Why are the odds now 2 in 3 if we change doors and only 1 in 3 if we
don't? There are only TWO doors. Where does the other door come from?-הסתר טקסט מצוטט-

-הר�ה טקסט מצוטט-

In the Monty Hall Problem, we are shown three identical doors, one of
which conceals a car and two of which conceal a goat. When we chose
any one of the doors our chance of winning the car is 1 in 3.

After we have chosen a door, game show host Monty Hall opens one of
the doors to reveal a goat and asks us if we would like to change our
choice of doors. We are told that we will increase our odds of
choosing the car to 2 in 3 if we change our selection.

The easy solution lies in the rules that Monty Hall (MH) must follow
when he opens the door: obviously he cannot open the door we have
selected and he cannot open the door concealing the car. Therefore:

(1) If we have chosen the door to goat 1, MH must open the door to
goat 2.

(2) If we have chosen the door to goat 2, MH must open the door to
goat 1.

(3) If we have chosen the door to the car, MH must open the door to
either goat 1 or goat 2.

Thus, in the first two cases, we will choose the car by changing our
selection, and in the third case we will chose a goat. In other words,
our odds of winning the car are now 2 in 3 if we change doors and only
1 in 3 if we don't.

Gene Ledbetter

Why are the odds now 2 in 3 if we change doors and only 1 in 3 if we
don't? There are only TWO doors. Where does the other door come from?-הסתר טקסט מצוטט-

-הר�ה טקסט מצוטט-

maybe the following could help:
in the first choice, there is a probability of 1/3 that the door with
the prize was chosen, and a probability of 2/3 that the prize is in
one of the other doors.
Now the host opens one of the other doors, but it did not change the
probability whatsoever,
so there is still a probability of 2/3 that the prize is in one of
the 2 doors that were not chosen, which now means it's in the last
door.

A Fuller Explanation of the Monty Hall Problem

It is given that game show host Monty Hall (MH) knows which of three
doors conceals a car, a goat, and another goat. The problem would
otherwise not be a problem.

The problem consists of three events. Firstly, we are asked to guess
which door conceals the car. With no special knowledge, we have one
chance in three of guessing the right door.

Secondly, MH opens one of the doors, revealing a goat. He then asks if
we would like to change our guess. Authorities inform us that by
changing our guess we can increase our odds of getting the car to two
in three.

Thirdly, we must choose between the two remaining closed doors, and if
we are reasonable we will be outraged by the claim that we can
increase our odds by changing our choice. In the absence of any
special knowledge, we now have one chance in two no matter which door
we choose.

But we do have some special knowledge, as we shall see when we think
about what MH did during the second event. In order to keep the game
going, MH had to open a door that concealed a goat and he could not
open the door we had chosen. If he had revealed the car or the object
behind the door we had chosen, it would be absurd to ask if we would
like to change our guess.

When we made our choice of doors during the first event, we had two
chances in three of choosing a goat. And if our choice was a goat, MH
would have to open the door to the other goat. In other words, in two
chances out of three our choice of doors would force MH to open the
only other door that did not conceal the car. Thus, in two chances out
of three, MH would be forced to provide the door that conceals the car
as the “other” remaining door.

Therefore, during the third event we know now that we are not choosing
between two doors at random, as we first thought. There are two
chances in three that the “other” remaining door conceals the car. So
we have two chances in three of choosing the car if we change our
choice of doors.

Gene Ledbetter

We don't have special knowledge because we know the goal-posts were
altered. We simply enter a new puzzle with a new set of guidelines and
rules.

J Jones <jonescardiff@aol.com> writes:

We don't have special knowledge because we know the goal-posts were
altered. We simply enter a new puzzle with a new set of guidelines and
rules.

Right.

Consider the following situation. There are 100 doors. Behind one
door is the prize and behind the other 99 doors are goats. You select
a door and hope the prize is behind it.

Of course, your chance of being correct at this point is 1/99.

Now, I tell you that I will open up 98 of the doors that have goats
behind them and then I do so. There are two doors left. Do you
really believe that you have a 50/50 chance of having picked the right
door at the start?

In any case, there's no sense in arguing about it. Try it for
yourself at http://math.ucsd.edu/~crypto/Monty/monty.html and see
whether not switching is as good a strategy as switching.

J Jones <jonescard...@aol.com> writes:

If I reveal one of a hundred, and you reveal 98 of a hundred, and if the
puzzle is dependent on the conditions of revelation then, again, we have
two entirely different puzzles - even if they are the same goats and the
same door-knobs.

No idea what you're talking about or what you mean by "dependent on
the conditions of revelation", but you can check the simulation I
mentioned and see if the odds are 1/2 or 2/3 when you switch.

Jesse F. Hughes wrote:
J Jones <jonescardiff@aol.com> writes:

We don't have special knowledge because we know the goal-posts were
altered. We simply enter a new puzzle with a new set of guidelines and
rules.

Right.

Consider the following situation. There are 100 doors. Behind one
door is the prize and behind the other 99 doors are goats. You select
a door and hope the prize is behind it.

Of course, your chance of being correct at this point is 1/99.

Now, I tell you that I will open up 98 of the doors that have goats
behind them and then I do so. There are two doors left. Do you
really believe that you have a 50/50 chance of having picked the right
door at the start?

In any case, there's no sense in arguing about it. Try it for
yourself at http://math.ucsd.edu/~crypto/Monty/monty.html and see
whether not switching is as good a strategy as switching.


If I reveal one of a hundred, and you reveal 98 of a hundred, and if the
puzzle is dependent on the conditions of revelation then, again, we have
two entirely different puzzles - even if they are the same goats and the
same door-knobs.