Monty Chooses

I want to revisit Monty Hall again, for reasons that hopefully will become apparent.

Suppose there was a slight variation to the standard Monty Hall scenario.  Consider the following:

Monty Hall is the compere on a game show.  In one special round of that game show there is the familiar set-up.  There are three doors, behind one of which is a car and the other two doors each hide a goat (placed randomly).  The contestant is invited to select a door, with the idea that if the car is behind her car, she will win it.  The contestant selects a door (we’ll call this “door X”, meaning “the door that the contestant chose”).

The main observable difference is that Monty now selects a door (we’ll call this “door Y”, meaning “the door that Monty chose”).  He goes up to that door and looks through a peephole.  Then Monty walks to the other door (which we’ll call “door Z”, meaning “the door that neither chose”), and opens that door, revealing a goat.

Given that we know nothing else, can we work out the likelihood that the contestant will benefit by swapping to door Y, if given the opportunity?

It looks very similar to the standard Monty Hall problem, but is the difference significant?

It’s tempting to assume that Monty will never reveal a goat, so he would have opened door Y if there was no goat there, but he didn’t.  He walked to door Z and opened it, revealing a goat.  Why would he have done that?  The hints are that if he never reveals a goat, then he must have known that the goat was not behind door Z, and the fact that he peeped at what was behind door Y indicates that he hasn’t been informed as to the distribution of goats and car, so the car must be behind door Y.  Therefore, the contestant should, in order to win the car (with an apparent likelihood of 100%), swap from door X to door Y.

Now consider instead, holding all other factors equal, a slightly different scenario in which Monty peeps behind his door, door Y, and then opens it.  What is the likelihood that the contestant will benefit from swapping to door Z?  Is it the same as Monty Falls, in which he accidentally opens the first door he passes and it just happens to reveal a goat, or does this situation mirror the standard Monty Hall problem?

To work it out, we have to remember that nothing was stated about Monty’s intentions nor about what he knows and doesn’t know – and it was specifically stated that we know nothing else.  We weren’t even told that the contestant would get the chance to swap, which she wouldn’t have if Monty had revealed a car (this possibility was not excluded, if you thought it was, this was just an assumption on your part).

Sticking with the scenario in which Monty opens door Y, all the contestant knows is that a goat is behind door Y and that Monty knew this when he opened the door.  She does not know how Monty selected the door to peep through, but we are tempted to assume it was random (or that it doesn’t matter because the distribution of car and goats is random).

If we make a different assumption, we get a different likelihood that the contestant would benefit from swapping.  Say that Monty is restricted from revealing the car and he knows where it is.  This does not prevent him from selecting a door to peep through at random.  There is a 1/3 chance that the contestant’s door hides the car, in which case he can choose to open either door Y or door Z.  There is a 1/3 chance that the door he peeps through hides the car, in which case he must open the other door (we’re assuming that he can’t open the contestant’s door).  There is a 1/3 chance that he has no choice and must open the door he peeps through (again assuming that he can’t open the contestant’s door).

Monty’s subterfuge

A general assumption in the standard Monty Hall problem is that, when given a choice as to which door to open, Monty will simply select one at random.  This is predicated on the assumption true that Monty doesn’t care whether the car is won.  However, if he does care and he gets the opportunity to peep behind a door, despite knowing the distribution of car and goats, then he has motivation to make thoughtful rather than random selections.  In this case does it matter which decision he makes and what decision should he make to minimise the likelihood of the contestant winning the car?

I suspect that it does matter and that his best option is to preferentially open door Z.  The reason for this is that the contestant does not know what is known to Monty and is likely (more on the basis of psychology than of statistics) to assume that there is a greater likelihood that Monty is forced to open door Z than that he had done so by choice.  (This would be the case if he chose at random when given the opportunity and also if he didn’t know the distribution of car and goats.)  Therefore she will use the logic laid out above to conclude that the car is behind door Y.  But Monty will choose door Z more frequently than the contestant thinks, 2/3 of the time – when the car is behind door Y and when the car is behind door Z.  So, she’ll only win the car 50% of the time by swapping to door Y.

If, on the other hand, Monty opens door Y after peeping behind it, then contestant might easily conclude that she is in a Monty Falls type situation (based on the assumption that Monty selects the door at random and, lo, it doesn’t have the car, so he opens it) and that therefore she doesn’t benefit statistically from swapping (or indeed from staying).  There are psychological factors that make us prefer things we possess, so if the contestant thinks that it’s 50-50 as to where the car is, she’s more likely to stick with her choice, meaning she’ll lose.  If she leaves the choice up to chance, tossing a fair coin to decide, then there’ll be a 1/2 chance that she’ll swap, and thus a 1/2 chance of winning.  If she simply ignores the extra information implicit in this scenario, Monty’s peeping behind door Y, she’ll win.

