Monty Hall, except Omega replaced the car with a million dollars if it predicts you won’t switch doors
then you open the door and BOOM it’s a fuckin’ trolley
Omega Hall Trolley Problem
Before the Problem begins, Omega uses godlike predictive power to find out whether you are a Switcher or a Stayer.
Omega presents you with three doors.
Two of the doors, if picked, will cause a trolley is to run over five people.
Behind the third door, there is another trolley.
-If Omega predicted you are a Switcher, it will run over ten people.
-If Omega predicted you are a Stayer, it will run over one person.
Having picked a door, Omega now reveals that one of the doors you didn’t pick would have been a Five-Person door.
Should you switch or stay?
this is horrible and I love it
switch: (-10 * 2/3) + (-5 * 1/3) = -8.333…
stay: (-5 * 2/3) + (-1 * 1/3) = -3.666…
Somebody fell into my trap!
In this variation it is 50%, not 2/3rds.
…rereading the problem, i think it is unclear. could you restate the problem with particular care for your pronouns and antecedents?
and then, could you explain how it is not 2/3rds, because i’m not even sure that’s where my confusion is. i don’t see it.
You got the 2/3rds from the Monty Hall problem (MHP).
This, despite looking an awful lot like the MHP, is not the MHP
I’ll remove the superfluous trolley/omega bits and write the not-MHP down.
- The Nonty Hall Problem
- On a game show, a contestant was offered a choice of three doors. Behind one door was a car, behind the other two were goats.
- After the contestant had selected a door, Nonty opened a different door, revealing a goat. Nonty offered the contestant a chance to change his selected door to the other, unopened, door.
- Would a door switch have improved the contestants chances of selecting the door behind which a car was hidden?
This is not the Monty Hall problem. In the Monty Hall problem, the answer is “Yes, from 1/3rds to 2/3rds.” In the Nonty Hall problem, the answer is “No, there is a 50% chance of receiving a car no matter which of the two un-opened doors that is selected.”
what makes the difference between ½ and 2/3 is whether the door that the host opens is selected randomly. if the host opens the door randomly, no information is given and there is nothing to condition. if the host always selects the goat over the car, that gives information to condition on.
this does not specify the host’s selection method, and being shown a goat is evidence that biases the credence of his selection method to be one that prefers goats. i can’t do the math here right now, and i might be wrong, but lacking information of the selection process is not the same as knowing the host always chooses randomly, so i will not assume that the resulting probability is the same. this is an interesting problem to consider. if you want to do the math for this, go right ahead.
ironically, in your statement of this problem (particularly with the ambiguity of the host’s selection process), you are transforming your NMHP into the MHP. you specify in the rules that the door the host opens contains a goat. to this effect, you are acting as a meta-host, who is conditioning the problem itself such that the car will never be selected.
i will call this interesting result the Meta-Monty Hall Problem. within the context given in the problem, the player does not have this information, and cannot assume the car would never be revealed. outside the context of the problem, the hidden rule itself gives information to the scenario, removing the possibility that the car will be selected. anthropically, the player who sees a car behind the opened door does not exist.
with specific respect to credence, no, the player should decide based on the results of the host-informationless problem specified in my first two paragraphs. however, because of this anthropic conditioning you accidentally baked into the problem, the actual chances of selecting the car still increase by switching, and we as outside observers know that the player should switch upon being given this anthropic information.
so. two very interesting results came out of this. neither of us was right! i was conditioning on meta-information, and so obtained the EVs for a problem the player was not knowledgeable of, and you did not give the player proper information of the host’s strategy!
if you know for a fact that the host-informationless problem (conditioning credence of the host’s strategy given a goat is revealed) results in the same credence as the host-random problem, ½, please let me know. i might do it myself, but i can’t for the time being. otherwise, this was fun.
Very very close.
First: You caught the exact important difference: The hosts selection mechanism is not explained. Since probabilities are in the mind, “unknown” is equivalent to “perfectly random” – we do not know Nonty’s algorithm for picking doors – or even if he has one – we just know that right now, at this moment, one of the other doors is open and has a goat behind it.
Second: I have done the math. It is 50% after Nonty has revealed a goat. Your mixup is near the very top:
if the host opens the door randomly, no information is given and there is nothing to condition. if the host always selects the goat over the car, that gives information to condition on.
In the original MHP, you gain no new information to condition on, therefore you cannot update, therefore you keep your prior odds that you picked a car. Formally, if P(Car)=1/3 and P(Car|GoatReveal) = 1/3, then the GoatReveal event must have given you zero bits of evidence.
It is exactly because you already know Monty’s algorithm that you gain no new information about your door – you know before he opens the door that he will open a goat door. Therefore opening a goat door is nothing new.
Nonty, on the other hand, gives you information by opening a goat door. New information allows you to update from “there are two chances in three that I have a goat” to “there is one chance in two that I have a goat.”
If you’re red-green color blind, the chart shows all nine combinations of your choice and Nonty’s random pick, Three are removed (Monty did not reveal a car), two more are removed (Monty did not reveal a goat that you had selected) leaving four possibilities. In two, you stand at a car door, in two others you stand at a goat door.
My math has been impugned and so:
P© The probability of selecting car on the first try. 1/3
P(G) The probability of Nonty revealing a goat. 2/3
Since P© and P(G) are independent:
P(G|C) The probability of Nonty revealing a goat, given that I have selected a car on the first try. 2/3
P(C|G) The probability of having selected a car on the first try, given that Nonty has revealed a goat:
P(C|G) = P(G|C)*P©/P(G) = (2/3)*(1/3)/(2/3) = 1/3.
But notice that this is not the problem as stated: This is Nonty opening a door at random and getting a goat. It is NOT Nonty opening a door at random, picking a different door from mine, and revealing a goat.
P© The probability of selecting car on the first try. Still 1/3
P(G) The odds of opening a different door from mine P(Different) = 2/3 and the odds of finding a goat behind a random door P(Finding) = 2/3, so P(G) = P(D and F) = 2/3 * 2/3 = 4/9.
P(G|C) The probability of Nonty revealing a goat behind a different door, given that I have selected a car on the first try.
Since I have the car, the odds of finding a goat if you don’t open my car door are 100%, so P(D and F) = 2/3*1 = 2/3.
P(C|G) The probability of having selected a car on the first try, given that Nonty has revealed a goat behind a different door:
P(C|G) = P(G|C)*P©/P(G) = (2/3)*(1/3)/(4/9) = 18/36 = ½.