[Not to beat a dead horse, but ...] Monty Hall redux

[bonanova]

Here are the rules; pick a strategy and follow it. Three doors; one car, two goats. You point to a door but do not open it. Monty discloses the contents of a different door. You choose a door and win its contents.

Prove or disprove: p[car] = 2/3.

Did I omit the part about Monty knowing where the car is? Yes. Do you need that to know your winning chances? That's the question.
_________________
_{Vidi, vici, veni.}

[Trojan Horse]

My answer:

Sneaky wording. Apparently, after Monty opens a door, I get to switch to ANY of the doors, including the one Monty just opened. So if he opens the door with the car, I'll just choose that door! Bwa ha ha!

If I know nothing about Monty's procedure for picking a door to open, then my best strategy does indeed give me a probability of 2/3 of winning the car. I just make sure to never stick with my original choice. If Monty reveals the car, I grab it. Otherwise, I switch from my door to the door that Monty didn't open. The only way I lose is if I pick the car at the start, which happens 1/3 of the time. So I have a 2/3 chance of winning.

Now, if I DO know something about Monty's behavior, then my chances could be even better. If I know that Monty always will open the door with the car whenever I don't pick it, then my chances rise to 100%. (If he shows the car, I grab it. Otherwise, I stick with my original door.) But without knowing anything about Monty's behavior, I better stick with the 2/3 strategy.

By the way: I'm currently reading Jason Rosenhouse's book on the Monty Hall problem. So this was already on my mind when you posted this puzzle.

[Zag]

LOL. Nice move, TH!

I assume we are not allowed to pick the car when Monty picks it, we are just stuck.

If we know that Monty is picking as randomly as we are, and we are immediate losers in the cases where he opens the car, then it doesn't matter if we switch.

Say we always pick A. Before we picked, it was equal chance of there being a car behind A, B, or C. Monty open door B and it is a goat. We have eliminated only one case of our original 3 cases, the case where B is the winner; and the remaining choices still have equal probability. In one case, switching helps, in the other case, it doesn't, and neither has a preference.

[Trojan Horse]

Under your assumptions, I agree with your conclusion, Zag.

Now we just have to wait and see which scenario was the one bonanova intended.

[bonanova]

Yeah, that's most of it.

TH, you must pick a single a priori strategy and stick with it. None of this "If Monty knows ..., then I'll ..." stuff. You don't know anything about Monty. That's what the tag asserts.

Final time that I edit this response. You got it. Your 1st strategy is general.

Zag, don't add assumptions. The OP is sufficient. I think. I think... I've said that before when it wasn't, but I think this one works on its own.

OK, the OP needs this additional bit: Assume [only] that Monty acts as your opponent. Translation: He won't knowingly show you the car.
_________________
_{Vidi, vici, veni.}

[bonanova]

TH, since you're into this, FWIW let me pursue it a bit. You say that p[car|original door] =1/3, so you should open another one, just be careful which, altho it's obvious. A subtle point about p[car|original door] is that when Monty doesn't know where the car is and shows you a goat, it increases to 1/2. Accordingly, p[car|door3] drops to 1/2. Not that it matters. Fortuitously they're equal. Thus you can get p[car]=2/3 by applying a single strategy in every case. Open the third door unless you see the car.

Here's how I thought of it. If Monty does not know, 2/3 of the time he shows a goat, and your winning chances are only 1/2. But 1/3 of the time he unknowingly shows you the car. 1/3 x 1 + 2/3 x 1/2 = 2/3.
_________________
_{Vidi, vici, veni.}

[Changabooniggiwan*]

Surely, if Monty is running the show and knows where the car is, he isn't going to open a door knowing that the car is behind it. It's not good TV. That takes us back to the same-old Monty Hall problem that everybody knows about.

[Changabooniggiwan*]

Got it.

Monty either knows (k) or doesn't know (1 - k) where the car is.

If he does know, he will not reveal it. So your best bet is to switch for a k.2/3 chance of winning.

If he doesn't know and accidentally reveals the car, steal it, for a (1-k).1/3.1 chance of winning.

