Gambler’s Fallacy and the Regression to the Mean • Hacker News Board

Re: Gambler’s Fallacy and the Regression to the Mean

#81

Post by 8note » Sat, Jan 08, 2022, 7:56 PM UTC

RubyRidgeRandy wrote: post #50

LesZedCB wrote: post #31

Earlier quoted context omitted.

best explanation i've heard is the game has 100 doors. you choose 1 of 100 possible doors. the host then opens 98 doors, all with nothing behind them. at that point it's much easier to see that you chances improve greatly by switching.

I always thought of this as a bullshit sleight-of-language problem. Realistically, it doesn't make a difference if you switch, because the chance of the prize being behind any door is 50%. The choice doesn't 'carry over' it's probability. You make a decision on one door out of 100. The odds are 1/100. Then all doors are eliminated besides one door and the one you already chose. the probability of the other door is 50…

The door that's left over went through many trials of not being selected by the host.

The one you chose doesn't get those additional tests. If it's bad, that door had 98 chances where it could have been chosen to be opened by the host, and didn't get chosen. It's either a very lucky empty door, or the correct door.

You can take this all the way to the limiting case where you pick one out of infinite doors, where the probability of you getting it right is ~0. Then, you're presented with a second door, knowing that there's definitely a prize between the two. You're prior knowledge that there's no chance that it's behind the door you picked doesn't go away because you've been given a second door.

Hell, you can go even further, and say you pick between several doors known to be empty, then be presented with a second door and the knowledge that there's a winning door between the two. Your chances looking at the doors starting with two doors is still 50/50, unless you remember that your door is already empty

Re: Gambler’s Fallacy and the Regression to the Mean

#82

Post by mikenew » Sat, Jan 08, 2022, 9:10 PM UTC

joe_the_user wrote: post #61

mikenew wrote: post #49

Earlier quoted context omitted.

Whether or not the host knows where the prize is absolutely makes a difference. There's a reason why it's stated twice in TFA, very unambiguously, that the host is intentionally opening a door they know does not contain the prize. In the case of a host that doesn't know where the prize is, your odds don't improve by switching. Think of it slightly differently: Imagine you pick a door, and then the host, who has no id…

Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. Yes, if the host were to pick a door, not tell you which it was or reveal anything else and then offer to switch your choice for their choice, there would be no difference in the odds of each choice. That just h…

Here, I wrote a simulation in nim: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`.

I ran 1 million iterations, and here are the results:

Running the simulation 1000000 times where the host chooses doors randomly

500193 # of times the contestant's originally chosen door contained the prize

499807 # of times the remaining door contained the prize

Running the simulation 1000000 times where the host intentionally chooses empty doors

334432 # of times the contestan'ts originally chosen door contained the prize

665568 # of times the remaining door contained the prize

Re: Gambler’s Fallacy and the Regression to the Mean

#83

Post by mikenew » Sat, Jan 08, 2022, 9:12 PM UTC

amluto wrote: post #74

mikenew wrote: post #6

So the author presents the Monty Hall problem this way (very explicitly saying that the host knows where the prize is an will not reveal it): > You are given a choice of three doors, behind one is a prize. You can choose one door. The host of this game, who knows where the prize is, then opens one door without a prize (again – they know where the prize is and deliberately choose one of the unchosen doors without a pr…

> If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer. No. If the host chooses a door randomly and it doesn’t happen to contain the prize, the outcome is exactly the same as if the host does know where the prize is. Look up “conditional probability” for how to calculate this. The host’s knowledge has no effect other than to prevent the…

Here, I wrote some code to prove it: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`.

I ran 1 million iterations, and here are the results:

Running the simulation 1000000 times where the host chooses doors randomly

500193 # of times the contestant's originally chosen door contained the prize

499807 # of times the remaining door contained the prize

Running the simulation 1000000 times where the host intentionally chooses empty doors

334432 # of times the contestan'ts originally chosen door contained the prize

665568 # of times the remaining door contained the prize

Re: Gambler’s Fallacy and the Regression to the Mean

#84

Post by amluto » Sat, Jan 08, 2022, 9:43 PM UTC

mikenew wrote: post #83

amluto wrote: post #74

Earlier quoted context omitted.

