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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)

#231

Post by [deleted] »

lnl wrote: post #215
mathrando wrote: post #113

Earlier quoted context omitted.

Whether Monte Hall is counter intuitive is a function of how the question is phrased. When you phrase it in a way that underlines the mechanical nature of the host's decision, people get it right. When you phrase it in a way that suggests the host's choice is itself random, people get it wrong. I think the first formulation primes people to think of it from the perspective of the host, which is the right perspective…

> When you phrase it in a way that suggests the host's choice is itself random, people get it wrong. In other words, they still get it right; they get it right for the separate question that that phrasing implies. If the host's choice is random, so that when you initially picked wrong it's equally probable that the host open the door with the car and then say "sorry, looks like you lost" (which is what I assumed when…

[deleted]

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

#232

Post by gonehome »

megablast wrote: post #227
gonehome wrote: post #69

Earlier quoted context omitted.

Yeah - and at least for me this was part of the confusion. The problem is often stated incorrectly. Critically the host always removes a goat. If the person presenting the problem just says “the host opens one of the doors”, but doesn’t specify he’s always revealing a bad door then it’s not clear why that matters. The many door example makes it easy to intuit as well.

That’s obvious. If he reveals the car, he isn’t going to ask you if you want to switch. A goat is a goat. It doesn’t matter to you which goat you get.

It’s not obvious if you think revealing the car ends the game (and you lose).

I think when I was first told it the person said “takes a door away” which is even less clear.

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

#233

Post by listenallyall »

jVinc wrote: post #134
TchoBeer wrote: post #87

Earlier quoted context omitted.

It's great that you're so sure of yourself, just a shame that you're wrong. If the host picks randomly, your chances are in fact 50-50. Think of it this way: imagine every time the hosts picks the car the universe resets. Now imagine you see the host picking a goat. There's a 2/3rds chance in your universe you picked the car, and a 1/3rds chance you picked a goat, and so switching would seem like the bad option. This…

It's funny to me that there always seem to be an over representation of rude people among those who get this problem wrong. There is absolutely no need to attack my person. As for the problem, you are just plain wrong. We are talking about are regular old TV-quiz, there is not "universe resetting button" to save your logic. In the regular situation, the host always remove a goat. because otherwise the quiz show is ki…

I agree with you that "the universe resetting" is a dumb way to visualize or explain the problem. But if Mr. Hall opens the first door randomly (essential that it was a random choice), and it revealed a goat, then no, switching does not provide an advantage. You do not go from 33 to 67%. Instead, you are left with 2 options, both of which originally had a probability of 33% of containing a car, and which, now that the third option has been eliminated, are both currently 50%.

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

#234

Post by Talanes »

kahirsch wrote: post #148
6gvONxR4sf7o wrote: post #13

There are a number of things I like about the monty hall problem. There's the history, the unintuitiveness, the subtle easy-to-screw-up nature of probability problems, the calculation, the sociology, and the overconfidence of wrong experts. Most of all, it's the calculation vs intuition that I like. You can do the calculation, or run simulations and prove correctness. In fact, it would be much harder to be so widely…

The day this was published in Parade magazine, I was sure that she was wrong. Even though I was a computer programmer then (1990), I didn't own a computer. I drove across town to my parent's house to use my father's computer to prover her wrong. I didn't even have to run the program. Just the act of writing the program made me realize that she was right. When I was writing the part where the host picks which door to…

Because in 2/3 scenarios you had to make the host choose the door without a car?

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

#235

Post by listenallyall »

galaxyLogic wrote: post #222
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…

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 probability is not 50-50, and switching away from the initially-chosen door is advantageous.

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

#236

Post by glitchc »

Talanes wrote: post #234
kahirsch wrote: post #148

Earlier quoted context omitted.

The day this was published in Parade magazine, I was sure that she was wrong. Even though I was a computer programmer then (1990), I didn't own a computer. I drove across town to my parent's house to use my father's computer to prover her wrong. I didn't even have to run the program. Just the act of writing the program made me realize that she was right. When I was writing the part where the host picks which door to…

Because in 2/3 scenarios you had to make the host choose the door without a car?

The host can never choose the door with the car, because then the game is over. Ergo the host has to choose the door with goat, every time.

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

#237

Post by tomerv »

TchoBeer wrote: post #64
dllthomas wrote: post #34

Earlier quoted context omitted.

Interestingly, if the host picks randomly (and if he reveals the car, you... start over, or you get the car, or you get nothing, or ... it doesn't matter because it happens not to have happened in the time we're considering) then you are faced with a 50/50 chance.

I don't understand how this could be. If you start over when the host picks the car, then isn't that the same as the host picking the goat every time, i.e. the same as the host knowing.

At first I agreed with you, but then I thought about the variation with 100 doors (this is what originally helped me grasp the correct answer): you choose a door out of 100 (with very slim chance of choosing the correct one), the host opens 98 of the remaining 99 doors and there are goats behind all of them. Now it's obvious that the car is in the remaining door.

If the host is opening random doors, then: * If you chose a door with a goat, in 98/99 cases we would start the game over because the host opened the door with the car. * If you chose the door with the car, the game cannot start over because the host can only open doors with goats.

This means that when you choose a goat, the randomness gives you another chance to choose the correct door.

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

#238

Post by [deleted] »

lnl wrote: post #215
mathrando wrote: post #113

Earlier quoted context omitted.

