Rawls' veil of ignorance. Bradley's

*Appearance and Reality*. Philosophers love -- one might even say they depend upon -- the distinction between truth and semblance; all the more when truth turns out to be a surprising reversal of semblance. And what better than a reversal that involves an actual unveiling?

One of my favorite counter-intuitive assertions is the so-called Monty Hall paradox, named for the game show host who exploited it on Let's Make A Deal. Imagine a set of three closed doors. Two of them are known to conceal a worthless booby-prize (I think it was a goat on the TV show); one of them conceals a coveted prize of considerable worth. It is of course unknown to you which door hides the real prize. In the scenario we are considering, you are instructed to pick a door to open: A, B, or C. You do so (let us say you pick A). At this point, the host (who does know which door hides the prize you want) opens one of the unchosen doors (say, C) to reveal a booby-prize. The choice is then presented to you: Do you wish to stick with your original door pick, or would you like to switch your choice?

Usually, one's intuition says; what's the difference? You've shown me a goat, so I know that one of the doors hides the prize and the other one hides a goat. Fifty-fifty. It makes no difference if I switch or not.

The answer is: You would like to switch. If you switch, you in fact double your chances of winning the prize.

This solution can be very difficult for some people (myself included) to accept. If you are one of them, you are in good company. Besides my illustrious self, the famous and brilliant Paul Erdos refused to believe his own calculations (which showed that the above solution was correct) until a computer program ran many, many simulations and demonstrated the statistical reliability of the switching strategy.

Why is it so hard to believe? Well, you initially choose a door with a random 1-in-3 chance of being correct, and the opening of the other door obviously does not change that likelihood. And that being the case, why should you switch? You could have chosen the other closed door before, but that door too had a 1-in-3 chance of being right, and still does. Yes?

No. The door you chose before is just as likely as it was to hide the prize. But the other door now has a probability of 1-in-2.

But wait -- your own choice also has a probability of 1-in-2, surely?

Think of it this way. Initially you had a 2-in-3 chance of picking wrong. Now you have a 1-in-2 chance of picking wrong. You thus should re-pick, because now you chances of being wrong are less than they were before.

But -- but -- (my own mind sputters) -- why can't I re-pick the same door as I picked before?

Here is where logic starts to stammer and mathematics (in the form of a table of probabilities) steps in to convince. Let us take the case where you picked door A. There are three possibilities. Door A could hide The Prize. Or it could hide booby-prize 1 or booby-prize 2 (we will call them both goats). You have a choice of staying with your choice or switching. Here is a table of outcomes:

Even with this clearly laid out, the rationale is (typically) not easy to accept. But the solution is usually referred to as a "veridical paradox," that is, an assertion demonstrably true despite its

*prima facie*counter-intuitive nature. You can run a simulation over and over again, comparing switching to non-switching strategies, and see the former's success rate slowly but surely outdistance the latter's, and thus pile up empirical evidence of the unreliability of intuition. This sort of thing is frequently alleged to build up the case for the fallibility of our often-incorrigible prejudices; to show yet again that we have certain deep-seated fallacies built into our much-vaunted "rationality." There are schools of cognitive science and philosophy of mind which like this sort of thing because it (ostensibly) gives implicit sanction to other arguments that are less demonstrable but no less counter-intuitive.

I am not so sure. I don't dispute the mathematics. But there is a very simple variation that doubles-down on the counter-intuition and makes its resolution far less amenable to tablature. (I don't assume this variation is original to me, but I don't keep up with the literature.)

Rather than one contestant, imagine three: we will call them Xavier, Yorick, and Zane. These three decide, by whatever means they like, to each choose one door (A, B, or C), so that each one of them has selected a different door. Or alternately, Monty can assign them their doors without telling them which door is whose, until revealing their assignments all at once. Now the host reveals the unpleasant truth: Zane has chosen (or been assigned) poorly: He is matched with door C, but behind door C was only a goat. The question now is: should Xavier and Yorick switch doors? And here the paradox comes into full force. For of course both Xavier and Yorick, being no fools, have studied the Monty Hall problem, and they both know that it would be advantageous to switch.... but then, it would also be advantageous to the opponent to switch. But it can't be in

*both*their interests to switch! And here, no series of simulations will avail, no matter how many quintillions or googolplexes you run, for in any case, either Xavier or Yorick will win, but not both. Which should mean that their

*proportion of winning*will remain 50/50.