> If the host is choosing a door randomly and it doesn't happen to contain the prize, your odds don't improve if you switch your answer. No. If the host chooses a door randomly and it doesn’t happen to contain the prize, the outcome is exactly the same as if the host does know where the prize is. Look up “conditional probability” for how to calculate this. The host’s knowledge has no effect other than to prevent the…

Here, I wrote some code to prove it: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`. I ran 1 million iterations, and here are the results: Running the simulation 1000000 times where the host chooses doors randomly 500193 # of times the contestant's originally chosen door contained the prize 499807 # of times the remainin…

I can’t read your code because I’m not logged in. But you pretty clearly have a bug. If the host does not know which door has a prize and opens a door at random, there is a 1/3 chance the host opens the door with the prize, and the game stops (or otherwise deserves its own logged outcome). And there’s no conceivable way that the host’s knowledge, strategy, or anything else could make the probability that the contestant’s originally chosen door contains a prize be anything other than 1/3. The contestant originally guesses with clean slate and has a 1/3 chance of guessing right, full stop.

Re: Gambler’s Fallacy and the Regression to the Mean

#85

Post by Tarsul » Sat, Jan 08, 2022, 10:46 PM UTC

mikenew wrote: post #6

So the author presents the Monty Hall problem this way (very explicitly saying that the host knows where the prize is an will not reveal it): > You are given a choice of three doors, behind one is a prize. You can choose one door. The host of this game, who knows where the prize is, then opens one door without a prize (again – they know where the prize is and deliberately choose one of the unchosen doors without a pr…

Thank you. The articles about this problem constantly neglect those ambiguities of the problem. I'm not arguing that people are good with probabilites (many are not), I'm arguing that the Monthy Hall problem has been overstated as a problem about understanding probabilities when in reality it's more a "understanding/articulating the premise correctly"-problem. Which is why I'm of the opinion that article writers should leave Monthy Hall alone in the future and write about different things. Or maybe someone should make a case study where the participants have to play a Monty Hall game and afterwards answer in a questionnaire why they chose whatever they did. Which would probably depend A LOT on how well the experiment would've been explained at the start, but you could test if the participants understood it in the questionnaire. That way you could finally have a more sophisticated answer about whether this problem has any merit regarding behavioral biases.

Re: Gambler’s Fallacy and the Regression to the Mean

#86

Post by mikenew » Sat, Jan 08, 2022, 10:53 PM UTC

amluto wrote: post #84

mikenew wrote: post #83

Earlier quoted context omitted.

Here, I wrote some code to prove it: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`. I ran 1 million iterations, and here are the results: Running the simulation 1000000 times where the host chooses doors randomly 500193 # of times the contestant's originally chosen door contained the prize 499807 # of times the remainin…

I can’t read your code because I’m not logged in. But you pretty clearly have a bug. If the host does not know which door has a prize and opens a door at random, there is a 1/3 chance the host opens the door with the prize, and the game stops (or otherwise deserves its own logged outcome). And there’s no conceivable way that the host’s knowledge, strategy, or anything else could make the probability that the contesta…

Edit: Sorry, didn't realize that gitlab.com requires you to be logged in to view things (that's dumb). Here it is on github: https://github.com/michaelnew/monty_hall/blob/main/monty_hal...

That is my premise exactly. If the host chooses at random, then there is indeed a 1/3 chance of choosing the door that contains the prize. In which case the game is discarded (or restarted, if you like). If that is the behavior of the host, then in the event that the host randomly chooses a door without the prize (which should happen 2/3 of the time), then the choice between the original door and the remaining door is 50/50. If the host never chooses the door that contains the prize (and therefore the game never has to be restarted), then the host is not choosing randomly.

It's stated pretty clearly in the wikipedia article. The 2/3 probability is only true when these three conditions are met:

1. The host must always open a door that was not picked by the contestant.

2. The host must always open a door to reveal a goat and never the car.

3. The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

Re: Gambler’s Fallacy and the Regression to the Mean

#87

Post by joe_the_user » Sat, Jan 08, 2022, 11:32 PM UTC

mikenew wrote: post #82

joe_the_user wrote: post #61

Earlier quoted context omitted.

Think of it slightly differently: Imagine you pick a door, and then the host, who has no idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. Yes, if the host were to pick a door, not tell you which it was or reveal anything else and then offer to switch your choice for their choice, there would be no difference in the odds of each choice. That just h…

Here, I wrote a simulation in nim: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`. I ran 1 million iterations, and here are the results: Running the simulation 1000000 times where the host chooses doors randomly 500193 # of times the contestant's originally chosen door contained the prize 499807 # of times the remaining…

I'm not going to create an account on that site just for your thing but since you essentially haven't defined what "the host chooses randomly" means in this circumstance", I don't see how a print-out or whatever it gives would enlighten me.

Edit: plus your claim is senseless on it's face. Why would what the host knew about the door influence the probability of the contestant's choice being correct? The contestant picks first and the host doesn't influence the contestant.

