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Gambler’s Fallacy and the Regression to the Mean

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

#61

Post by joe_the_user »

mikenew wrote: post #49
joe_the_user wrote: post #42

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. I'm not sure what you mean here. If you pick 1 door in 3, the chance you've picked the door with the prize is 1/3 and so the chance the prize is behind another of the 2 remaining door is 2/3. If one of the doors you didn't pick is opened and reveals no prize and the situation is as-de…

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 happens not to be the statement of the situation.

The statement of situation is: you pick a door. The host picks a door, other than what you picked, then opens the door and reveals there's nothing there. This excludes the situation of the host picking the prize or the host picking the same door as you. How that exclusion happened doesn't matter. At that point, you get to choose your original pick or switch to the remaining door. At that point, what the host knew before irrelevant. You had a 1/3 chance of picking the door with the prize before regardless of the host, and now you can make a choice that gives every other possibility and so a 2/3 chance of getting the prize.

Edit: I think that the problem is stated as "the host knows" because that means the host is guaranteed to open a door without the prize. It's not that the host's knowledge matters to the strategy, it's that host's knowledge makes it certain the situation will happen as described. See mtlogstdo's comment. https://news.ycombinator.com/item?id=29846498

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

#62

Post by RubyRidgeRandy »

yayachiken wrote: post #52
RubyRidgeRandy wrote: post #50

Earlier quoted context omitted.

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…

But the probability is not 50%. It is 99% to win if you switch and 1% if you stick with the same door. Imagine you repeat the process multiple times, but the contestant chooses to pick door 1 every time. Would you still assume that staying with door 1 every time would give you better odds, that is you would win 50% of the time if you always pick door 1 out of 100?

Think about it this way: there are two doors as the starting condition, only one has the prize behind it. 50% odds. The host opens the wrong door, so only one remains. The prize MUST be behind the only remaining door.

The host gives you the option to select that door or he can start a new game where there is only one door and one prize.

Would anyone really argue that there is a difference in outcome here? But the logic I see is that people are saying you should switch because the probability in the first set is 50% even though its literally impossible in this scenario for it to be 50%.

-----

I know mathematically it's wrong. But I hate this problem so much. I am convinced is a philosophical / semantical issue.

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

#63

Post by dkjaudyeqooe »

RubyRidgeRandy wrote: post #62
yayachiken wrote: post #52

Earlier quoted context omitted.

But the probability is not 50%. It is 99% to win if you switch and 1% if you stick with the same door. Imagine you repeat the process multiple times, but the contestant chooses to pick door 1 every time. Would you still assume that staying with door 1 every time would give you better odds, that is you would win 50% of the time if you always pick door 1 out of 100?

Think about it this way: there are two doors as the starting condition, only one has the prize behind it. 50% odds. The host opens the wrong door, so only one remains. The prize MUST be behind the only remaining door. The host gives you the option to select that door or he can start a new game where there is only one door and one prize. Would anyone really argue that there is a difference in outcome here? But the log…

Think of it this way:

Each door has 1/3 chance of containing the prize, that never changes, so your choice has a 1/3 chance of being right and the unselected doors have a 2/3 chance. Again, these facts will never change, you'd surely agree.

So once one of those unselected doors is opened, the remaining door has a 2/3 chance because you now know which of the unselected doors you must chose to capture the 2/3 chance (because the opened door has been revealed to be empty). Those pair of doors still contain the 2/3 chance, you just know which to choose now.

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

#64

Post by tawan »

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…

I disagree with your conclusion that the odds remain 2/3 even if the host doesn’t know where the prize is. If that were the case, it would mean that if the same scenario is repeated many many times, then in 2/3 of the cases you would still pick the door with the prize with the same strategy. But what about the instances where the host picks the door with prize before you even get a chance to pick the other door? These cases happen with a probability of 2/3 * 1/2 = 1/3. Therefore, you only get to choose in 2/3 of the cases and in 1/2 of those cases you have already picked the correct door the first time.

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

#65

Post by goto11 »

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…

Yeah, the scenario is often stated ambiguously. The important points:

1) The prizes are placed behind the doors before the game begins.

2) The host will open a door regardless of whether the contestant have selected the door with the prize or not.

3) The door to open is always selected from the remaining doors. The host will never open the door the contestant selected.

4) The door is not selected at random. The host knows where the prize is, and will only open a door without a price.

The counter-intuitive part is not the probabilities, it is the rules the host operates. Some of the rules (2,3) depends on the host not knowing or ignoring where the prize is, while some (4) depend on the host using the knowledge of where the prize is.

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

#66

Post by galaxyLogic »

goto11 wrote: post #65
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…

Yeah, the scenario is often stated ambiguously. The important points: 1) The prizes are placed behind the doors before the game begins. 2) The host will open a door regardless of whether the contestant have selected the door with the prize or not. 3) The door to open is always selected from the remaining doors. The host will never open the door the contestant selected. 4) The door is not selected at random. The host…

The trick is that it looks like the host is doing the same thing as you, choosing one of the remaining doors. That trick us into thinking the host is operating under the same rules as the guest, it is like a game and in a game typically all players have to follow the same rules. So it is natural to assume this is so.

The situation would be perhaps easier to understand correctly if the host didn't open a door, but told you "I can guarantee to you the car is NOT behind this door"

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

#67

Post by galaxyLogic »

jodrellblank wrote: post #36
mhb wrote: post #29

Earlier quoted context omitted.

Your intuition is wrong and it doesn't matter what the host knows before showing you the door that doesn't have the prize. Here's some explanation: https://betterexplained.com/articles/understanding-the-monty...

How can they reliably show you the door that doesn't have the prize, if they don't know where the prize is? They can't, which means the host must know, which means it must matter what the host knows?

To a casual observer it looks like they picked the door without price by chance.

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

#68

Post by galaxyLogic »

gexla wrote: post #23
lloydgrossman wrote: post #8

Earlier quoted context omitted.

Funner fact, many even have a green double zero as well to tip it even further.

Nice points being made here. American roulette has double zero, European roulette has a single zero. The zero(s) represent the house edge and that's what the casino makes its money from. Even funner fact, even if there isn't a zero and you truly get 50/50 odds, you'll still hit ruin over the long run because the house has "unlimited money."

Why doesn't competition work here? Shouldn't somebody open a casino in Las Vegas with only a single 0 and thus attract more customers?

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

#70

Post by yayachiken »

RubyRidgeRandy wrote: post #62
yayachiken wrote: post #52

Earlier quoted context omitted.

But the probability is not 50%. It is 99% to win if you switch and 1% if you stick with the same door. Imagine you repeat the process multiple times, but the contestant chooses to pick door 1 every time. Would you still assume that staying with door 1 every time would give you better odds, that is you would win 50% of the time if you always pick door 1 out of 100?

Think about it this way: there are two doors as the starting condition, only one has the prize behind it. 50% odds. The host opens the wrong door, so only one remains. The prize MUST be behind the only remaining door. The host gives you the option to select that door or he can start a new game where there is only one door and one prize. Would anyone really argue that there is a difference in outcome here? But the log…

The problem formulation doesn't really work for nBut even then, I don't see your problem here. The invariant is that the chosen door has probability 1/n to contain the prize, and all other doors together have probability 1-1/n. This still holds for n=2: Both doors have a probability of 50% to contain the prize.

Also if you do not believe or grok the mathematics, you can just whip up a program to simulate the problem, and verify the probabilities empirically.

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