As
pointed out by KaySen,
there is a treatment of Jeff Rosenthal’s Monty Fall
scenario by Christopher Pynes which suggests, amongst other things, that in a Monty Fall
scenario the correct answer is still 2/3.
Interestingly enough, Pynes is making precisely the kind of error that I
was accusing everyone of making when I was first defending my arguments first
raised in The Reverse Monty Hall Problem – that is by treating a single instance,
one shot game as if it were part of a series.
I am
now persuaded that the intent of this argument was misguided in terms of both
the Monty Hall Problem and the Reverse Monty Hall Problem, but I consider it to
be entirely valid with respect to the Monty Fall scenario.
The
whole concept that Rosenthal is trying to convey is that the host accidentally
falls against a door, thereby revealing a goat, rather than deliberately
selecting a door to open. This entails
that the accident is a one-off. There is
no implication that the host repeatedly slips and falls against a door, and
that in each case a goat is revealed – as implied by Pynes:
In cases where the contestant
picks the door with the prize behind it in the first place, Monty never reveals
it, but reveals another door: in these cases one wins by staying and loses by switching—1/3
and 2/3 respectively. In the case where the contestant doesn’t initially pick the
prize door, Monty’s fall reveals the only door it can that isn’t the prize—win
2/3 of the time by switching. This makes Monty Fall and Monty Hall the same in
all instances; they are logically equivalent, and thus must have the same
probabilities.
There
is no “never”, there are no other instances required to make “all instances” a
meaningful concept. It happens once,
whoops, and it’s entirely possible that it never ever happens again. Pynes, I believe, is the one being seduced by
his familiarity with the Monty Hall Problem.
There
is, however, a way in which Monty Fall can be saved in so much as it can be
reliably repeated.
---
Monty
de Sade was continuing to cast about for a more devious variant of the three
doors, two goats and a car game. He
heard about the use of coins to discuss the Reverse Monty Hall problem and a
new idea formed in his evil little mind.
His
new game goes like this:
There are three doors, with
two goats and a car distributed behind them at random just like before.
The contestant is offered one
door to choose from, just like the classic Monty Hall Problem.
Monty then offers a more
complex continuation. If the contestant
agrees to proceed, Monty will toss a coin and then open one of the unselected
doors on the basis of the result. If the
car is revealed, the contestant will automatically lose. If the car is not revealed, then the
contestant will be offered the opportunity to switch doors.
What is the likelihood of
benefitting from a switch?
There
are three decisions here on the part of the contestant:
The selection of a door, which
we take to be random
Whether to risk losing the car
on the basis of a coin toss
Whether to switch
As far
as I can tell, noting that I have been wrong before, there is a 1/3 likelihood
of selecting the car from the outset.
This means that there is a 2/3 likelihood of the car being behind one of
the two unselected doors. However, when
Monty tosses a coin, there is 1/2 likelihood of revealing the car, if the
car is behind one of those doors, and a 0/1 likelihood of revealing it
if the car is behind the selected door.
This
means there is a 1/3 likelihood of having the car already and a 1/3 likelihood
that the contestant will lose the car in the coin toss and a 1/3 likelihood
that the car won’t be lost in the coin toss, but will be behind the one
remaining unselected and unopened door.
This
means that the likelihood of benefitting as a result of switching, after
surviving the coin toss, would be equal to the likelihood of
benefitting as a result of not switching.
In which case, there was no point in going through the charade, was
there?
This
seems counter-intuitive. It seems to me
that if the contestant bothered to go ahead with the coin toss then, once a
goat is revealed, if a goat is revealed, she is essentially
committed to switching. By why, if it’s
50-50? Some might suggest that it’s not
50-50, but the probability tree below indicates that it is – unless I’ve made
an error somewhere.
That
said, there are two perspectives one could take, and they seem to have equal
weight:
Perspective
One:
At the beginning the
contestant picks one door, the likelihood of having selected the car is 1/3, so
there is a 2/3 likelihood of not having selected the car. If she has the car, then she loses nothing by
having Monty toss the coin and open a door.
If she doesn’t have the car, then she gains nothing by sticking with her
selection. So she should proceed.
Then Monty tosses the coin and,
phew, doesn’t reveal the car.
Given that there was 1/3
likelihood of the car being behind the selected door and a 2/3 likelihood of it
being behind one of the other two doors, it’s more likely that the car is
behind the remaining unselected and unopened door and the contestant should
therefore switch.
Perspective
Two:
The contestant selects a
door. If she has selected the door with
the car, then she loses nothing by having Monty toss the coin and open a
door. If she hasn’t selected the car,
then she gains nothing by sticking with her selection. So she should proceed.
Monty tosses the coin and,
phew, reveals a goat. The likelihood of
Monty revealing a goat when there are two goats behind the unselected doors is
twice that of revealing a goat when there is only one goat that could be
revealed.
Given that Monty revealed a
goat, it’s more likely that the car is behind the selected door and the
contestant should therefore stay.
So, I
throw the question out there: what is the best strategy for the
contestant (not to switch or stay, but to take the coin toss or not)?
Note
that this time I have no firm answer of my own beyond a gut feeling that the rational
contestant should risk a coin toss to make use of the 2/3 likelihood of not
having selected the car and switch if a goat is not revealed, having dodged an
automatic loss entirely by chance. I do recognise
that the probability calculations don’t seem to support this gut feeling but perhaps
someone has some better probability calculations.
---
And
how, precisely, does this save Monty Fall?
If my 50-50 solution is correct, then it is possible to run the scenario
over and over again, as many times as you like, and then extract one iteration of
the game in which Monty reveals a goat.
This then represents the single instance involved in the Monty Fall scenario,
it’s just that the “luck” is now based on a coin toss rather than a humorously
placed banana peel.
