning by switching from your original choice to the remaining closed door is 2/3.

If he opens the door you chose, (and the car is not behind it), there is then an equal probability (1/2) the car is behind each of the remaining doors. Pick one.

Per above, if the door he opens doesn’t have the car behind it, and it’s the door you originally picked (probability 1/3), then your chances of winning the car are 1/2. If it’s one of the doors you didn’t pick (probability 2/3), then your chances of winning are (also) 2/3.

So your probability of winning the car is 2/3 (the probability of this scenario) x (1/3 x 1/2 + 2/3 x 2/3) (the probability you’ll win the car now) = 2/3 x (11/18) = 22/54, + 1/3 (which is the probability you won the car behind the door Monty opened) = 40/54 which is greater than 36/54, i.e. 2/3, which you would expect, since there’s a chance Monty may win the car for you. As the above are all mutually exclusive events.

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