If he doesn't know and doesn't reveal the car, switch anyway for a (1 - k).1/3.1/2 chance of winning.

k.2/3 + (1 - k).1/3.1 + (1 - k).1/3.1/2 = 2/3

Now for another. This time Monty has an assistant, Wilbur. Either Monty or Wilbur (but not both) know the location of the car. Either Monty or Wilbur (but not both) is instructed to lie at all times (the other is instructed to tell the truth). You select, but don't open, a door. Monty then selects a door, peeks behind it, and states its contents (or lies, if applicable). Wilbur takes the third door, peeks behind it, and states its contents (or lies, if applicable). You may then select any of the doors for your prize. What is your optimal strategy?

[ralphmerridew]

Optimal strategy: If Monty and Wilbur both claim to see the car, or neither claims to see the car, choose my original door. If one claims to see the car and the other doesn't, choose one of the other two doors at random.

I win 2/3 of the time.

[Lepton*]

If Monty and Wilbur both claim to see the car, that means that one of them actually does see the car, and so you should switch, no?

[ralphmerridew]

Yeah, I got things backwards.

[Chuck]

The original problem doesn't say that Monty must open a door, only that he does in this particular case. It might be that Monty doesn't have to open a door and does so only if there's a car behind your original choice. If this is the case then you should never switch if given the opportunity since you get the opportunity only if your original choice is the car.

Or maybe Monty gives you the opportunity to switch only if your first choice was a goat. Then you should always switch.

What's the best strategy if you don't know whether or not Monty was required to open a door and don't know his motive in any case?

[Changabooniggiwan*]

Ralph has it, despite the backwards bit.

Now, an apology in advance for the following monster that I call "The Full Monty"...

The usual setup - three doors; one car, two goats.

Monty and Wilbur are hosting the show – they are stuck in a strange contract in which they have to pay for the car themselves (thus it is in their interests for the contestant to win a goat).

Before the show starts one of the two hosts (picked at random) is told the location of the car. Whichever one of the two it is, he may not under any circumstances divulge this information to his co-host unless it is through the normal playing of the game (work on the assumption that this is impossible to do). Additionally, one of the hosts (again picked randomly) is told that he must lie about what he sees behind the door he will later pick. This may or may not be the same host who knows the location of the car.

By deduction, the host who is not selected to know the car’s location will know that the other host does. Similarly, the non-liar will know that the other host is the liar. Only the contestant is in the dark about all this, although he does know that one of the hosts knows the car’s location and that one of them (maybe the same one) is compelled to lie.

The game is played like this: First, Monty picks a door, peeks behind it, and tells the contestant what he sees (he may lie, if he is the selected liar). Second, Wilbur picks one of the two remaining doors, peeks behind it and tells the contestant what he sees (again, lying if required). This leaves a third door which nobody has looked behind. Finally, the contestant gets to choose to open any of the three doors and claim as a price whatever is behind it.

(1) Assuming the contestant is very canny, has a profoundly logical mind, and wants to win the car, what strategy should the hosts use to minimise his chance of winning?

(2) Assuming the hosts have a perfect strategy, what can the contestant do to maximise his chance of getting the car?

[Zag]

Interesting puzzle, Chang.

SPOILER ALERT: I discuss the results here, but didn't invis. On your own head be it. (I hate reading large blocks of invissed stuff.)

After working out all the cases just to get the simple strategy, I realized that the simple strategy is actually very simple. (But then it breaks down.)

First, ignore their actual choices and just consider if they same the same or different. By symmetry, this is all that matters, and it lets you ignore which is the liar.

If Monty and Wilbur say different things, then the contestant knows that the third one is the car. Therefore, suppose they always conspire to say the same thing. First, is this possible? (yes) Second, what is the player's win rate if he knows they are doing this? Third, can they improve on this? Fourth, what are the decision points for the hosts.

I'll answer #4 first. The only decision point for the hosts are if Monty knows which is the car. If Wilbur knows, then he should never pick the same as what Monty picked, because then the contestant always wins. (If Monty and Wilbur say different things, the contestant knows they both saw goats and the car is the third choice.)

