Monty Hall problem revisited

[EXTRAPOLATER]

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

the above reprinted from Wikipedia, here: http://en.wikipedia.org/wiki/Monty_Hall_problem

I've heard this a few times and, barring some psychological shit a la Let's Make a Deal, I can't,
for the life of me, buy arguments such as the following:

Vos Savant[edit]

The solution presented by vos Savant (1990b) in Parade shows the three possible arrangements of one car and two goats behind three doors and the result of staying or switching after initially picking door 1 in each case:


behind door 1 car goat goat

behind door 2 goat car goat

behind door 3 goat goat car

result if staying at door #1 car goat goat

result if switching to the door offered goat car car


A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.
---

...appears to be B.S. disregarding extraneous factors, as mentioned.
(copy-paste from same wiki article)

Anybody else been head-fucked with this before?
I've heard it before, but only became h-f'ed over the past 24 hours, with little willingness
remaining to change my initial considerations.

 

[JOSHNAUDI]

Yes

Your initial odds are 1 in 3, therefore sticking with your initial pick wins 1 out of every 3 picks.

When you change your pick, you have a 1 in 2 chance of being correct since they eliminated a wrong choice. You win half of the time.

The 2 out of 3 is misleading because it really is only a fiddy fiddy proposition

---

I saw it on the myth busters

 

[lostinamerica]

The fallacy is that while your initial odds of being right may have been 1 in 3, those odds going forward for sticking with your original pick just got sliced to 1 in 2 when the one "zonk" was revealed, the identical 1 in 2 odds you have if you decide to switch.

Since the host knew where the goats were, there was nothing random or "equally likely" about his revealing one of the "wrong choices" . . . And if the option being revealed first by the host was totally random, then the 1 in 3 times when you would find out initially that you made the wrong (or right) choice would account for the initial odds discrepancy between a 1 in 3 chance and a 1 in 2 chance, and your odds going forward after the 2 in 3 truly random times when your losing (or winning) choice was not the first one revealed would still come back to 1 in 2 chances whether you stick with your initial choice or change your choice.

GL

 

[EXTRAPOLATER]

Yes

Your initial odds are 1 in 3, therefore sticking with your initial pick wins 1 out of every 3 picks.

When you change your pick, you have a 1 in 2 chance of being correct since they eliminated a wrong choice. You win half of the time.

Your initial odds are 1 in 3, but after a door is eliminated, and you are offered the switch,
your odds--at that point--are 1 in 2 if you stick, same as if you switch.

Apparently this is controversial.
I confess to not consuming the mega-long wiki article (due to beverage impatience) but I
think I perused enough of the arguments to conclude that there is some erroneous mind-game
number shit being submitted in lieu of objective probability theory.

 

[EXTRAPOLATER]

The fallacy is that while your initial odds of being right may have been 1 in 3, those odds going forward for sticking with your original pick just got sliced to 1 in 2 when the one "zonk" was revealed, the identical 1 in 2 odds you have if you decide to switch.

Since the host knew where the goats were, there was nothing random or "equally likely" about his revealing one of the "wrong choices". And if the option being revealed was totally random, then the 1 in 3 times when you would find out initially that you made the wrong (or right) choice would account for the initial odds discrepancy between a 1 in 3 chance and a 1 in 2 chance, and your odds going forward after the 2 in 3 truly random times when your losing (or winning) choice was not the first one revealed would still come back to 1 in 2 chances whether you stick with your initial choice or change your choice.

GL

The first part is sorta what I was saying in first response.

I confess to being lost on the 2nd paragraph, after reading a couple of times. It either sounds like
you're suggesting that an illusion is involved (as I surmise) or else trying to explain what I don't
get, which remains ungot. I'll re-read after a moment of silence, fresh air, and beverage.

 

[Jord20]

ALWAYS change if you want the best odds. Real quickly...

Imagine you had a million doors and you picked one. Then the host eliminated all of them except 1 (in addition to yours). NOW would you change doors? OF course you would, because you were most likely wrong to begin with. Same applies to there being 3 doors (on a lesser scale obviously). Since you will be wrong 66.6% of the time with your initial choice, over the long run you will killed with sticking with your original.

 

[rrc]

Can't say I agree, even with the million door scenario.

At this point there are 2 doors available...which gives each door at this point in time a 50/50 chance.

Don't see the impact of the fact that in the beginning that my door only had a one in a million chance has on the current scenario of 2 doors remaining.

 

[JOSHNAUDI]

At this point there are 2 doors available...which gives each door at this point in time a 50/50 chance.

But those weren't the odds when you made the selection

In both cases

That's why you always change your pick, You were a 33% dog (1 in a million) to pick the right one the 1st time and now have the ability to increase your odds to 50% by changing doors

 

[EXTRAPOLATER]

ALWAYS change if you want the best odds. Real quickly...

