"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?"The problem is trivial and I said immediately, 'You switch, of course'. However, then I added, 'it is irrelevant whether the host knows what is behind the doors'. This is a statement that has not been agreed with by some of the smartest people in the world. In a private message with Garth Zietsman and Rick Rosner, I found I could convince neither of them of my assertion. In a way that is good, because it required me to work very hard at creating a proof for what I saw as obvious, but difficult to express convincingly.
The argument for its necessity goes like this. If the host knows what is behind each door (and as Rick correctly pointed out wasn't inclined to thwart your attempt to win a car), by opening a door with a goat behind it, the host has inserted information. Essentially he has collapsed the Bayesian prior and changed the probability to 50% that there is a car behind the unchosen and unopened door. The Wikipedia article does a pretty good job of explaining this if you aren't unfamiliar with the problem.
If the problem said that you were going to play the game 100 times and asked what strategy would be best, then whether the host knew what was behind the doors would be relevant. Either, in that sequence of 100 games, the host would or would not open doors with cars behind them based upon whether he knew what was behind the doors and was inclined to open only goat doors. Again, Rick made the very correct point that it matters what was done in the cases when the host did open a car door, if we are in the scenario where that can happen.
But, of course, that is not the problem as stated. One is presented with a unique event and, as such, the problem surfaces one of the peculiarities of probability. I recognized this when I was about 8 and I started asking people, 'what is the probability that, if I flip a coin, it will land on heads?' Everyone, of course, answered 50%. I then replied, 'It is not. It is either 0% or 100%'. To my amazement, nobody 'got it'. In essence, you really can't average one event. The Monty Hall problem is of that type. It is played once.
PROOF:
Consider two sets. One is a set of n games where the host always opens a goat door. In the other set of n games, the host opens a car door 1/3 of the time. In the first set of n games, switching will win a car 1/2 of the time and staying with the door first chosen will win a car 1/3 of the time. One should switch. In the second set, one will win a car 1/3 of the time if one switches or if one stays with the door first chosen.
Here is the key. The result is not dependent upon the value of n. If each set contains 100 games, it will be 1/3 and 1/2 for the goat door only set and 1/2 and 1/2 for the goat and car door set. If n=200, it will be the same. If it is 50 it is the same. In the first set, one should switch and in the second, it doesn't matter.
Also, we will note that in the first set, it doesn't matter if the host is opening only goat doors by happenstance or by artifice. One can argue, convincingly, that if the host opens only goat doors 100 times and never opens a car door, he almost certainly knows what is behind the doors. However, whether he knows or just got insanely lucky doesn't matter. The answer is based upon the elements in the set, not why the elements are as they are.
In the problem, as stated, n=1 and it belongs to the set where only goat doors are opened. In that set, one should switch doors. As we see, it doesn't matter whether a goat door was opened because of chance or because the host knew it was a goat door.
I invite questions or refutations.
So I have read this a few times, and it's an interesting question and series of points. The point about the quarter is correct, if one is only flipping the coin once then it can't come up 1/2 heads and 1/2 tails so the probability has to be 0% or 100%, but of course we don't know which it will be if it's a fair coin flip. So when we say 50% we are acknowledging that we cannot predict the future so it is also a correct answer, in my opinion.
ReplyDeleteNow to the doors. This one is a bit more difficult. I agree with your assertion as to the opening of the car door 1/3 of the time, in that case the guesser will choose the car 1/3 of the time. However, I disagree with the assertion that always opening a goat door changes the odds. Look at it this way with a slightly different scenario.
Let's say the host showed me three doors and said pick one and try to pick the car. I answer no, that's too hard. Instead of sending me on my way, the host says ok, I'll help you out and reveals one of the goats and says now you have only two to pick from, pick one and try to pick the car. I think you would agree that I will get it 50% of the time, given that this is series of events and not the 0% or 100% scenario.
So, how is it any different if I preselect the door I want?
Looking at it another way, I pick door 1, Monty reveals a goat through door 3 and says ok, the car is either behind door 1 or door 2. This is a fact, it's either in door 1 or door 2. Monty shows no bias, I play this game 100 times and every time he says and does the same things, and the correct door is random and fair just like the coin flip. That means half the time it's behind the door I've already picked and half the time it isn't. So if I switch every time, I have the same odds as if I stand pat every time.
I enjoyed this exercise and perhaps you have thoughts on my assertion here.
You are incorrect. However, this article is not about the overall Monty Hall problem. That is not a matter of controversy among mathematicians. The result is based upon Bayesian probability and the short explanation is that by opening a goat door, the host is collapsing the Bayesian prior that caused the 1/3 probability of selecting a car.
ReplyDeleteThis article is about why the proviso that the host knows what is behind the doors is unnecessary. That is a harder problem. It has to do with whether this problem is better assessed with Bayesian probabilty or set theory. Because, in the problem, the Bayesian prior has collapsed, I do not believe that it is appropriate to consider the problem as stated by considering it before the Bayesian prior has collapsed. Using set theory is novel, I suppose, but I think it provides better understanding.