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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

#51

Post by compiler-guy »

bryanlarsen wrote: post #19
chx wrote: post #5

> “I know that the fact that the roulette wheel has come up red 10 times in a row tells me NOTHING about spin #11. On the other hand, I know that over time, there will be just as many black spins as red spins, so at least intuitively, a black spin seems at least a little more likely to come up next in order to push that ratio back towards 50/50. Are these two principles actually in tension with each other? If not, ho…

On the gripping hand, coming up red 10 times in a row is evidence that the odds of red and black aren't even.

Strangely, ten reds in a row is not particularly good evidence of an unfair wheel.

For an American wheel, the odds of red in a single spin is 18/38= 0.4737. So ten in a row is 0.4737^10, which is about 0.0005689.

So one should expect five or so such sequences in ten thousand spins. Or one in every two thousand spins.

If a casino spins its roulette wheel 200 times a day, you would expect to see a sequence of ten reds in a row every ten days. If the casino runs ten roulette wheels, it's an every day occurrence.

Low probability events happen regularly with enough samples.

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

#52

Post by yayachiken »

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…

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?

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

#53

Post by chrismcb »

mikenew wrote: post #22
gerdesj wrote: post #14

Earlier quoted context omitted.

"which to me makes intuitive sense" Your intuition is either very good or complete bollocks, or at least worryingly odd 8) Monty Hall is a really clever problem and worth studying in some depth. Whenever I've encountered it, the rules are always given without ambiguity. Even so, it is very hard to get to the bottom of the probabilities. You can reason your way through it and possibly get to the right answer, unaided.…

Maybe I'm just misunderstanding something then? I'm not trying to be dismissive or act like I think I have some special intuition here. It really does just seem straightforward. Imagine the problem this way: The host of a game show presents you with 100 doors, behind one of which is a prize. You pick one, and you know that your odds of having chosen the correct door are 1 in 100. The host, who knows where the prize i…

YOU aren't missing anything. But a lot of people can't give over "two doors means fifty fifty"

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

#54

Post by tzs »

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…

That statement of the Monty Hall problem is still not quite correct. It needs to also include that you are told beforehand that the host will be opening a door and giving you a chance to switch.

If the host is not bound to make the offer, they could do something like only make the offer if you have picked the right door. Switching would then always lose.

People would learn that is the case, and no one would switch which would not be as interesting for the audience. To avoid this the host could give the option to switch half the time when the player picks an empty door and all the time when they pick the prize door. On average under this approach the player wins 1/3 of the time regardless of their choice.

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

#55

Post by chrismcb »

almostdeadguy wrote: post #35
yata69420 wrote: post #27

Earlier quoted context omitted.

I'm with you. Usually when it comes up in pop culture, it's not mentioned that the host can't open a door with a prize. That's the only reason it's confusing.

And that they can't open the door chosen by the contestant! Also perhaps more subtly, that Monty picks at random when given the option. The original problem prompt is _highly_ underspecified (and makes assumptions that are counter-intuitive to how a game show host might behave).

The problem has posed by von Savant specified the host knew where the prize was. And claims the host reveals a non prize for. It is fairly obvious the host won't reveal the winning door, as it would end the problem. Not sure how the assumptions are fighter intuitive.

Of course Hall himself said maybe I open a door because I want you to switch.

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

#56

Post by bopbeepboop »

mikenew wrote: post #22
gerdesj wrote: post #14

Earlier quoted context omitted.

"which to me makes intuitive sense" Your intuition is either very good or complete bollocks, or at least worryingly odd 8) Monty Hall is a really clever problem and worth studying in some depth. Whenever I've encountered it, the rules are always given without ambiguity. Even so, it is very hard to get to the bottom of the probabilities. You can reason your way through it and possibly get to the right answer, unaided.…

Maybe I'm just misunderstanding something then? I'm not trying to be dismissive or act like I think I have some special intuition here. It really does just seem straightforward. Imagine the problem this way: The host of a game show presents you with 100 doors, behind one of which is a prize. You pick one, and you know that your odds of having chosen the correct door are 1 in 100. The host, who knows where the prize i…

[dead]

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

#57

Post by chrismcb »

coldtea wrote: post #30

> That’s a great question, and the answer is a definite no – they are not in conflict. Again, the pressure to think that the past influences future independent events is powerful. Regression to the mean is not a power in the universe that ensures that statistics work out in the end, it is purely a probability. I find TFA's argument about the gambler fallacy not being associated with regression to the mean quite hand…

After successive black streaks about the only thing you can say is maybe the wheel is biased toward black. Assuming no green, and a perfectly unbiased wheel, black if a fifty fifty chance. Which means the next infinity sounds half will be black. If there were three blacks in a row, it doesn't mean red is more likely, as the rest of the spins half will be black.

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

#59

Post by gexla »

rout39574 wrote: post #33
coldtea wrote: post #30

> That’s a great question, and the answer is a definite no – they are not in conflict. Again, the pressure to think that the past influences future independent events is powerful. Regression to the mean is not a power in the universe that ensures that statistics work out in the end, it is purely a probability. I find TFA's argument about the gambler fallacy not being associated with regression to the mean quite hand…

But "streaks" are irrelevant. Reversion to the mean doesn't tell you things across trials, it tells you things about individual trials. Try this on: What is the "mean" red-black on the roulette wheel? There isn't one. The only way you can reasonably expect the past streak to have an impact on the next spin, is if you've concluded the streaks are sufficiently unlikely to cause you to judge the wheel not to be fair. If…

Right, I have never heard "mean" being used in the context of casino games.

I get what they are trying to say though, at some point the variance should start evening out to reflect the odds of the game. But we should be practical and remember we're in a casino. A red/black bet on roulette is one of the few games where you're going to get that low of variance. With slots, you have to play a ton of games before you can get an idea of what the variance is just from your game play. In poker, the rule of thumb is 10,000 hands before the player can be confident of having an idea of how effective a change of strategy is.

Overall, whoever wrote this article is out of their element. Like this bit...

> So we think we can use our amazing powers of pattern recognition to determine that black is “due” and use that power to win big. Casinos love this delusion, because they know that math wins out in the end.

This is contradictory. They're saying the player is wrong, because of math. By that line of thinking, the casino would also be wrong to love it, because of math. In a typical casino game, gambler's fallacy won't make your choice more or less correct, unless the choice is to keep playing or to walk away. The author doesn't specify why the casinos would love this though. As an aside, gambler's fallacy could affect more decision making in poker, but poker is a game of skill and the casino doesn't have a house edge on poker (they take a rake.)

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

#60

Post by alecst »

LesZedCB wrote: post #31
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…

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.

Or think about it in reverse. While you’re blindfolded the host opens 99/100 doors. While you’re still blindfolded he then gives you a door to choose from one of those 100. If you pick a different door than the one closed door, fine. If you pick the same door as the host, the host will choose at random a different door to close.

Now you’ve done all this and taken off your blindfold. If you guessed any door but the closed one, you chose wrong. The host asks you what you would like to do — stay with your choice or switch. So do you switch?

I actually meant to respond to the person below you who said that this explanation was bullshit but I’ll just leave this here.

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