By engaging in this subterfuge, Monty can possibly minimise the likelihood that the car will be won, but only if it remains a secret.  If the contestant becomes aware of Monty’s subterfuge, she’ll know that if door Y is opened then the car is definitely behind door Z, but if door Z is opened then it’s 50-50 as to where the car is.  Her best strategy would then default to that of the standard Monty Hall, always swap.  In other words, she should ignore an element of information available to her to maximise her chances of winning.

Helpful Monty

Monty could, however, have a different set of intentions.  He might desperately want to give away a car, but be obliged to remain within the rules of the game (an equivalent to this appears to be a standard assumption within intelligent design theory).  If so, he merely swaps his strategy to encourage a consideration on the part of the contestant that would maximise her chance of winning – by preferentially opening door Y, only opening door Z if the car is behind door Y.  In this case, the contestant will have the same likelihood of winning whether she pays attention to Monty’s peeping or not.  It also doesn’t matter whether she believes that Monty has no idea where the car is, or if she thinks he’s trying to help.  In this case, she can ignore the information or try to make use of it, it doesn’t really matter either way.

There’s Something about Mary

Alternatively, Monty might not care about the car at all and merely prefers to reveal Mary (one of the goats).  If so, he’ll open the door Z if Mary isn’t behind door Y when he peeps, even if that might reveal the car, because it maximises his chance of revealing Mary.  If the contestant knows this to be the case, then she would know that there’s a benefit in swapping to door Y, if door Z was opened, but it’s 50-50 if he opens door Y.  There’s a 1/3 chance of Mary being behind door Y.  There are two equally likely distributions in which Mary could be behind door Y, one which has the car behind door Z, to which the contestant might consider swapping.  There is a 2/3 chance that Mary will not be behind door Y in which case Monty will open door Z.  There is a 1/2 chance that he did so after seeing the other goat and if so, there is a 1/2 chance that when he opens door Z, he’ll reveal the car – nullifying the game.  This gives the contestant a 2/3 chance of winning if Monty opens door Z and she swaps.  The overall likelihood of winning as a result of swapping is 3/5 rather than the standard 2/3 because of the 1/6 chance that Monty will reveal the car.

However, yet again, the contestant can obtain the same result merely by ignoring Monty when he peeps behind the door.

Now for the key question: So frigging what?

Whoa, that was unnecessarily aggressive!  The point is that Monty is an intelligent agent, with a psychology that the contestant must intuit.  The same sort of consideration applies in Hawthorne’s prisoners, only in that case the (ultimate) psychology being considered was that of a putative god.

What I’ve tried to show here is that the intent of the intelligent agent can matter sometimes, when working out the likelihood of desired result, but it doesn’t always matter – sometimes it has little or no apparent influence at all.  The trouble is that it’s not exactly obvious as to what matters.  It matters if Monty is committed to not revealing the car, and it could matter if we are persuaded that he doesn’t know where the car is, but we can maximise the likelihood of a good outcome by simply ignoring his intent.  Similarly, for the purposes of working out whether Miss Justice is responsible for a release decision, it matters how she would choose a prisoner to release and whether their location matters, but so long as location doesn’t matter it won’t really matter much what her criteria actually are – meaning that our ability to detect her influence will not vary if she prefers to release the shortest innocent prisoner rather than one who is most worthy of clemency.  An implication that we can draw from this is while it certainly does matter what an intelligent agent knows, what they intend is somewhat less likely to matter.

This has some bearing, I think, on the fine tuning argument for the existence of some sort of god.  Fundamentally, the argument derives from an observation that the universe has a suite of constants and initial conditions that could not be varied much without making life impossible.  A designer is then posited as the mechanism by which the values for these constants and initial conditions were selected – with the assumption that the designer intends that (intelligent) life should manifest.

While there is a danger going from the specific to the general, let alone from one specific to another specific, what the consideration above (together with Weisberg’s prisoners) tells us is that we should at the very least be careful when basing an argument on an assumed intent.  The fact that theists base their best argument * on such foundations should give them pause.

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* Admittedly this is the theists’ best argument according to Christopher Hitchens.  Perhaps theists themselves think that they have a better argument.