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:
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.
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?
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.