Gambler’s Fallacy and the Regression to the Mean

https://theness.com/neurologicablog/index.php/gamblers-fallacy-and-the-regression-to-the-mean/

by allthebest • 5 years ago

68 90 5 years ago

roenxi2• 5 years ago

It is very annoying in business. Arguing that two outcomes are equally likely when one has happened many more times than the other is an exercise in frustration. It takes a certain level of courage, charisma and stubbornness to argue that there is no evidence that 2nd best is just as good as best when there is a measurable difference between two options.

This is a big contributor to why markets perform so well in my humble opinion. People who see past the gambler's fallacy to the actual odds of things working get rewarded, and the people who assume the current leader must be doing something different and better than the competition get punished.

When your variable disease is expressing its worst symptoms, you are likely to feel better in the future

This is the influence of time & the healing process, not a regression to the mean. Symptoms in a given sickness' progression aren't random variables in a practical sense. It goes no symptoms -> bad symptoms -> lingering symptoms -> better (or permanent symptoms, bad luck :[ ).

jldugger2• 5 years ago

Arguing that two outcomes are equally likely when one has happened many more times than the other is an exercise in frustration

Are they actually equally likely though? Perhaps there comes a point where one must consider if observed outcomes match with assumed probabilities.

Or are you just having arguments about whether the unfavored outcome is 'overdue.'

roenxi2• 5 years ago

Funnily enough, I might have been thinking of the exact opposite of the Gambler's fallacy. My mistake, sorry. But it still seems related to me and I'll explain what I was thinking anyway:

For reference: https://www.omnicalculator.com/statistics/coin-flip-probabil...

There is a 20% chance that in 7 coin flips there will be >5 heads. By symmetry, 40% chance that there are >=5 heads or >=5 tails.

If you look at a series of events (say, annual results) and one has exceeded the other in 5/7 years, that doesn't tell us that one is consistently better than the other. But people aren't good at accepting that.

contravariant• 5 years ago

This is a big contributor to why markets perform so well in my humble opinion. People who see past the gambler's fallacy to the actual odds of things working get rewarded, and the people who assume the current leader must be doing something different and better than the competition get punished.

The problem with markets is that rewarding the people who can predict the market correctly isn't a particularly useful end result. What you wanted to happen is for the people who are best able to predict the market to determine the 'fair' price, but that's usually only guaranteed as a steady state which fails as soon as a market is even remotely volatile.

chx• 5 years ago

“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, how do we resolve the apparent tension? If yes, isn’t it the case that there must be “some” validity to the gambler’s ‘fallacy’?”

To this my answer would be: if you are doing millions and millions of spins then you should see roughly 50/50 but that just means over a few million spins it will balance out -- and even then it's not usable for predictions because it's not exact. Also, in a similar long sequence you would expect that in 2046 spins you see 10 of a single color in a row. That's not such an outrageously big number...

zerof1l• 5 years ago

Fun fact, the actual roulette wheel in casino has green zero. When betting on any color, you will always have less than 50/50 odds.

lloydgrossman• 5 years ago

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

bryanlarsen2• 5 years ago

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

thom• 5 years ago

Always happy to see this Motie idiom still in use!

Kranar• 5 years ago

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.

compiler-guy• 5 years ago

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.

alan-crowe• 5 years ago

I find it interesting that saying "a black spin is needed to push the ratio back towards 50/50" feels like a sensible thing to say. What role do context and framing play?

Consider the first one hundred spins breaking 60 red, 40 black. That is a little odd, but not outrageously so. The ratio is 60/40 = 1.5

Suppose that the next one hundred spins break 50/50. There are no surplus blacks to push the ratio towards 50/50. So what is your intuition? It seems clear enough that the totals are 110 red, 90 black, for a ratio 110/90 = 1.22...

Having imagined the next hundred spins vividly, we are comfortable saying "Unless the surplus reds keep coming, the ratio is going to drift back towards 50/50 all by itself.". There is a striking clash between different "things we might say" seeming sensible, depending on the vividness and detail with which we imagine the situation.

mikenew9• 5 years ago

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 prize), and then ask if you want to change your choice to the other unopened door. If you change your choice your odds of winning go up from 1/3 to 2/3.

which to me makes intuitive sense. Every other time I've encountered this it's usually presented in a very ambiguous way, where it isn't clear if the host has any information about what's behind the doors. 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 guess I just want to say that I think the Monty Hall problem is dumb. It's not some mind blowing truth of probabilities that human brains can't grasp; it's just a poorly worded problem that causes you to make incorrect assumptions.

You could argue that if you don't know whether or not the host has any knowledge about where the prize is you might as well switch, because if they do you've improved your odds and if they don't you haven't made things worse. But still, the "problem" itself doesn't seem that interesting.

syki• 5 years ago

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.

davemp• 5 years ago

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.

mtlogstdo• 5 years ago

I don't really understand how you think the game could work if the host didn't know where the prize is? They'd keep accidentally revealing the prize, and then asking you whether you want to switch to the other remaining door, at which point... what's the point? The game is over.