#1. Can they conspire to always say the same thing. Yes, they can.
If Monty knows, he picks the car.
If Wilbur knows, he picks the opposite of what Monty picked. Note that Monty picks the car 1/3 of the time, because he doesn't know.

#2. What is the player's win rate against this strategy?
The contestant (assuming the two have said the same thing), should always pick what Monty picked, and he will win 2/3 of the time. (1/2 the time Monty knew, and 1/6 of the time he didn't know but picked the car.)


#3. Can Monty and Wilbur do better than 1/3 win, by sometimes risking saying the opposite?
The only time that this is relevant is when Monty knows, because clearly Wilbur should always choose to make their answers the same. Should Monty sometimes pick a goat and risk the 1/2 chance that Wilbur will also pick a goat (since he knows that the contestant is going to take his pick as long as the two declarations are the same)?

Let's say he does so x fraction of the time. Of those, half are a guaranteed loss and half are a guaranteed win for the contestant, but they were all going to be a guaranteed win, so this is a gain for the hosts.

What if x = 1. That is, Monty, when he knows, never picks correctly, then if the player sticks with the strategy of always taking Monty's choice (when they are the same), his win rate drops to 5/12. (1/6 of the time Monty doesn't know and picks the car, plus 1/4 of the time Monty knows and picks a goat but Wilbur also picks a goat.)

However, if this is the strategy the hosts use, the player can always pick Wilbur's choice (when they say the same), and he arrives at a whopping win rate of 5/6! He only loses in the cases where Monty doesn't know and happens to pick the car.

#3b. Are there other strategies?
No, since we showed with #4 above that the only interesting factor for the hosts is what they do when Monty knows.

So, the correct strategy for the hosts is to choose randomly between those two strategies, where they choose the first one x (fraction) of the time. We select x to be the Nash equilibrium point, where the contestant can not improve his odds by taking one strategy over the other.

I leave the calculation as an exercise for the grad student. However, my guess is that the best approach for the hosts is if Monty, when he knows, picks the car 2/3 of the time. I'm curious if my gut guess is right, but I'm out of time to work on this.

[Trojan Horse]

Whew. When I saw that you'd posted, Zag, I thought you'd completely beaten me to the punch. What a relief that you didn't. Allow me to fill in the gaps.

Here's my independent solution. Looks like I'm repeating a lot of things Zag said already.

First, a couple of facts.

Fact 1: If neither of the hosts peeks at the door containing the car, then their answers will differ; one will claim to have a car, the other will claim to have a goat. The contestant will then pick the 3rd door, and win. So the hosts need to avoid this happening at all costs.

Fact 2: If one of the hosts peeks at the door containing the car, then the hosts' answers will match; they will either claim to both have cars or both have goats. Then the contestant will need to choose one of those two doors. Whether the hosts both claim "car" or both claim "goat" is irrelevant, since each host has a 50% chance of being the liar.

Now, let's talk about the strategy for the hosts.

If Monty does not know where the car is, then he will choose a door at random. If he chooses a goat, then Wilbur will choose the door with the car; it's the only way for the hosts to have a chance. If Monty stumbles onto the car by chance, then Wilbur might as well choose one of the other two doors at random.

If Monty does know where the car is, he must decide whether to pick the door with the car or a door with a goat. Seems like he should just play it safe and pick the door with the car, but let's not assume that yet. Say Monty decides to play a mixed strategy, and picks the car with probability p. Regardless, Wilbur will pick one of the other two doors at random.

Given these constraints, let's find the probability of the car being behind Monty's door, Wilbur's door, or the third door. (In each of these calculations, the first term represents the possibility that Monty knows where the car is. The second term represents the possibility that Wilbur knows where the car is.

Car behind Monty's door: (1/2)(p)+(1/2)(1/3)=p/2+1/6

Car behind Wilbur's door: (1/2)(1-p)(1/2)+(1/2)(2/3)=-p/4+7/12

Car behind neither door (so contestant wins): (1/2)(1-p)(1/2)+(1/2)(0)=-p/4+1/4

Now let's look at things from the contestant's side. The contestant just needs to decide whether to choose Monty's door or Wilbur's door, when the car is not behind the 3rd door.

