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The Time Everyone “Corrected” the World’s Smartest Woman (2015)

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Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#321

Post by 6gvONxR4sf7o »

KALSTR wrote: post #245
dorgo wrote: post #241

Earlier quoted context omitted.

No. Intention does matter. If the host chooses the goat by chance ( with the possibility to choose the car instead ) then my chances to win are 50% ( switching or not ). But we are talking about Monty Hall after all. So I could be wrong :P

Nope. If the host shows the goat _for any reason_, then the 2/3 chance that the doors you didn't choose contain the car now applies to the one unchosen door remaining -- and you should switch.

Perfect case study right here. It’s trivial to simulate and y’all aren’t budging. In this case, simulating, I see 2/3 for switching after the host shows a goat on purpose, and 1/2 for switching when the host happens to have shown a goat on accident.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#322

Post by TchoBeer »

stale2002 wrote: post #313
TchoBeer wrote: post #282

Earlier quoted context omitted.

You could also consider a case where they send in a new contestant every time the Random Monty Hall picks a car, and the result is the same.

> You could also consider a case where they send in a new contestant every time the Random Monty Hall picks a car In the actual show, do you believe they are sending in a new contestant every time Monty hall picks are car? Do you think that is what is happening in the actual game show?

No, obviously not. We're talking about expected values though, so it's a useful thought experiment.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#323

Post by dec0dedab0de »

megablast wrote: post #258
dec0dedab0de wrote: post #199

The problem with the Monty Hall problem is that it's based on implicit rules. Every explanation I've ever hear never says that no matter what they will never remove the winning choice.

They show all the doors. Why would they ask you to switch doors, if they already shown you the car?? Think about it.

Ahh yeah that does make sense. I didn't realize they actually show you what the prize would have been.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#324

Post by stale2002 »

TchoBeer wrote: post #322
stale2002 wrote: post #313

Earlier quoted context omitted.

> You could also consider a case where they send in a new contestant every time the Random Monty Hall picks a car In the actual show, do you believe they are sending in a new contestant every time Monty hall picks are car? Do you think that is what is happening in the actual game show?

No, obviously not. We're talking about expected values though, so it's a useful thought experiment.

> We're talking about expected values though

No, what people are talking about is the actual game show.

They are doing a statistical analysis of how the actual game show works, in real life.

So nobody is talking about a different, hypothetical situation, where the universe is reset, or the game show host brings out a new contestant every time.

Instead people are talking about how the actual show works. And your scenarios that you bring up, are not relevant, and therefore wrong.

> so it's a useful thought experiment.

No, it is a different thought experience that applies to a different situation.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#325

Post by haberman »

jml7c5 wrote: post #318
haberman wrote: post #303

Earlier quoted context omitted.

I don't think this actually matters. Suppose Monty doesn't actually know which door has the prize, and just picks randomly between the two. Now the only difference is that Monty might accidentally pick a prize. Suppose he does. What are the options now? 1. The game goes on, and you can just switch your guess to the door he picked. (WIN) 2. The game resets due to Monty's error and you play again. (REDO) 3. Monty just…

>Now the only difference is that Monty might accidentally pick a prize. This a mistaken conclusion, as actually there is a difference if a goat is revealed by devious Monty (who makes sure he doesn't reveal the car) and dumb Monty (who picks at random), because the rules of the game really are different. The apparent outcome (you pick door, Monty reveals a goat) is the same. But the process is different. In the latte…

TL;DR: I was wrong (but I did understand the distinction you are drawing)

I'm aware that there is a difference between devious Monty and dumb Monty. The question I was trying to answer is whether this difference affects the actual outcome of the game, or the choice you should make.

I ran a Monte Carlo simulation to test my analysis, and this proved that my previous analysis was in error. If Monty doesn't know to avoid the door with the car, then it makes no difference statistically speaking whether the player switches or not after Monty reveals a goat.

If the rules are such that you win when (dumb) Monty picks the car, then your overall chances are now 2/3 instead of 1/3, since you effectively get two guesses (your guess and Monty's guess). But it makes no difference whether you switch after Monty picks a goat.

On the other hand, if (dumb) Monty choosing the car results in either (1) you losing or (2) a reshuffle and redo, then your overall chances are only 1/3. In this case, it also makes no difference whether you switch after Monty's guess.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#326

Post by galaxyLogic »

listenallyall wrote: post #235
galaxyLogic wrote: post #222

Earlier quoted context omitted.