I've not usually seen it explicitly stated like this that the host knows, but it's always seemed obvious to me that they must.

seoaeu• 5 years ago

There's an alternative version where the host opens a door at random. If it is a goat, they proceed the same as the normal version. But if it reveals the car, the host says, "Sorry, the car was behind door C. You lose."

And a third version where the host only asks you to switch if you initially picked the car, and otherwise always ends the game early.

The best part is that during the game you can't actually tell which rule set you are playing with!

HWR_142• 5 years ago

you change your choice your odds of winning go up from 1/3 to 2/3 which to me makes intuitive sense

Well, my guess is that's because you've fixed your intuitions either by prior exposure to the problem or education. Certainly I think it's intuitive - now. The point is most people's intuitions think it changes from 1/3 to 1/2, regardless of whether you switch or not.

But hey, maybe Bayesian inference is intuitive to you without any math education.

Meanwhile, when this was published in 1990 and thousands of people (10% with PhDs) wrote into the magazine that it was an error. So some people have a very strong contrary intuition.

gerdesj2• 5 years ago

"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. In which case you are far brighter than the followers of a maths column that introduced this problem back in the day. Ah, let's get to the canonical source of knowledge: https://en.wikipedia.org/wiki/Monty_Hall_problem (para 1)

Years (>30) ago I ran a sort of Monte Carlo simulation on this thing. It took a while to encode the rules in GWBASIC and luckily I had a copy of Peitgen's (both Beauty and Science of Fractals) and I was able to get roughly or reasonably random numbers to do the simulation by following his algorithm for generating them! After a while of running on my 286 I got a result. Then I spent a few hours fixing bugs. It took a few more iterations until I got an answer that looked right or at least was between 0 and 1 and didn't end up in syntax error 8)

Dr Ian Stewart gives a great description of the problem in one of his books too.

mikenew9• 5 years ago

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 is, then opens 98 other empty doors, leaving one. You know that the odds were 99 in 100 that the prize was behind one of those other doors, and the host just eliminated 98 of them.

...what am I missing here?

almostdeadguy2• 5 years ago

The Monty Hall problem absolutely becomes more obvious when the host's behavior is explicitly documented. And as this article points out, many key aspects of the host's behavior are implausible, which is a large part of what makes it counter-intuitive: https://ima.org.uk/4552/dont-switch-mathematicians-answer-mo...

zeroxfe• 5 years ago

Wow... okay, I'm pretty sure you don't actually understand the Monty Hall problem.

bhelkey2• 5 years ago

I'm pretty sure you don't actually understand the Monty Hall problem

What did you disagree with? Their understanding of the problem seems pretty solid to me.

WoahNoun• 5 years ago

I've encountered this it's usually presented in a very ambiguous way, where it isn't clear if the host has any information about what's behind the doors. If the host is choosing a door randomly

Why would the host reveal the car when the contestant has a chance to switch doors?

mhb3• 5 years ago

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...

jodrellblank• 5 years ago

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?

bhelkey2• 5 years ago

Your intuition is wrong and it doesn't matter what the host knows before showing you the door that doesn't have the prize.

This is not correct and your source doesn't support your point

From your link:

Monty helps us by “filtering” the bad choices on the other side

Monty knows which door holds the car and which holds the goat

LesZedCB• 5 years ago

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.

RubyRidgeRandy2• 5 years ago

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.

alecst• 5 years ago

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.

joe_the_user4• 5 years ago

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.

mikenew9• 5 years ago

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.

chrismcb5• 5 years ago

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.

tzs• 5 years ago

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.

goto11• 5 years ago

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.

galaxyLogic4• 5 years ago

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"

amluto4• 5 years ago

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 awkward outcome in which the host opens the door with the prize.

Even if the host did open the door with the prize on occasion, the outcomes would still be the same if the rule was that the whole game starts over if the host accidentally reveals the prize. (Well, the game would take longer to finish, but that has no effect on the probability of winning.)

mikenew9• 5 years ago

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

Tarsul• 5 years ago

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.

windows2020• 5 years ago

For an honest roulette wheel (ignoring 0 and 00):

What would you say about the logic of betting the next 100 spins will be 50% red and 50% black?

Would you say something different about betting on black after 50 consecutive reds?

gexla4• 5 years ago

Look up Nassim Taleb on Fat Tony about flipping coins.

If 50 consecutive reds come up on a roulette wheel and you bet on black each time, then you might be a sucker. Someone was lying to you about that wheel being honest.

gexla4• 5 years ago

Yet the illusion of streaks or certain outcomes being “due” is powerful, and such thinking is almost ubiquitous among gamblers. It derives from our tendency toward pattern recognition (apophenia), seeing illusory patterns in random noise.

And yet, pattern recognition is also what you would draw on to see that this is an illusion? Maybe you just need to know the way the game works rather than guessing?

Then continues droning on about patterns.

No mention of volatility or what "regression to the mean" actually looks like (the casino's house edge.)

