Monty Hall Bsuinss

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Post Post #25  (ISO)  » Wed Jan 16, 2019 6:32 am

Monty knows where the car is, I mean, and he uses the information in deciding what door to open. I guess the problem isn't exactly misleading, but many popular explanations are.

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Post Post #26  (ISO)  » Wed Jan 16, 2019 6:35 am

Okay I think I've come back to my basic issue with this problem.

If I don't pick a door to begin with what are the odds for getting a car when picking one of the remaining two doors, after Monty picks a door?

How does my initial decision influence the outcome?

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Post Post #27  (ISO)  » Wed Jan 16, 2019 7:02 am

Your initial decision forces Monty to not open the door you picked. This may seem obvious, but it's crucial.
If you pick first, then there is a 2/3 chance that you pick a goat door. All of a sudden, he is forced to open the other goat door; he has no choice in the matter. By picking a goat door, you are basically forcing him to tell you where the car is.
1/3 of the time, you pick the car and he chooses which door to open. Similarly, if Monty opens a door first, he always has a choice. In these cases you technically gain no information about the remaining doors.

Incidentally, imagine if when you pick a door Monty was allowed to open your door, to show you it's a goat. You would know for a fact that you want to 100% switch doors, but not which door to switch to.
You don't have ambiguity; you have options.

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Post Post #28  (ISO)  » Wed Jan 16, 2019 7:55 am

Hmmnyeees. Monty is constrained not only by -not- opening the door I've picked, but *also* by not opening the car door.

Spoiler: "choice tree (image)"

Monty's choice is always narrowed down to two possibilities - a random pick or an instruction. 66% of the time he gets an instruction.
Monty has no choice if "we" need (as a supervisor controlling the rules of the game) to tell him to pick door notCar(B) or notCar(C). These are equivalent (non)choices as far as the original pick is concerned.
He can pick notCar(A[1] A[2]) if I as the player pick the car.

The information he actually gets is whether my choice was the car, and if it wasn't, he gets an instruction to pick notCar(B) or notCar(C)
If I picked the car, his choice means nothing.
If I didn't pick the car, he reveals an option which is not the car, which to me is equally likely to be the car or not.

The information I get is that he's picked a door that I didn't choose which is a goat.

There has to be a very plain statistical or quantum calculation for this if it's mechanical; I can see that his choice is narrowed by me, and I can see that my second choice is constrained by him, but I'm struggling to understand the difference between Car(B) and Car(C) because in one state I choose B and in the other I choose C, and both seem quite equal.
Last edited by talah on Fri Jan 18, 2019 5:46 am, edited 2 times in total.

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Post Post #29  (ISO)  » Wed Jan 16, 2019 10:39 am

I like the explanation where there are not 3 doors, but 100, and monty is forced to open 98 doors. Now it's much clearer that you should switch

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Post Post #30  (ISO)  » Fri Jan 18, 2019 5:41 am

Hey. Stop trolling :P
(I actually liked that explanation too and discussed it with several of my friends a couple of years ago, coming to the conclusion that is was an extremely solid idea of how the statistics "sorta" worked. Now I'm interested again but in more specific terms.)

Spoilered the random too-large image
No closer to finding a specific mathematical equation on this | (someone else has done the work) (x)
No closer to shimmying up some actual code myself explaining the logical choice tree rather than the rationalisations | (I didn't do the work) (true)

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Post Post #31  (ISO)  » Mon Jan 21, 2019 9:43 am

My thing is, why doesn't Monty want the goat?

I would stay, higher chance of getting a goat.
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Also all of my limbs fell off, I must of tripped oh no, hope I don't fall in a bowl of boiling water, then I would drown.

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