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Tuesday, February 19, 2013

An Elaboration on the Monty Hall Problem

Many, many years ago I breezed into a Mensa meeting and saw then Chancellor of the Triple Nine Society, Cyd Bergdorf, and Ron Hoeflin working busily on some kind of problem.  I knew Cyd personally, but I knew Ron only by reputation.  Being the gadfly that I was, I walked over, sat down and asked what they were working on.  It was the now famous 'Monty Hall Problem.'  They explained it to me.  It goes like this:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

I immediately said, 'You switch, of course.  And, by the way, the qualification that the host knows what is behind the door is unnecessary.  I'm going to go get a drink.'  This remark that the qualification is not needed has led to a never ending dispute with other high IQ people, most recently Garth Zietsman and Rick Rosner.  So, I am going to put the argument here where it may be able to garner a larger audience.

To make it very clear, I am stating the the reservation is unnecessary precisely in the context of the problem as stated.  Thee are a whole lot of Bayesian assumptions that, I argue, do not apply because there is nothing in the problem that states that it should be taken as one is a series of similar games.

First, I will reproduce verbatim what I found to be the most clearly stated positions of Rick Rosner and Garth Zeitsman.  If you want to read the full dialogue, it is available here

Rick Rosner
Three doors, you pick one, then Monty randomly opens one. 1/3 chance the game is wrecked, because Monty prematurely revealed a car - he's never supposed to reveal the car until you make your final choice of door. 2/3 chance the game isn't wrecked. Two equal possibilities among the 2/3 of games that aren't wrecked - car is behind your first choice, or car is behind the door Monty didn't open. Random door-opening seems to either wreck games or leave an equal probability among the remaining unopened doors.

Garth Zeitsman
There is an urn with one blue and two green balls. A selects a ball followed by B (randomly) who notes his ball is green. If A had selected a blue ball B would have had two ways to select a green ball. If A had selected a green ball B would have had one way in which to select a green ball. However A had two ways in which to select a green and one way to select a blue ball. So the chance of A
having selected a blue given that B's was green is 1*2/(1*2 + 2*1) = 1/2.

When B can look into the urn and deliberately pick out a green then his probability of selecting green is always 1 and the probability of A selecting blue given that B has green is simply the initial probability of his selecting blue.

Michael Ferguson
Now both Rick and Garth are correct if we assume that the game described is one element in a string of games where n>>1.  However, that is not stated in the problem.  We have no reason to assume that this game has ever been played before or ever will be played again.  If that was the problem, they would be correct.  We could expect to improve our odds from one third to two thirds if Monty knows what is behind the doors and chooses to always open a door with a goat behind it.  We would expect that our odds would be the same by either staying or switching if Monty doesn't know what is behind the doors.  However, that is not the problem as stated.

When we flip a coin, the natural odds of a fair toss coming up heads is 1/2.  If we choose one door out of three the natural odds of choosing a car is 1/3.  One might ask how this natural odds of 1/3 could somehow change to 1/2.  The answer is that it does not.  Something very different is the cause.

Suppose we play 12 games.  We would expect that we will choose a door car 4 times.  That gives us the 4/12=1/3 natural odds.  If Monty always opens a goat door we didn't choose, we will have 8 unopened car doors and the odds if we switch is 8/12 or 2/3.  However, if Monty doesn't know, we will have 4 games where we chose a car door, 4 games where the unchosen door has a car and we will have four games where Monty, as Rick puts it, wrecks the game.  So, now, switch or not, we should expect to win 4 cars.

So the odds to not change from the natural 1/3 to 1/2 actually.  Rather, we have created a subset that doesn't include four unopened car doors.  This artificially decreases the probability that the unchosen door has a car.  In other words, and this is absolutely the crux of the issue, the lowering of the odds does not reside in the individual games but rather is a characteristic of the subset that is created.  The changing odds is directly a result of opening car doors.

In other words, by resorting to Bayesian reasoning, Garth and Rick are calling into existence games where car doors are opened when, in fact, there is no indication in the language of the problem that there ever will be played any other games, with or without Monty knowing what is behind the door.  In the only case of the game that we know to exist, Monty revealed a goat.

This game naturally belongs to the set of games string of n length in which no car doors are opened.  By naturally, I mean that there is no way to take this game out of that set.  If I could state that this game belongs to a set of n games in which Monty Hall knows what is behind the doors.  I could take it out of that set by stating that Monty Hall doesn't know what is behind the doors.  But it irrevocably belongs to this set.

I then state that a common factor of all game strings, regardless of the value of n, is that we expect that there are twice as many cars behind the unchosen door than there are between the chosen door.  To illustrate the significance I use the following thought problem.

Suppose we play twelve simultaneous games. We choose one door in each of twelve sets of three doors. Our expectation is that we will choose a car four times and will choose a goat the other eight times. Then Monty opens one door in each set of three doors and reveals a goat. Now, we have chosen a car door four times and the twelve unchosen doors contain the other eight cars. Clearly, we should switch doors when asked.

Now after all this transpires, Monty tells us that he didn't know what was behind all the doors. We are surprised because the odds against him choosing only goats in twelve straight games is (2/3)^12 or approximately 130:1. Yet, it does not change the proper strategy; there are still only four cars behind the chosen doors and eight behind the unchosen doors.

Now suppose that we play three simultaneous games. The logic is the same. Our chosen door has only one car and the unchosen door has two. We should switch. We are not so surprised when Monty tells us that he didn't know what was behind the doors because the odds against have fallen to (2/3)^3 or just a little over 3:1. In fact, for n games in which no car doors are opened, no matter the value of n, switching doubles our chances.

This happens because there is nothing magical about Monty knowing or not. The advantage falls from 2:1 to 1:1 by the process of games with opened car doors being eliminated from the pool. In other words, the odds do not change one iota UNTIL a car door is opened. When we are presented one game in which a goat was revealed and n=1, there are no car door openings that can change the odds of having chosen a car door from 1/3 to 1/2. Consequently, in one game in which a goat was revealed, my initial probability of 1/3 of having chosen a car door remains. The unchosen door still contains 2/3 of the probability.

For some reason, even people at the highest IQ level can have difficulty in grasping this.   I want to make this clear that I am not arguing against Bayesian probability.  I am only arguing that it is applicable only in those cases where multiple 'runs' of the game are taking place.