This

*does*seem to be a paradox, because the veridicality works both ways.

*Both*Xavier and Yorick "should" switch (and therefore, equally, should not). You can't resolve it by calling the problem a positive-sum deal, for the terms of the problem are that one will lose and one will win.

One option for trying to resolve it is by saying that the probability enters in when X, Y, and Z make their original choices, since one of them must have chosen first, leaving the other two with only two options. This could mean that the one who chose first (either X or Y) ought to switch, while the one who chose second ought not.

There is an objection to this attempted solution: the question is not whether you had three choices initially, but rather whether what your initial chances were, at the outset of the problem, of being right in your choice.

*That*variable is not altered by the order in which the choices are made.

I'm not sure what I think of this objection, which may just show up my slow-learning curve when I try to think about probability. But the scenario in which the host arbitrarily assigns the doors to X, Y, and Z and only reveals the assignments after all assignments have been made. makes the issue irrelevant, because in that situation, the three contestants are not aware of the order of the assignments; but they are aware that the mathematics indicates that they should switch.

It is possible that I am missing something here, in which I will be grateful for a statistician to step in and offer some

What makes this more than just a logic puzzle is the fact that the intuitively "obvious" solution, which is shown to be wrong by patient application of probabilities, is the fact that it is a kind of

*re-instatement*of intuition. What's involved here isn't just probability; it's the very essence of the experience of being right or wrong. Wittgenstein said that there were never surprises in logic (not in logic

*qua*logic); but this is the sort of logical question in which the very premise involves a surprise. Yet the surprise is in having one's expectations momentarily suspended and then fulfilled. It reminds me of Nietzsche's remark on "Kant's joke," which was (he said) to demonstrate "in a way that would astound the common man," that the common man was right. Not so far from "bad reasons for what one believes on instinct," to come back to Bradley here at the end. But of course this "rightness" of intuition -- if one even grants it -- is itself puzzling. Which perhaps should highlight the puzzlingness of intuition

*per se*.

But your modification is a totally different experiment and the original version's calculation just doesn't carry over.

ReplyDeleteIn the original version Monty Hall has a clear method for choosing which door to open. If there's a goat behind your originally chosen door, he will always open the door with the other goat. If the grand price is behind your originally chosen door he opens one of the other two but never yours. But in your modified version he

can'tbe following that method forbothXavier and Yorick. That's easy to see in the case where Xavier and Yorick have the two goats, because then the rules of opening a goat door and not opening the door the contestant has chosen would contradict. So the question is what method Monty Hall is following now and the math and the simulation will both depend on that.In the simplest case his new method was to just open one of the two goat doors at random. So Xavier and Yorick both already survived a 1/3 chance of loosing on the first opening. This totally extinguishes the difference between the two remaining doors in the original version. For in the original version you know that your door wouldn't have been opened under any circumstances, while the opening of the other remaining door depended on what was behind it. And if you do the math, this changed version gives a 0.5 probability for the price being behind either of the remaining doors, both in the analytical calculation and in the simulation.

Of course the result can become different again if Monty Hall followed different rules again.

Gilbert,

ReplyDeleteThanks, this articulates something I was groping towards but having a hard time putting numbers to. Yes, the question cannot presume that any one player (X, Y, or Z) will be left standing after the first door is open. (I eliminated Z for convenience's sake, but I realize that extending the experiment into infinite iterations makes that illegitimate.)

Is it possible then that the counter-intuitive nature of the original puzzle arises because we are thinking that it is really

thispuzzle instead?