Monty Hall Bsuinss

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talah
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Post Post #0  (ISO)  » Fri Jan 11, 2019 9:53 am

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 #1  (ISO)  » Fri Jan 11, 2019 10:34 am

In post 0, talah wrote:Why is or isn't that a thing?

human psychology

but yes, you should always switch

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Post Post #2  (ISO)  » Fri Jan 11, 2019 11:00 am

but then i could predetermine my switch always, guaranteeing that my initial random choice was bad

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Post Post #3  (ISO)  » Fri Jan 11, 2019 11:16 am

how

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Post Post #4  (ISO)  » Fri Jan 11, 2019 11:17 am

choose anything, it's guaranteed to be bad, because switching is preferable.

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Post Post #5  (ISO)  » Fri Jan 11, 2019 11:25 am

its guaranteed to be half as likely
not bad

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Post Post #6  (ISO)  » Fri Jan 11, 2019 11:31 am

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 #7  (ISO)  » Fri Jan 11, 2019 11:39 am

I'm not sure what the point this thread is trying to make is
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talah
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Post Post #8  (ISO)  » Fri Jan 11, 2019 11:45 am

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 #9  (ISO)  » Fri Jan 11, 2019 12:47 pm

Or you could look at it as "I'm picking the door the prize is NOT at" with your first pick. Therefore if the prize isn't behind your first pick (which it probably isn't) then you picked CORRECTLY (since you're switching and were trying to find a door without the prize) and it was in fact a good pick, not a bad pick. Does that help at all?
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Post Post #10  (ISO)  » Sat Jan 12, 2019 10:06 am

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 #11  (ISO)  » Sat Jan 12, 2019 2:31 pm

easier way to think of it is switching is like picking two doors
not switching is like picking one door

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Post Post #12  (ISO)  » Sun Jan 13, 2019 12:23 pm

@talah: But the probabilities are at the heart of this problem. That part is inescapable.

Your initial choice does affect the outcome. If you pick the wrong door (2/3 options, so good chance of that), you are basically forcing the presenter to tell you which door is the good one by not opening it. The other 1/3 of the time, you are letting him do his WIFOM.

Another way to look at it is as follows: Imagine that right as you are about to pick a door magical lightning strikes you, and there are now three near-identical universes. The only difference in each of them is the arrangement of your "select door" neurones, so you pick a different door in each universe. Now suppose that you actually are aware of this arrangement, but nothing else. Only one of those yous picks the right door. The other two don't. But you don't know which one got the right door. Two of the yous need to switch, the other one has to not do so, but they don't know who they are. But since there's three of you, and only one benefits from not switching, it's more reasonable for all 3 to assume they should switch. Then 2 of them win, and only 1 loses.
You don't have ambiguity; you have options.

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Post Post #13  (ISO)  » Mon Jan 14, 2019 5:16 am

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 #14  (ISO)  » Mon Jan 14, 2019 9:07 am

But he is giving you information: He opens the door he opens, and not the other one. If he had a choice, then whatever, there is nothing of interest to be gained, but this only happens if you chose the car (1/3 probability).
You don't have ambiguity; you have options.

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Post Post #15  (ISO)  » Tue Jan 15, 2019 3:49 am

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  (ISO)  » Tue Jan 15, 2019 5:50 am

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  (ISO)  » Tue Jan 15, 2019 5:59 am

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 #18  (ISO)  » Tue Jan 15, 2019 8:53 am

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.

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Post Post #19  (ISO)  » Tue Jan 15, 2019 10:01 pm

In post 9, mhsmith0 wrote:Or you could look at it as "I'm picking the door the prize is NOT at" with your first pick. Therefore if the prize isn't behind your first pick (which it probably isn't) then you picked CORRECTLY (since you're switching and were trying to find a door without the prize) and it was in fact a good pick, not a bad pick. Does that help at all?


good explanation
wryyy

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Post Post #20  (ISO)  » Wed Jan 16, 2019 3:37 am

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/questions/15524/riddle-simple-arithmetic-problem-illusion

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Post Post #21  (ISO)  » Wed Jan 16, 2019 5:35 am

This has brought up a question for me that I’m struggling to answer. I’m probably either overthinking it or missing something obvious.

Normally in the Monty Hall problem, Monty specifically chooses a goat door, and you are made aware of this. Does anything about the answer change if, instead, he chooses a door randomly, regardless of whether or not it’s a car or goat?

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Post Post #22  (ISO)  » Wed Jan 16, 2019 6:11 am

if monty chooses the door randomly, it doesn't matter whether you switch

switching only makes sense because you're able to use the information monty has about the game

this aspect is typically not emphasized in the problem nor in the solutions, which is why many people consider the problem to be misleading

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Post Post #23  (ISO)  » Wed Jan 16, 2019 6:19 am

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  (ISO)  » Wed Jan 16, 2019 6:32 am

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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