Monty Hall Bsuinss

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Monty Hall Bsuinss

Post Post #0 (isolation #0) » Fri Jan 11, 2019 3:53 am

Post by talah »

So I get it, probabilities determine an ultimate choice being better or worse.

But the question I have is, why not have a predetermined switch then, and why wouldn't that always give you a better outcome?

So my algorithm for 1 of 3 choices, is pick, have monty reveal a dud, then automatically switch.

Why is or isn't that a thing? The solution to the problem is elegant but doesn't explain why determinism does not equal choice.
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Post Post #2 (isolation #1) » Fri Jan 11, 2019 5:00 am

Post by talah »

but then i could predetermine my switch always, guaranteeing that my initial random choice was bad
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Post Post #4 (isolation #2) » Fri Jan 11, 2019 5:17 am

Post by talah »

choose anything, it's guaranteed to be bad, because switching is preferable.
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Post Post #6 (isolation #3) » Fri Jan 11, 2019 5:31 am

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Implying that the initial choice had some kind of effect on the result - in that the purely random initial choice is guaranteed to be worse than swapping the choice afterward - even though Monty is guaranteed (and indeed forced) to reveal a goat.
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Post Post #8 (isolation #4) » Fri Jan 11, 2019 5:45 am

Post by talah »

Never mind I just rationalised it.
The result is 66% success for swapping purely because you're wrong 66% of the time.
The mindfuck is that you think you have a choice that changes anything.
When in reality it was always 33/66.
Got it.
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Post Post #10 (isolation #5) » Sat Jan 12, 2019 4:06 am

Post by talah »

Not really, because I'm trying to understand the mechanic behind switching necessarily being a better choice than not switching. It seems to me that there should be a reason that one is better than another (switching, not switching) which is a bit more eloquent than pure statistical result. This doesn't seem a superposition, rather a result which on the face of it is unexpected given expectations. Nothing changes mechanically based on the decision to switch or not. Statistically results can be given saying that switching is "better" than not switching, apparently meaning that the decision which has nothing to do with the result has an unexpected effect.

However it seems that the switching or not switching action does nothing and is a statistical artifact, because the result is in fact in line with your initial chance to guess correctly anyway.
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Post Post #13 (isolation #6) » Sun Jan 13, 2019 11:16 pm

Post by talah »

That absolutely makes sense in a quantum superposition, if you have knowledge of the superposition (which is your magic lightning, and is fine, I acknowledge that this is potentially possible in several senses including in the sense that we know the mechanic and it's a physical property of the plane we exist upon). The glazed over part, then, that I'm trying to find meaning for, is the WIFOM decision (he has 2 goat choices he can reveal) vs the Known decision (he's forced to reveal the only remaining goat) from Monty.

So some combination of myself deciding to switch and Monty knowing something is the key of the matter. Or is it only one of the two decisions/observations which is important?
Do we share something by decision and knowledge? It certainly doesn't seem that I have any pre-information which would make a switch important, and that Monty doesn't pass me any information by revealing -a goat- because he was going to either way.

Ed: (As an addendum - is it really just that we get 2 choices and taking the same choice twice amounts to only one choice?) /afterthought
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Post Post #15 (isolation #7) » Mon Jan 14, 2019 9:49 pm

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He either has a choice or he doesn't - either way, he opens a goat door (car/goat/goat - I only pick one, so there is always a goat door for Monty to open). That conveys no information to me except that he opened the door that he did.

This is actually quite interesting. I guess it is a superposition. There should be a calculation around if it is.

--
Ed: reading this, not sure if it's even correct, but it's in Times New Roman and starts like this:
:P
In the classical Monty Hall game the banker (“Alice”) secretly selects one door of three behind which to place a prize. The player (“Bob”) picks a door. Alice then opens a different door showing that the prize is not behind it. Bob now has the option of sticking with his current selection or changing to the untouched door. Classically, the optimum strategy for Bob is to alter his choice of door and this, surprisingly, doubles his chance [9] of winning the prize from 1/3 to 2/3.
https://arxiv.org/pdf/quant-ph/0109035.pdf
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Post Post #16 (isolation #8) » Mon Jan 14, 2019 11:50 pm

Post by talah »

Here's some super-clumsy pseudo-code.. Im' not quite sure what it means but it's how I'd write a program for this for testing.
----------------


//i'm using plain integers starting as 1 as array indexes and RND counters, not common 0 starting arrays here
//pseudocode

// create a 3-option array
define initialThreeArray[3]

// set each of the members of the array to equal 0.
// 0 will indicate goats, 1 will indicate cars.
initialise initialThreeArray(0)

// at this point all 3 array members = 0
// pick a random member of the array to equal 1
car = rnd[3]

// update the array so that the randomly picked number of 1 to 3, equals 1; so one member equals 1 and the others equal 0
// the randomly picked number is the index of the array, and will be changed from 0 to 1
update initialThreeArray[car] = 1

// expected result = 0,0,1 / 0,1,0 / 1,0,0
//

// have the player pick something at random
playerChoice = rnd[3]

///// ((we know at this point that if initialThreeeArray[playerChoice] == 1 then they have won, but we're doing nothing with it))
///// ((at this point, initialThreeArray[playerchoice] equals 1 sometimes (precisely one third of the time?)))