Re: Gambler’s Fallacy and the Regression to the Mean

#88

Post by mikenew » Sun, Jan 09, 2022, 12:00 AM UTC

joe_the_user wrote: post #87

mikenew wrote: post #82

Earlier quoted context omitted.

Here, I wrote a simulation in nim: https://gitlab.com/mikenew/monty_hall/-/blob/main/monty_hall... If you have nim installed you can run it yourself with `nim c -r monty_hall.nim`. I ran 1 million iterations, and here are the results: Running the simulation 1000000 times where the host chooses doors randomly 500193 # of times the contestant's originally chosen door contained the prize 499807 # of times the remaining…

I'm not going to create an account on that site just for your thing but since you essentially haven't defined what "the host chooses randomly" means in this circumstance", I don't see how a print-out or whatever it gives would enlighten me. Edit: plus your claim is senseless on it's face. Why would what the host knew about the door influence the probability of the contestant's choice being correct? The contestant pic…

Didn't realize that gitlab.com doesn't allow you to view things if you're not logged in. Here it is on github: https://github.com/michaelnew/monty_hall/blob/main/monty_hal...

By "the host chooses randomly", I mean the host picks from the remaining two doors without any knowledge of whether or not the contestant chose the prize door, or which of the two remaining doors contains a prize. The host flips a coin and if it's heads, they choose the remaining door to the left. Tails and they choose the one to the right. Which means that in 1 of 3 games the host will accidentally reveal the prize.

Whether or not the host is acting with knowledge to filter out incorrect choices or they are just randomly revealing doors is what makes the difference between a 50/50 probability between the two remaining door, or a 1/3 vs 2/3 probability.

Re: Gambler’s Fallacy and the Regression to the Mean

#89

Post by joe_the_user » Sun, Jan 09, 2022, 1:28 AM UTC

mikenew wrote: post #88

joe_the_user wrote: post #87

Earlier quoted context omitted.

I'm not going to create an account on that site just for your thing but since you essentially haven't defined what "the host chooses randomly" means in this circumstance", I don't see how a print-out or whatever it gives would enlighten me. Edit: plus your claim is senseless on it's face. Why would what the host knew about the door influence the probability of the contestant's choice being correct? The contestant pic…

Didn't realize that gitlab.com doesn't allow you to view things if you're not logged in. Here it is on github: https://github.com/michaelnew/monty_hall/blob/main/monty_hal... By "the host chooses randomly", I mean the host picks from the remaining two doors without any knowledge of whether or not the contestant chose the prize door, or which of the two remaining doors contains a prize. The host flips a coin and if it…

Tails and they choose the one to the right. Which means that in 1 of 3 games the host will accidentally reveal the prize.

And what happens then? I mean, I think people have said from the start, that this behavior is outside the specification of the problem - which is that the host opens a door and reveals nothing.

The situation is about only the situation where the host choose the door with nothing. The host "opens one door without a prize". The key detail is this, "opens one door without a prize", not "knows where the prize is". If the host doesn't know where the prize but still, by chance, "opens one door without a prize", then, in this situation the contestant's information remains the same and the odds remain the same.

Re: Gambler’s Fallacy and the Regression to the Mean

#90

Post by mikenew » Sun, Jan 09, 2022, 2:29 AM UTC

joe_the_user wrote: post #89

mikenew wrote: post #88

Earlier quoted context omitted.

Didn't realize that gitlab.com doesn't allow you to view things if you're not logged in. Here it is on github: https://github.com/michaelnew/monty_hall/blob/main/monty_hal... By "the host chooses randomly", I mean the host picks from the remaining two doors without any knowledge of whether or not the contestant chose the prize door, or which of the two remaining doors contains a prize. The host flips a coin and if it…

Tails and they choose the one to the right. Which means that in 1 of 3 games the host will accidentally reveal the prize. And what happens then? I mean, I think people have said from the start, that this behavior is outside the specification of the problem - which is that the host opens a door and reveals nothing. The situation is about only the situation where the host choose the door with nothing. The host "opens o…

I don't know what to tell you at this point. From the wikipedia article:

> Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence does not hold.

Random means there's a possibility the host opens the door containing the prize. The Monty Hall problem specifies that that is not a possible scenario, and that the host does not choose randomly.

> If the host doesn't know where the prize but still, by chance, "opens one door without a prize", then, in this situation the contestant's information remains the same and the odds remain the same.

The host cannon open a door by chance if there's no chance of opening the door with a prize behind it.