Yes. The probability that car is behind the door NOT selected by the host is 1/2 - because the host had only two choices which are equally good for him. Probability that car is behind the door initially selected by the user is 1/3 because that is what happens when you randomly choose one out of three. We have to think in terms of two different probabilities: a) That user selects the correct door initially and b) That…

People come up with the incorrect result because with 2 doors remaining, it would initially appear that they both have equal chance of containing the car. Because it is unclear HOW we arrived at the final two doors. If we arrived there randomly, the probability is in fact 50-50. If there was some manipulation - i.e. if the host knows where the car is AND deliberately opens doors that do NOT contain the car - then the…

I came up with a slightly different reasoning why you should switch: When the host is about to open a door there are 2 possibilities, he can either choose between 2 goats, OR he can choose between a car and a goat.

In more than 50% of the cases the host encounters the situation where he must choose between a car and and goat. Why because in > 50% of the cases the user's initial choice does NOT have car behind it, so the host must choose between a goat and a car in > 50% of the cases.

And of course he chooses the goat. So in > 50% of the cases the remaining door will have the car behind it,

So by switching, you have > 50% chance of winning the car. Is this the correct answer? It would seem to me your chance to win if you switch is 1/2 x 2/3. Is this correct?

What is counter-intuitive is why the action by the host should give us any information about which door has the car. And it does not do so always. It only gives us information in 2/3rds of the cases. If the host chooses between two goats that should not give us any information at all should it? But that is good enough for us since we play the odds, and switching seems to increase our odds.

But, now I have my doubts. This kind of probabilistic reasoning would seem to apply only if we repeat the experiment many times. But no contestant gets to play this game more than once. Probability theory does not tell us anything about what happens in a single experiment. So is it rational for the player to think they should switch if they can play this game only once?

How much should they be willing to pay for the option of being able to switch the door, if they know they can only play this game once?

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#327

Post by admax88q »

ikt wrote: post #244
admax88q wrote: post #196

Earlier quoted context omitted.

Here's another thought experiment. Forget the opening of the door. Start with picking a door at random, you have a 1/3 chance of having picked the car. On that I think we all agree. Now let's say that the host offers to let you switch from the door you picked, to the other _two_ doors combined. He hasn't opened any doors, they are all closed, you're allowed to stick with your initial guess of one door, or switch to a…

Intuitively it makes sense it would be half, because there was a 33% chance of winning but now it's 50/50 because 3-1 = 2, but when you read it logically it makes sense. Yours is the first explanation I 'got', thanks!

Glad it helped, I just came up with it and it made it click more for me.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#328

Post by atdrummond »

listenallyall wrote: post #314
atdrummond wrote: post #312

Earlier quoted context omitted.

He never chooses randomly in the Monte Hall problem. He’s always going to reveal a goat. That’s why the maths are so straightforward - and yet so counterintuitive.

> He never chooses randomly in the Monte Hall problem. But you never state that, which is why I took exception with your explanation. You also state that probabilities never change, which is also untrue as more information is revealed.

No. All I said is that the probability of the first door winning doesn't change. Which is true - it remains 1/3. The change in probabilities impact the other two doors, whose terminal values I listed (0 and 2/3).

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#329

Post by dllthomas »

tpush wrote: post #317
dllthomas wrote: post #309

Earlier quoted context omitted.

(Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.) The intentions of the host do matter. Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door.…

So, thinking about it really hard and reading about it online: My comment was definitely wrong: If Monty could have opened a car door, but just didn't, then duh the probabilities for the car to be behind the doors are different than if Monty always opens a goat door. So in that way, the intentions of Monty, meaning how he chooses, definitely matter. But I think your example here doesn't show that? Are you trying to i…

Sorry it took me so long to get back to this; real life intervenes sometimes.

I think what I was trying to do was frame the original Monty Hall problem as a variant of Monty Fall, in a way that (I hoped) makes it clear where Monty is doing work to convert some outcomes into other outcomes (and therefore producing different likelihoods).

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#330

Post by CRConrad »

listenallyall wrote: post #273
wahern wrote: post #251

Earlier quoted context omitted.

That he always opens a door that is not the contestant's choice is a necessary implication of saying that he opens "another" door. That he opens a door containing a goat is similarly implicated. It's not even an implication, really; it's just definitional in vernacular English. And in any event, any of the academics writing in would have been perfectly familiar with the semantic structure of such logic puzzles. Also,…

> That he always opens a door that is not the contestant's choice is a necessary implication of saying that he opens "another" door It really isn't. It's possible Monte Hall always opens the doors in reverse numerical order, so if you picked Door #3, that would get opened first. Or that if Door #3 contained the car, that would be revealed before any goats. The way the scenario was initially described by Ms. Vos Savan…

> > That he always opens a door that is not the contestant's choice is a necessary implication of saying that he opens "another" door

> It really isn't. It's possible Monte Hall always opens the doors in...

It really is. The "other" in "another" means, quite literally, "not the one already mentioned". And since the only door mentioned before that was the one the contestant had chosen, "he then opens another" not only implies but quite literally and explicitly says that this "another" door the host opens absolutely isn't, cannot be, the one the contestant chose. That's quite simply what "another" literally means.

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