// check Monty's options by removing the player's choice from the array
// create a new 2-member array and skip the player-chosen index

define montyArray[2]
initialise montyArray(0)

// both Monty's choices are initialised to 0, now to update the array
// we need to cycle through each member of the initial array and ignore the player choice index

if playerChoice = 1 then
montyArray[1] = initialThreeArray[2]
montyArray[2] = initialThreeArray[3]
end if

if playerChoice = 2 then
montyArray[1] = initialThreeArray[1]
montyArray[2] = initialThreeArray[3]
end if

if playerChoice = 3 then
montyArray[1] = initialThreeArray[1]
montyArray[2] = initialThreeArray[2]
end if

// Monty now has two choices, and there is 100% a goat available because the player can only choose one of the 3 options (car/goat/goat)
// Monty needs to check what he has and only reveal a goat
// Monty has a choice here if both options are goat

// logic gates here would be pretty good i imagine, like a simple nand

// if both results match they can only both be 0
if montyArray[1] = montyArray[2] then
// pick one at random to reveal as a goat
reveal = rnd[2]

else if montyArray[1] = 1 then
////////**above is the first time we're measuring the result**/////
// pick the non-car result
reveal = 2

else
// this means that montyArray[2] is the car, so we reveal [1] instead
reveal = 1

// print the revealed index (ie - a goat)
print "There is a goat behind door number " montyArray[reveal]
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Post Post #17 (isolation #9) » Mon Jan 14, 2019 11:59 pm

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Actually the above is wrong because I''m not keeping track of doors that the player picks and outputting it as a door number.
But Monty only ever reveals a "0" meaning a goat.
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Post Post #20 (isolation #10) » Tue Jan 15, 2019 9:37 pm

Post by talah »

In post 18, shaft.ed wrote:Might be the wrong way of thinking about it, but what about a simpler explanation.

When you picked your door it had a 1/3 chance of being correct. Yes Monty opens a goat door, but that goat door doesn't change your odds because he can always open a goat door after your pick. Thus that doors odds remain 1 in 3.

However, for the switch door, you have new information. The goat door being opened does impact that door's odds since you are selecting it after the reveal. So it is a 1 in 2 chance of being the correct door.
We--eeell, this is the potential statistical artefact I'm thinking about.
I'm just trying to figure out if there's any result that changes based on the reveal - and it sort of seems that it does. Because from my own perspective I don't know if my pick fell into the 33% or the 66% bracket. Monty revealing a door reveals exactly one thing: of the two doors Monty could have picked, he picked -that door-. If he had both doors to choose from, then he took a 50/50 himself, otherwise he was forced. And he's forced 33% of the time (because 33% of the time, I picked correctly). That leaves a 66% chance that Monty picked something inconsequential. Granted, I no longer get to choose the option that Monty picked after he reveals - but then, he couldn't choose out of 3 either if I picked first. (So Monty gets information about the door I picked, although he always has the option to open a door with a goat because he knows which one not to open, if it exists between his choices.) Monty's expected choices are more logic-gate-like, as in 0|0, 0|1, 1|0. Perhaps there's some symmetry there?

I was just thinking about it in a different way - in terms of order of operations.
Let's say you are told that there are 3 options you will be able to choose from, but Monty reveals a goat *first*.
Monty can only pick from 2 of the 3, because one is a car. Monty is *forced* into 50/50 decision which is reduced by the fact he can only select from 66% of the options if he selects first.
This is definitely information because Monty *cannot* pick the car door. However, it doesn't give information on the remaining choices' probability for success because either of the remaining options has nothing qualifying it.
The information revealed amounts to "Monty picked door [x]" which you already know and provides nothing extra.

So...
You ignore that information and pick randomly of the 3 doors.
Then you look at whether you picked Monty's goat (which is no longer an option) and pick out of the remaining two doors if you did. If you did pick Monty's door, it seems like the other two options have the same probability of success because Monty was constrained similarly to you and your own decision is null. If you didn't - you're left with only one option for a switch (literally the "switch"), one of which Monty didn't pick.
You know that there are 3 states, some of which are already collapsed because you know what Monty picked, and you know if you picked the same.
You did not pick what Monty picked and Monty picked (A) door (goat) | can switch
You did not pick what Monty picked and Monty picked (B) door (goat) | can switch
You picked what Monty picked and it was a goat | must switch (to A or B)

Yeah I dunno, I'm searching for the extra information here and I'm not finding it. Something either mechanical or concrete that links Monty's choice to my own choice without resorting to analogies like "Imagine Monty had 999 doors and told you that 998 of them were wrong".
It reminds me of the whole "where did the extra dollar come from" puzzle that I read years ago:
https://math.stackexchange.com/question ... m-illusion
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Post Post #23 (isolation #11) » Wed Jan 16, 2019 12:19 am

Post by talah »

The only information Monty has about the game is whether he gets to pick one of two goats or whether he is forced to pick the goat instead of the car.

ed: I'm actually going to look up the code others are using
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Post Post #24 (isolation #12) » Wed Jan 16, 2019 12:32 am

Post by talah »

Posting this here because it seems interesting (it's a numberphile comment from youtube)

2 years ago
This is an EASY explanation.
You have 3 ways to play.
1: You pick Goat A, Monty shows Goat B, switch and you WIN
2: You pick Goat B, Monty shows Goat A, switch and you WIN
3: You pick the car, Monty either shows Goat A or B, you switch and LOSE

2 times out of 3 you win.

The comment below it contains

most 50/50 people say that the door Monty opens doesn’t matter

egh, anyway
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Post Post #26 (isolation #13) » Wed Jan 16, 2019 12:35 am

Post by talah »

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 #28 (isolation #14) » Wed Jan 16, 2019 1:55 am

Post by talah »

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)
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 Thu Jan 17, 2019 11:46 pm, edited 2 times in total.
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Post Post #30 (isolation #15) » Thu Jan 17, 2019 11:41 pm

Post by talah »

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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