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Many of us know the Monty Hall Problem

But the other day I was asked a variation of this riddle.

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 he does not know where the prize is, and after you make your 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 :

I read this on wikipedia and if I understand it correctly it 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.

What are your thoughts?

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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. $\endgroup$ – ShreevatsaR May 28 '11 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 '11 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 '11 at 13:09
  • $\begingroup$ Dont the exact options as listed above repet themselves even when rhe host knows? It is the exact same situation as agen the host knows which leads to 66% $\endgroup$ – Jason May 28 '11 at 16:36
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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. $\endgroup$ – ShreevatsaR May 28 '11 at 19:02
  • $\begingroup$ I think I get it now thanks $\endgroup$ – Jason May 28 '11 at 19:11
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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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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 '17 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. $\endgroup$ – mathreadler Aug 7 '17 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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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.

  • There is a prise behind the door you selected, so Monty shall certainly reveal a goat and you shall loose if you switch. $\mathsf P[P\cap G\cap S^\complement](p)=\tfrac 13$
  • There is no prise behind the door you selected and Monty reveals the remaining goat, so you shall win if you switch. $\mathsf P[P^\complement\cap G\cap S](p)=\tfrac 23p$
  • There is no prise behind the door you selected and Monty doesn't reveal the remaining goat, so you shall loose whatever you do. $\mathsf P[P^\complement\cap G^\complement\cap S^\complement](p)=\tfrac 23(1-p)$

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

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

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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