Benutzer:Gerhardvalentin/m

Effects of switching, all possible configurations

Supposition

The Guest, having no knowledge of the location of the Car, selects one of three doors.

There are two Goats, but there is only one Car. So the non-selected pair of two doors inevitably hides at least one Goat.

The Host, knowing the location of the Car, will always open one of those two non-selected doors to show you a Goat.
This event occurs every time, not only if the Guest has selected the Car.

The Host, if he has two Goats to choose from, will not be opening one special door "above average" nor "whenever he can".

Car
behind
door
Guest
selects
door
The unselected
pair of two doors
hides always
at least one Goat

but in 2/3
also the Car
Loss by switching only in 1/3
(only in 3 of 9 cases):
Only if, by chance = 1/3,
the original choice was the only Car

Switching
wins in 2/3

in 6 of 9 cases:
Always if one
of the two doors
hiding
a Goat was
selected
Host opens a
door showing
a Goat.
If he has got
two Goats:
in 50 % door
X
and
in 50% door
Y
Effect of swapping to the second still closed Host's door
that he always must offer as an alternative:
1 1 GoatGoat Loss, no matter whether Host opens door 2 or door 3 2  or  3  Car door 1 was selected, switching hurts.
1 2 GoatCar   Prize 3  Goat door 2 was selected, switching to door 1 will win the prize.
1 3 GoatCar   Prize 2  Goat door 3 was selected, switching to door 1 will win the prize.
2 1 GoatCar   Prize 3  Goat door 1 was selected, switching to door 2 will win the prize.
2 2 GoatGoat Loss,  no matter whether Host opens door 1 or door 3 1  or  3  Car door 2 was selected, switching hurts.
2 3 GoatCar   Prize 1  Goat door 3 was selected, switching to door 2 will win the prize
3 1 GoatCar   Prize 2  Goat door 1 was selected, switching to door 3 will win the prize
3 2 GoatCar   Prize 1  Goat door 2 was selected, switching to door 3 will win the prize
3 3 GoatGoat Loss, no matter whether Host opens door 1 or door 2 1  or  2  Car door 3 was selected, switching hurts.
Chance
to win:
${\displaystyle {\tfrac {1}{3}}}$ ${\displaystyle 0}$           ${\displaystyle {\tfrac {2}{3}}}$ Distribution of chances 1/3 : 0 : 2/3 is in effect from the
start until the end of the game, according to the rules
Chance to win the Car by swapping to the other door offered is not 1/2 : 1/2.
By staying I will win the Car only in 1/3, if – by chance – I selected the Car.
Risk
of loss:
${\displaystyle {\tfrac {2}{3}}}$ ${\displaystyle 1}$           ${\displaystyle {\tfrac {1}{3}}}$ Distribution of risk  2/3 : 1 : 1/3 is in effect from the
start until the end of the game, according to the rules
Swapping to the other door offered doubles my chance on the Car from 1/3 to 2/3,
it will give me the Car for sure if – in 2/3 – I have selected one of the two Goats.
Regardless of what door the Guest selected and what door was opened by the Host.
The standard scenario of the paradox forever firmly excludes any closer information.

Someone asks you the following question:

If, not knowing about the location of the Car, you pick one of these three doors, for example let's say #1, and then the Host, who knows what's behind the doors, opens another door, for example let's say #3, which has a Goat. He then says to you, "Do you want to pick door #2?" –  Is it to your advantage to switch your choice?

Quite easy, your question says "picked #1, opened #3, offered #2".  –  Just have a look on the table above.
When having picked one door, then my risk to have picked a Goat will be 2/3, and my chance to have picked the Car 1/3 only.
The unselected pair of two doors however will hide at least 1 Goat, but it will hide the Car in considerable 2/3. So my probability to win by switching will rise from 1/3 to 2/3.
And now to the exact door numbers you asked for in your question ("picked #1, opened #3, offered #2"):
See line 1 (1/9 of all possible constellations): By picking door #1,  I will have picked the Car, and doors #2 as well as #3 both do hide Goats.
In only 1/2 the Host will have opened door #3 to show the Goat. That's in 1/18 in this subset of all possible constellations that I will loose the Car f I selected door #1 and the Host opened door #3.
And then see line 4 (1/9 of all possible constellations): I will have picked a Goat, the Car being behind the offered door #2, so the Host cannot open door #2 but was forced to open door #3, and then I will win with double chance of 2/18 (1/9) in this subset of all possible constellations if I selected door #1 and the Host opened door #3.

And yes, it's to my advantage to switch my choice from door 1 to door 2, because my chances will double from 1/18 to 2/18 in this subset of all possible constellations. Quite a simple solution. And as to your question ("picked #1, opened #3, offered #2"): The probability to win by switching will double and rise from 1/18 to 2/18 in this subset of all possible constellations, you see. So, independent of door numbers, it doubles from 1/3 to 2/3 in any case.

But please note: This scenario is exactly about the world famous paradox only, and not about quite a lot of other different scenarios where the Host, e.g. not knowing the actual location of the Car, just randomly opens one of his two doors that hides a goat and by chance just not the Car, and by that devastated one full 1/3 of the chance on the Car, showing a goat only in a subset of 2/3 of all possible constellations. And by that reduced the chance to win the Car by swapping to the other door to only 1/2. Nor is it about other imaginable scenarios quite outside the world famous Monty Hall paradox.