I get the sense the person who wrote this never gambles. For a much better explanation, find writing from poker players. Also excellent on this subject is Nassim Taleb.

I recommend everyone put up some tuition (losses at the table) to learn poker at a low stakes table. The game will teach you a ton about life.

coldtea4• 5 years ago

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 wavy.

"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."

Who said the opposite?

The gambler making the gambler's fallacy basically says that after successive black streaks it's more likely to see a red. They aren't saying that there's some "power in the universe" ensuring it. They're saying it's just more probable.

And in a way it is: assuming a non-biased roulette, each successive repeat N-same-streak should be increasingly less likely.

Taking all the groups of N rolls and showing that any N-same-streak going is just one of the possible permutations same as any other (and thus any N+1, N+2, N+3, etc. would be too) doesn't really cut it.

If you happen to actually see a 1000th black roll streak in a casino, you'd better suspect the roulette is rigged/biased, than naively think that "well, even with random independent events, there are bound to be large streaks".

rout39574• 5 years ago

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 it's random, then you have no recourse but to post-hoc description. There is -no- predictive weight.

... And editing your comment to refer to groups of throws is disingenuous. It would have been more honest to modify your case in a reply.

gexla4• 5 years ago

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.)

coldtea4• 5 years ago

If it's random, then you have no recourse but to post-hoc description. There is -no- predictive weight.

Well, a "post-hoc description" of empirical observations is how we make predictions. We gather cases (and their probabilities) post hoc, and predict future results based on that.

Why shouldn't it be applied in this case, especially when here not only we can calculate the probabilities post-hoc, but even use probability to pre-calculate those post-hoc probabilities for every series of streaks?

The only way you can reasonably expect the past streak to have an impact on the next spin

Regression to the mean is a reason to expect past streaks to have an impact on future spins. Doesn't have to be some force that specifically affects the individual next spin, but it's a probability that affects (and moves toward the mean) the outcome of next spins.

Intuitively speaking, why shouldn't the gambler use it?

If they shouldn't, "because spins are independent" seems too hollow as an answer, because we have independent spins but also a statistical property for their aggregation.

I guess we could test it empirically: in a fair sequence of coin tosses, would a gambler always betting on tails after seeing N (for e.g. N > 4) consequtive heads have an advantage, over one always betting randomly on the same N+1 toss?

But regardless of that, my point is that intuitively is quite justified, and the explanations why it's not seem more hand-wavy.

Try this on: What is the "mean" red-black on the roulette wheel? There isn't one.

Well, (ignoring non red-black case), there's not a mean value, but for the distribution there's an expected long term tendency of a balanced red and black outcomes.

... And editing your comment to refer to groups of throws is disingenuous. It would have been more honest to modify your case in a reply.

I didn't modify my case, I expressed it more clearly. What exactly was disingenuous? I more often than not post a quick comment (HN habbit, as sometimes old edit forms expire), and then re-read it and quickly improve it as I re-read it, find typos, better wording, etc. Didn't even think anybody would read it and answer in the time it took me to refine it.

And, how about it, you can always answer to my final, improved, formulation. Does it have any historical interest if you answer to a worse worded case?

psyc2• 5 years ago

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.

coldtea4• 5 years ago

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.

Well, kind of. There are 2^6 patterns with 6 tosses, and only a few of them have some recognizable pattern (are compressible). HHHHHH is the worst of them. It might be "equally likely" to any other 6-toss pattern seen from a fair coin, but it's also a great example to make one suspect a totally rigged (same sided H) coin.

Now, on average a fair coin would have roughly balanced H and T. So the (much more numerous and thus probable) series where H and T are mixed (e.g. HTTHHT..., HTHTHH..., etc...), are more probable than a large streak of H.

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.

A, my sentiments exactly!

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.

That's my argument: you don't have zero information.

You know that the next fair flip is random - sure.

But you also have the information that large series of tosses end up balanced in the long run, which is a kind of push (given fairness) towards making a T after tons of Hs increasingly likely.

Why should this be discarded?

chrismcb5• 5 years ago

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.

coldtea4• 5 years ago

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

Well, the point is:

"black if a fifty fifty chance" and "the next infinity sounds half will be black"

doesn't sit well with:

"if there were three blacks in a row, it doesn't mean red is more likely".

The roll is independent, but in some sense either the roulette is unfair or red will increasingly be more likely.

In other words, since there is a statistical property working across many/infinite rolls (regression to the mean, black/red equally likely), this property makes sense to also somehow affect the possibility of individual rolls (as individual rolls are still rolls within an aggregate).

tbp105• 5 years ago

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.

intrasight• 5 years ago

Remember that the Monty Hall example even tripped up Paul Erdős

chrismcb5• 5 years ago

ESPN used to talk about the regression to the mean a lot. And it always bugged me, as you are talking about skilled players. Perhaps a player's skill improved

galaxyLogic4• 5 years ago

Gambler's Fallacy is caused by the artificiality of roulette table. In nature and for the hunter-gatherers there were no roulette tables.

Gambler’s Fallacy and the Regression to the Mean

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