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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)

#241

Post by dorgo »

fourthark wrote: post #217
LanceH wrote: post #187

Earlier quoted context omitted.

It's right depending on your assumptions. If the host has rules of only picking the right most door the contestant doesn't pick, and it happens to be a goat then the answer is 50%. Basically if the host is just as willing to show a car as show a goat and it is a goat which is shown, the answer is 50%. There is a confirmation bias that people expect people who are aware of the Monty Hall Problem expect others to be wr…

Intention doesn’t matter. The problem states that the host opens another door and reveals a goat . It doesn’t matter that the host meant to do that. Now that the information is revealed, the probabilities have changed. Obviously if the host revealed the car, that would also affect the probabilities!

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

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

#242

Post by 6gvONxR4sf7o »

megablast wrote: post #221
6gvONxR4sf7o wrote: post #121

Earlier quoted context omitted.

I wrote out a hopefully helpful intution down below [0]. Here's a terse part of it: > Or just imagine monty hall with infinite doors. I'm thinking of a number between 1 and infinity (secretly, it's 19083412039102388171230123). You pick a number, and then I'll narrow down your choice to two options. Do you think you just happened to pick my number, or do you switch? You're right that it's absolutely the narrowing down…

Ok. But you picked a very very very very small number between 1 and infinity. That can’t be random.

Tons of common distributions have infinite support, but the details don't matter for this bit of intuition. If you want to formalize it in a simpler way, just consider it as an arbitrarily large finite number of doors with a uniform distribution.

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

#243

Post by defterGoose »

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 find it a little more intuitive in a visual way with the 100 doors. If the 98 doors that the host subsequently eliminates leaves one in the middle of the pack (e.g. #52), most rational people would look at that door and say "hmm its probably in there rather than in the one i picked, #1"

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

#244

Post by ikt »

admax88q wrote: post #196
lachstar-x wrote: post #102

Earlier quoted context omitted.

Can you help me understand - in my view the choice to switch doors or keep the same door is irrelevant, because even if you keep the same door you're making a choice that is now 2/3 of the right answer. If you switch or keep, you're still choosing from two doors that contain a car and a goat. The other door is no longer relevant and doesn't affect the new state at all. It's your perspective (narrowing the choice down…

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!

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

#245

Post by KALSTR »

dorgo wrote: post #241
fourthark wrote: post #217

Earlier quoted context omitted.

Intention doesn’t matter. The problem states that the host opens another door and reveals a goat . It doesn’t matter that the host meant to do that. Now that the information is revealed, the probabilities have changed. Obviously if the host revealed the car, that would also affect the probabilities!

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.

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

#246

Post by thaumasiotes »

defterGoose wrote: post #243
listenallyall wrote: post #235

Earlier quoted context omitted.

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 find it a little more intuitive in a visual way with the 100 doors. If the 98 doors that the host subsequently eliminates leaves one in the middle of the pack (e.g. #52), most rational people would look at that door and say "hmm its probably in there rather than in the one i picked, #1"

Eh, the formulation I like goes:

You pick a door.

The host runs up and knocks on one of the other doors, without opening it or saying anything. He just indicates a door.

Now you have the option to stick with your one original door, or to take both of the other two doors.

Which is better?

And then: how is that scenario different from the one where he opens the special door and reveals a goat?

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

#247

Post by RogerL »

saeranv wrote: post #9

I guess Monte Carlo simulations weren't well-known or easily accessible back then? It would have been the perfect way to shut up her critics.

You don't need Monte Carlo, you just need a table with three rows. GGC, GCG, CGG. Pick the first door, cross out the other goat, and switch. you will lose once and win twice. if for some reason you think the door you pick at first matter somehow, make nine rows, and you will lose three times in win 6. Easy peasy.

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

#248

Post by mercer »

javajosh wrote: post #23

So this is at least in part about the "Monty Hall Problem" and why it's solution not intuitive. The article missed an important angle: when the host opens a door, he's giving you more information , which explains why it's better to switch. If you're the host, you need to know which door the car is behind to do your job 2/3 of the time, to avoid revealing it. It's this quality of unexpected information exchange that I…

The explanation that best helped me 'intuitively' understand things is to rephrase the game as being two separate games: choosing one of three doors, and after that having the option to go for a coin toss instead. Obviously you'd go for the coin toss.

Now I understand that the reason why that worked for me is precisely because it becomes much more clear that it's really about new information, and what tripped me up was trying to maintain a kind of 'narrative', I suppose?

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

#249

Post by ncallaway »

FabHK wrote: post #195
mkl wrote: post #185

Earlier quoted context omitted.

The 1990 Parade Magazine description is almost identical (and actually more explicit, since "say #3" is a removable parenthetical): "the host, who knows what’s behind the doors, opens another door, say #3, which has a goat" -- https://web.archive.org/web/20130121183432/http://marilynvos... The host's knowledge is explicitly mentioned, and the only purpose this could have is that so he can use it to avoid giving the g…

That is literally not explicit (where by "literally" I mean literally, not figuratively, and by "explicit", I mean explicit, not implicit). It is sort of hinted at, but it is not explicitly said that the host will mechanically reveal a door that has a goat. It is quite conceivable that the host picks a door randomly, or in fact that he picks the door with the car with a certain probability (saving the show quite some…

> It is literally explicit in the 1990 rendition.

> You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat

Remove the parenthetical example "say #3", and the parenthetical "who knows what's behind the doors" and that sentence reads: "...and the host opens another door which has a goat".

The "has a goat" is not a hypothetical example. It's a (both literal and explicit) declaration of the rules statement. The rules seem very clear that the host will open a door with a goat.

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

#250

Post by h0l0cube »

utopcell wrote: post #174

You pick the car with 1/3 probability and a goat with 2/3. If you picked the car and switch, you lost. If you picked a goat and switch, you win because the only other door left has the car. The problem is simple to reason through. The hard part is convincing yourself that you need to think through it given what seems "obviously correct".

This is the same explanation I arrived at too. Crucial to answering the problem correctly is understanding that the host doesn't reveal a random door, but a door that you have not chosen, and also is wrong. And then the reasoning follows naturally to the correct answer. It's amazing to be so prejudiced that someone would be adamant on the incorrect answer, when it can be explained in 3 simple lemmas.
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