Whether Monte Hall is counter intuitive is a function of how the question is phrased. When you phrase it in a way that underlines the mechanical nature of the host's decision, people get it right. When you phrase it in a way that suggests the host's choice is itself random, people get it wrong. I think the first formulation primes people to think of it from the perspective of the host, which is the right perspective…

> When you phrase it in a way that suggests the host's choice is itself random, people get it wrong. In other words, they still get it right; they get it right for the separate question that that phrasing implies. If the host's choice is random, so that when you initially picked wrong it's equally probable that the host open the door with the car and then say "sorry, looks like you lost" (which is what I assumed when…

[deleted]

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

#239

Post by JoshuaDavid »

npinsker wrote: post #78
jVinc wrote: post #61

Earlier quoted context omitted.

> vs "the treadmill moves backwards to keep plane airspeed at 0". You might be correct in thinking people misunderstand it as that, but that's physically impossible, so I don't think you can argue that people have some sort of alternate understanding under which they are actually correct. That's just one type of wrong reasoning people might apply to the problem. Add to that, in your hypothetical understanding of Mont…

It does change your odds -- you're now living in a world in which the car didn't get revealed, and by Bayesian reasoning that means it's more likely that you live in a world where you picked the correct door.

The problem is that there are two interpretations of the problem, and is is not obvious that the incorrect one is wrong unless you know that Monty knows what's behind all the doors, and chooses never to open a door with a car behind it. The question isn't "does Monty showing a goat change the odds that it's behind the door that neither you nor Monty picked", it's "does it change that probability from 33% to 50% or from 33% to 66%".

Scenario 1: The first, incorrect, interpretation of the problem is "You choose a door, which has either a goat or car behind it. Monty then chooses one of the other two doors, which will also have either a car or goat on it, and opens that door. You then have the choice of whether to switch doors or stay with your original choice".

Scenario 2: The second, correct interpretation of the problem is "You choose a door, which has either a goat or a car behind it. Monty then looks behind the other two doors, and chooses the one that has a goat behind it. If both have goats behind them, Monty chooses randomly. Monty opens his chosen door. You then have the choice of whether to switch doors or stay with your original choice".

    +-----+------+--------+---------+--------+--------+------------+------------+
    | Row | Car  |  Your  | Monty's | Result | Result | Frequency  | Frequency  |
    |     | Door | Choice | Choice  | Stay   | Switch | Scenario 1 | Scenario 2 |
    +-----+------+--------+---------+--------+--------+------------+------------+
    |   1 | #1   |     #1 |      #2 |    Car |   Goat |       1/18 |       1/18 |
    |   2 | #1   |     #1 |      #3 |    Car |   Goat |       1/18 |       1/18 |
    |   3 | #1   |     #2 |      #1 |   Goat |   Goat |       1/18 |       0/18 |
    |   4 | #1   |     #2 |      #3 |   Goat |    Car |       1/18 |       2/18 |
    |   5 | #1   |     #3 |      #1 |   Goat |   Goat |       1/18 |       0/18 |
    |   6 | #1   |     #3 |      #2 |   Goat |    Car |       1/18 |       2/18 |
    |   7 | #2   |     #1 |      #2 |   Goat |   Goat |       1/18 |       0/18 |
    |   8 | #2   |     #1 |      #3 |   Goat |    Car |       1/18 |       2/18 |
    |   9 | #2   |     #2 |      #1 |    Car |   Goat |       1/18 |       1/18 |
    |  10 | #2   |     #2 |      #3 |    Car |   Goat |       1/18 |       1/18 |
    |  11 | #2   |     #3 |      #1 |   Goat |    Car |       1/18 |       2/18 |
    |  12 | #2   |     #3 |      #2 |   Goat |   Goat |       1/18 |       0/18 |
    |  13 | #3   |     #1 |      #2 |   Goat |    Car |       1/18 |       2/18 |
    |  14 | #3   |     #1 |      #3 |   Goat |   Goat |       1/18 |       0/18 |
    |  15 | #3   |     #2 |      #1 |   Goat |    Car |       1/18 |       2/18 |
    |  16 | #3   |     #2 |      #3 |   Goat |   Goat |       1/18 |       0/18 |
    |  17 | #3   |     #3 |      #1 |    Car |   Goat |       1/18 |       1/18 |
    |  18 | #3   |     #3 |      #2 |    Car |   Goat |       1/18 |       1/18 |
    +-----+------+--------+---------+--------+--------+------------+------------+
In scenario 1, before any door is opened, you chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. Monty then opened a door which happened to have a goat behind it, which eliminates rows 3, 5, 7, 12, 14, and 16. Now you have a 6/12 chance of winning the car if you stay, and a 6/12 chance of winning the car if you switch, and this is how you come to the conclusion that there is no advantage in switching.

In scenario 2, you still chose the correct door in the scenarios corresponding to rows 1, 2, 9, 10, 17, and 18, for an aggregate probability of 6/18 == 1 / 3 of the time having picked the door with a car. However, this time Monty's door-opening doesn't eliminate any rows with nonzero probability, since rows 3, 5, 7, 12, 14, and 16 have zero probability to start with. As such, you still have a 6/18 chance of winning the car if you stay, and a 12/18 chance of winning if you switch, so you should switch.

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