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Gambler’s Fallacy and the Regression to the Mean

theness.com

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

#41

Post by tbp105 »

Thankfully, the Monty Hall example here is explained well. Oftentimes it is explained in a way to make you think just the opposite. If you assume the House knows where the prize is, they'd have picked it (and won) when they reveal the door. So you'd be a sucker to switch. They went out of their way in this example to say that the host always opens a door without the prize rather than trying to win themselves, which gets you to the "surprising" answer that you should always switch. Maybe if you were familiar with the format of the show (I am not), you'd understand. But the question usually just states that the host knows where the prize is and you are expected to come to the unintuitive conclusion that they don't actually want to win.

The gamblers fallacy is also amusing in that it is almost the exact opposite of what people often think it should be. If you have an extremely unlikely string of "red" on the roulette wheel, rather than think that "black" is due, you should start to consider that the wheel isn't fair. So a gambler who thinks he is due for a win is far more likely to be just getting scammed.

Another way of looking at regression to the mean is basically the law of big numbers... infinity+1 is still infinity so no matter what the starting offset is (streak of improbable events), given enough rolls, it'll be irrelevant and you'll end up with the statistical probability.

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

#42

Post by joe_the_user »

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.

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-described, the chance of the prize being behind one of the non-picked doors remains the same, 2/3. And since you know now which of these door doesn't have the prize, the chance of the final unpicked door having the prize is now 2/3.

The host has to pick a door with no prize for the described situation to happen but if the situation happens, your knowledge doesn't depend on the host's knowledge.

Maybe there's some world where the host picks at random and so 1/3 of the time, the door with the prize open and you then know with 100% certainty where the prize (though whether anyone gets a prize is ambiguous, to say the least). But that revealed-prize situation isn't what the problem describes.

Edit: The only way that host knowledge matters is if the host can choose whether or not to offer this particular deal based on whether the contestant picked the correct door originally. If we assume some optimal game theoretic counter-play on the part of host, then I would guess the host's action give no information and so there's no reason to switch. But that's a bit far from your comment.

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

#44

Post by nkurz »

thom wrote: post #34
bryanlarsen wrote: post #19

Earlier quoted context omitted.

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

Always happy to see this Motie idiom still in use!

I was familiar with the phrase, but not the origin. Since "motie" is somewhat hard to search for (Ministry of Trade, Industry and Energy?), here's a reference page in case others are interested:

In the science fiction classic The Mote in God’s Eye, we are introduced to the Moties, a species that is trapped near a star without interstellar travel, but that has evolved extraordinary abilities. Anatomically, the Moties have two normal hands … and a third one, which gives them the ability to make and manipulate tools with great dexterity. This leads to the idiom of “on the gripping hand,” following “on the one hand or the other,” and suggesting another way, a better alternative.

https://ellingtonlab.org/blog/2014/12/1/on-the-gripping-hand

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

#45

Post by davemp »

syki wrote: post #9
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…

In the Monty Hall problem the host always knows where the prize is. It’s a mind blowing problem because usually a person’s intuition is wrong. For almost everyone they need to carefully analyze the problem in order to understand why their intuition is wrong. There are lots of such examples in math.

Most peoples’ intuition is that the host wouldn’t always offer the choice unless it’s more likely that you’ve selected the correct door.

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

#46

Post by chrismcb »

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…

It isn't the Monty Hall problem if the host doesn't know where the prize is. The problem is generally stated, up funny, the host knows where the prize is. The issue isn't poorly worded, it is that a lot of people can't understand how two doors don't represent half and half. You are correct, in that if that just randomly selects one of the two doors, occasionally he will reveal the prize, changing to odds. Which is why it is usually started that the host knows and selects she empty door.

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

#47

Post by psyc »

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…

HH is less likely than H (.25 vs .5), but HT is also less likely than H and by the same amount. HH and HT are equally likely. This extends to HHHHHH and HHHHHT, and so on. There’s no way to frame the gambler’s fallacy that makes it a better bet.

The probability that there will be one T somewhere in the sequence is a lot higher than the probability that there will be no T. That's only because there are a lot of somewheres for the T to be, but only one way for there to be no T. But once you have already seen 99 H, you still have precisely zero information about the next fair flip, and there's no way around it.

If you must make predictions about the streak as a whole, you have to lock them in before the first flip. Otherwise you piss off both Claude Shannon and Tony Soprano.

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

#48

Post by Kranar »

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.

Coming up red 10 times in a row is expected to happen once every 2000 rolls. Given 45 spins an hour that means a table will spin red 10 times in a row once a month. Given a casino with 20-30 tables you can expect to see it once a day. Hardly a reason to shake your faith.

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

#49

Post by mikenew »

joe_the_user wrote: post #42
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. 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 idea what's behind the doors, also picks one. Then the remaining door opens and reveals that there's nothing behind it. There's really no difference between your choice and the hosts.

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

#50

Post by RubyRidgeRandy »

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.

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 of choosing the same door is 50% as well.

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