Chance of winning by choosing Monty's door or the 3rd door: (p/2+1/6)+(-p/4+1/4)=p/4+5/12
Chance of winning by choosing Wilbur's door or the 3rd door: (-p/4+7/12)+(-p/4+1/4)=-p/2+5/6

Basically, the hosts need to choose p so that the larger of those two probabilities is as small as possible. And that will be when they are equal.

p/4+5/12=-p/2+5/6 -> 3p/4=5/12 -> p=5/9

So, if Monty knows where the car is, he should choose the car with probability 5/9. Then it doesn't matter what the contestant does; he can choose at random when the car is not behind the 3rd door, and he wins with probability:

(5/9)/4+5/12=5/36+15/36=20/36=5/9

WHEW!

[Zag]

5/9, eh? My gut was close, but not quite there.

BTW, I didn't bother to invis because there is really nothing that jumps out at you as the answer, so someone won't 'accidentally' see it, and I hate reading large blocks of invissed stuff. I think it kicks my dyslexia into high gear. Hmmm. I'll make a note, at least.

[Changabooniggiwan*]

Nice work, gentlemen.

I love the way that the problem, at first sight, seems to obviously favour the hosts. Then a quick analysis shows the fatal flaw in giving different answers, and that, when adjusting the hosts' strategy to avoid that the contestant gains another advantage. At the end, the very 'perks' that the hosts have conspire to give the contestant a better chance of defeating them.

[ralphmerridew]

I was just looking this over, and I think the hosts have a better strategy.

Before learning where the car is, Monty and Wilbur flip a coin and remember the result.

If Wilbur knows where the car is, then Monty opens a door at random. Wilbur then chooses the car if Monty didn't; otherwise, he chooses a goat. (2/3 of the time Monty finds a goat and Wilbur finds the car; 1/3 of the time Monty finds the car and Wilbur finds the goat.)

If Monty knows where the car is, then Wilbur's strategy is:
- if the flip was heads, choose the door right of Monty's; if it was tails, choose the door left of Monty's. (Imagine the doors being in a circle.)

2/3 of the time, Monty chooses the car. 1/3 of the time, Monty chooses a goat so that Wilbur will find the car. (1/3 of the time Monty finds a goat and Wilbur finds the car; 2/3 of the time Monty finds the car and Wilbur finds the goat.)

Overall, 1/2 of the time Monty finds a goat and Wilbur finds the car; 1/2 of the time Monty finds the car and Wilbur finds the goat.

Player's strategy is: If Monty and Wilbur say the same thing, pick one at random. If they say different things (which they won't under my strategy), pick the third door. Player has a 50% chance of winnng.

[Zag]

Hmmm. I'm trying to figure out if the player can glean any more information out of that, but it seems not.

The 'moderator' will have to decide if the flipping the coin technique is fair. If it isn't, then they have to agree in advance that Wilbur will, in cases where he doesn't know, always pick the one to the left of Monty's pick. If that's the plan, then it clearly gives the player more information, because sometimes (1/6) he will have to pick the one to the right to avoid picking the other goat after Monty had to guess and picked the wrong goat.

I'm inclined to say that them flipping a coin before the game is equivalent to them having a secret strategy that the player doesn't get to know, which disallows it. However, I'm pretty sure I agree that the probability has dropped to 1/2 for the player if they are allowed to do it.

[ralphmerridew]

I took the bit about restricting communications to mean that the game can be treated as:

1) They arrive at the studio for that game.
2) They are taken to isolated rooms.
3) Monty is informed whether he knows the location of the car, and if so where it is, and whether he is truthful or a liar.
4) Monty must pick door A, B, or C.
5) Wilbur is told whether he knows the location of the car, and if so, where it is, whether he is truthful or a liar, what door Monty chose and what Monty said.
6) Wilbur must choose A, B, or C, not matching Monty's choice.
7) The player is told what doors Monty and Wilbur chose, and what they said they saw.
8) The player must choose door A, B, or C.