So,
some might be thinking, how on Earth did I get in this state with The Reverse
Monty Hall Problem?
Especially when, only two short weeks or so ago, I was a definite 2/3 answer
person. How did I manage to spend more
than two weeks convinced that the answer was 1/2?

It
started with balls. Well, it started
with a little thought experiment associated with the Anthropic Principle, but
it involved balls. There are some who
think that “Fine
Tuning” is an argument for the existence of a god. Even some closer to normality think that it is
something that ought to be explained (or explainable). I go along with the anthropic principle,
because I agree wholeheartedly that, given we are here to observe the universe,
the universe must be suitable for beings like us to observe it. In other words, the posterior likelihood that
the universe being suitable for beings like us to observe the universe is 1/1,
not 1 in whatever ridiculously large number some apologists come up with.

So, I
started thinking about balls. What, I
wondered, would someone say if, after a sequence of 999,999 black balls from a
barrel, having been taken at random one by one, I stopped them and asked them
to assess the likelihood of the last ball, the one millionth, being white. It seems ridiculously unlikely, right? This would mean that every single ball before
then had a chance of being a white ball, but they selected a black ball every
time. So, it would have to be 1 in a
very large number. Well, it’s not that
large, I thought. It’s one in a
million. Say that there’s a black ball
in the barrel, one black ball, and the balls are removed at random. It’s a sequence and, if the black ball’s
position in the sequence is random, that ball is equally likely to be in the
1,000,000

^{th}position as it is to be in the 456,978^{th}position, or the first, or any other specific position. So, it’s one in a million.
However,
I thought, that’s not quite right. Balls
aren’t limited to black and white. The
millionth ball could be a red ball, or a green ball, and so on. So, it’s more accurate to say that the
likelihood of that last ball not being white is one in a million. So, I came up with another scenario, which is
at Two
Balls One Urn. This
was a little too complicated, so I adapted it at Two
Balls One Urn, Revisited.
In this scenario, there are two million balls in a barrel which is in a
pitch black room. The balls are all
various shades, colours and patterns including two that are just white. Someone goes into the room with
an urn, picks a ball at random from the barrel and puts it in the urn. Then they pick another ball at random from
the barrel and puts that in the urn as well.
Then they pick a ball from the urn at random and hold it in their
hand. Finally they leave to the room,
they look at the ball in their hand and see that it is white. What is the likelihood that the ball in the
urn is also white?

It
works out to be 1/(n-1) = 1/1,999,999 where n=2,000,000 is the number of balls
in the urn.

One of
the commenters, Anonymous B, suggested doing this with n=3 and pointed out that
the answer in this case must be 2/3. In
my response I indicated that, at the time, I didn’t think that an n=3 version
of Two Balls One Urn and the Monty Hall Problem can’t be equivalent scenarios,
because one comes up with an answer of 1/(N-1)=1/2 and the other comes up with
2/3.

Then I
thought about it some more and cognitive dissonance kicked in.

So,
because I still find this difficult, I shall go through the Two Balls One Urn
with three balls, just to make it perfectly clear what I was doing:

I have an enormous barrel in a
pitch black room and I know that in the barrel there are three balls, two of
which are entirely white, the other being black. I take an urn into the
pitch black room and, completely at random, I take out two balls from the
barrel and place them in the urn. Because it is so dark in the pitch
black room, I cannot see either of the balls when I do this.

Then, while still in the pitch
black room, I reach into the urn and, completely at random, I draw
out one ball. Because it is so dark in the pitch black
room, I cannot see either of the balls when I do this.

Finally I walk out of the
pitch black room and I look at the ball I drew out of the urn. It is
white.

What is the probability that
the second ball - the one still in the urn - is white?

Now,
following precisely the same logic as when n=2,000,000, the answer should be
1/2. The white ball in my hand means
that either:

the first ball I took out of
the barrel was white, and when I took the second one out, I was randomly
selecting from two, one white and one black, or

the second ball I took out of
the barrel was white, and when I took the first one out, I must have taken one
at random from the other two, one of which was white with the other being
black.

Therefore,
the ball in the urn has a 1/2 likelihood of being white and a 1/2 likelihood of
being black.

Now,
if the white balls are goats, the black ball is the car, the barrel represents
the three doors, the urn represents the selected door, and the ball taken at
random from the urn is the opened door, then we have a scenario that is analogous
to a variation of the Monty Hall Problem.
And the answer in the Monty Hall Problem is widely accepted to be 2/3
rather than 1/2.

This was
a problem. I thought that it must be
wrong, and then I looked up the history of the Monty Hall Problem and saw that
prior to 1990, pretty much everyone who thought about it was convinced that the answer was
1/2 (despite the fact that Steve Selvin had
written a paper giving the correct answer as far back as 1975, a fact I
discovered only quite recently). I also
stumbled across the Monty Falls variation of the Monty Hall Problem in which
the answer is 1/2.

There
is clearly something strange going on.
So, as is my wont, I looked to see if there was a middle path. Is it possible that the answer is 2/3 in one
sense and 1/2 in another sense? I thought
that it was quite possible, and an answer came to me when I was thinking about
the white ball in my hand.

In my
scenario, I had "forced" the situation. In a real situation, it
wasn’t guaranteed that the white ball in my hand was going to be white. It could have been black. However, in my scenario it quite explicitly
states that the ball in my hand is white.

In the
Monty Hall Problem, in the treatments of it I had seen, this was not so
explicitly stated. There is plenty of
talk about how, before the door is opened, the host might have opened another
door, or the contestant might have selected different doors, or the goats and
car might have been arranged differently.
But, I reasoned, this is not the situation that the contestant finds
herself in once the door has been opened.
Once the door is opened, the goats and car have been placed, the doors
have been selected and the host has already opened the door. Therefore, I thought, this makes the Monty Hall
Problem, as posed by Craig F. Whittaker, analogous to me standing there with
one white ball in my hand.

Some
might argue that in my situation, the decision of the host is not taken into
account. However, I tried to be very
explicit in my wording of the Reverse Monty Hall Problem, the host is forced to
reveal a specific goat in those circumstances where by not doing so he would
reveal the car and otherwise the goat revealed is selected entirely at random.

This,
I thought, was directly analogous to my selection of the ball from the
urn. If there is a black ball left in
there, then the white ball I removed was the only white ball I could have
selected. If there is a white ball left in
the urn, then I could have equally likely have selected that one. I don’t know which situation I am in, of
course, but it is (or rather was) apparently equivalent to the situation that
the contestant is in a Reverse Monty Hall game.

And
then I wrote about it …

So,
now you know how I got to the point I was at when I posted The Reverse
Monty Hall Problem, being pretty much convinced of an answer that
the vast majority of mathematicians are convinced must be wrong.

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