Imagine you had a million doors and you picked one. Then the host eliminated all of them except 1 (in addition to yours). NOW would you change doors? OF course you would, because you were most likely wrong to begin with. Same applies to there being 3 doors (on a lesser scale obviously). Since you will be wrong 66.6% of the time with your initial choice, over the long run you will killed with sticking with your original.

I saw this as part of the elaboration at the wiki site.
Yes, you were most likely wrong with your initial choice but also were as equally probable as being
a winner as you would have with a single, provided, alternative.

This part of the 'million doors' example I find particularly misleading:

On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat ? the chance that the player's door is correct has not changed. A rational player should switch.

Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. It's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens 999,998 for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.
---

I am still lost, barring some coercion factors or whatever, as to why this is an issue.
I'll continue to try to grasp the same, through responses or a better reading of the wiki-article,
though I'd sure prefer another beer and a poker game right now.
Guess I can solve one of those, at least.

 

[Jord20]

Can't say I agree, even with the million door scenario.

At this point there are 2 doors available...which gives each door at this point in time a 50/50 chance.

Don't see the impact of the fact that in the beginning that my door only had a one in a million chance has on the current scenario of 2 doors remaining.

Yeah, you are missing the point. I promise that I am right though

You do NOT have anywhere near a 50/50 shot once he closes all the remaining doors.

Think about my million door scenario. You pick 10 different times. DO you think in 10 tries you will have picked, with your initial choice, the 1 in a million shot!?!? No, most likely not. If, however, he eliminates all the doors except 1 (the correct door, assuming you were wrong), your odds of it being your door remain the same (1 in a million)... however, he knows the correct one, and has to pose it as an option to you. 999,999 out of 1 million times, you would be right to keep your pick. The other 999,999 times you should have changed.

This seems very obvious to me, but maybe I'm not explaining it perfectly.

Maybe later. But, again... I'm right.

:toast:

 

[JOSHNAUDI]

My 50% was wrong, they show how it is 66% here

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[Jord20]

But those weren't the odds when you made the selection

In both cases

That's why you always change your pick, You were a 33% dog (1 in a million) to pick the right one the 1st time and now have the ability to increase your odds to 50% by changing doors

ACtually, you are now a 67% favorite if you change... you have doubled your chances of being correct.

 

[EXTRAPOLATER]

But those weren't the odds when you made the selection

In both cases

That's why you always change your pick, You were a 33% dog (1 in a million) to pick the right one the 1st time and now have the ability to increase your odds to 50% by changing doors

Your odds on your initial selection need to be modified.

You're speaking of the 1 in 3, here.

Sure, you only had a 33% chance of being right on your initial decision, but once a door is
eliminated then you have 1/2 of the doors picked (50%) so you are not increasing your odds but
leaving them the same if you stick or switch.

 

[Jord20]

Your odds on your initial selection need to be modified.

You're speaking of the 1 in 3, here.

Sure, you only had a 33% chance of being right on your initial decision, but once a door is
eliminated then you have 1/2 of the doors picked (50%) so you are not increasing your odds but
leaving them the same if you stick or switch.

nah

 

[JOSHNAUDI]

ACtually, you are now a 67% favorite if you change... you have doubled your chances of being correct.

:0008:0008:0008

 

[EXTRAPOLATER]

My 50% was wrong, they show how it is 66% here

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This is actually a good video, which explains the argument quite well.

However, I think it also makes quite clear the mistakes in reasoning that I'm having a problem
with. Quite simply, there is no adjustment on your odds of winning after a door is revealed
(left at 33%, as opposed to the adjusted 50%). Forget the 66% chance of winning mentioned as
that is total B.S. after a door is eliminated.

Anywho, good video for anyone (me included) who wants a straightforward listen to the argument.

 

[Jord20]

This is actually a good video, which explains the argument quite well.

However, I think it also makes quite clear the mistakes in reasoning that I'm having a problem
with. Quite simply, there is no adjustment on your odds of winning after a door is revealed
(left at 33%, as opposed to the adjusted 50%). Forget the 66% chance of winning mentioned as
that is total B.S. after a door is eliminated.

Anywho, good video for anyone (me included) who wants a straightforward listen to the argument.

The odds do not adjust on your original pick. They only increase IF you make the change.

 

[EXTRAPOLATER]

The odds do not adjust on your original pick. They only increase IF you make the change.

Of course the odds adjust. You are making a new wager. Originally you had a 33% chance and
with the 1/2 option now offered you're odds increase to 50%.

You are not betting if your original pick was correct, but on what might be correct with 2 remaining
options.

 

[Jord20]

Of course the odds adjust. You are making a new wager. Originally you had a 33% chance and
with the 1/2 option now offered you're odds increase to 50%.

You are not betting if your original pick was correct, but on what might be correct with 2 remaining
options.

Not true

 

[rrc]

I guess my question would be that with 2 doors left...I have door #1...are my odds at THIS POINT not 50%?

Also..hear me out here...2 doors left...I have door #1 which has a 50% of being the correct door...why would I switch to door #2 which only had 1 in a million chance in the beginning?

Interesting discourse...