Gambler’s Fallacy and the Regression to the Mean
theness.com
theness.com
> 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.
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.
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!
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.
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.
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?
Of course Hall himself said maybe I open a door because I want you to switch.
To get the subtlety here, you probably need to model it which is what I did. That's why I'm an IT bod these days, with a Civil Engineering degree and not a mathematician or statistician.
Even when relayed the right way and the wording isn't necessarily unclear or misleading per se, it is still framed in a particular way and context that pushes you to believe your choice is inconsequential, less about probability and math, and more for psychological games and dramatic effect in the context of a TV show.
In your example, the fact that the presenter opens all 98 doors and not just one is telling. If they only opened 1, and then told you to stick to your original or open one of the remaining 98 doors, that he picks for you, would you change?
This is what you're missing, I think. Most people don't even get to the point where they reframe the "would you change" question to "what are the chances my original choice was correct vs not". Most people get stuck on "He knows something; he clearly demonstrated very theatrically that he knew the car wouldn't be behind door No.2. And now he's asking me to swap my choice for door No.67. Why 67? Is he trying to trick me? Is this a psychological trick to get me to forfeit my door, which may be the right one? Should I trust him? After all, HOW IS MY DOOR ANY DIFFERENT THAN ALL THE OTHERS?"
(the fallacious thinking, obviously, lying in the last part here)
Does the host _always_ do this, or does the host have discretion about _whether_or_not_ to do this? This has not been stated in any of the problem descriptions here, and I believe it's highly significant. If the host can choose whether or not to do this, then there is the potential for mindgames. If the host cannot choose whether or not to do this, that must be stated.
What did you disagree with? Their understanding of the problem seems pretty solid to me.
Why would the host reveal the car when the contestant has a chance to switch doors?
Here's some explanation: https://betterexplained.com/articles/understanding-the-monty...
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
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.
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.
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?
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%.
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I know mathematically it's wrong. But I hate this problem so much. I am convinced is a philosophical / semantical issue.
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.
But 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.
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
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.
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.
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.
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
Even if you could, you already know that you're losing, unless he doesn't show you what's behind it.
If the game is still going, you know that he picked an empty door regardless of whether he knew he was doing it
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
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.
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.
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.
> 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.
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.
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.
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"
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.)
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
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.
I’m not convinced this has anything to do with the host’s _knowledge_ per se.
For example, consider a different version. The contestant chooses a door. The host then peeks behind a door the door contestant didn’t choose (without showing anyone else what’s there) at random. If the prize isn’t there, the host reveals that there is no prize behind the door. Otherwise the host forgets they peeked and repeats the peek-and-choose step until they randomly get an empty door. Once an empty door has been revealed, the contestant may choose to switch. In this version, the host has no secret knowledge, but the 1/3 vs 2/3 probabilities still happen.
That isn't what happening in the first simulation. The game doesn't end if the contestant chooses the prize door initially. The contestant simply chooses a door. The host randomly chooses between the remaining doors. If they reveal the prize, the game is restarted. If they reveal an empty door, then the game is logged as either "the original door was correct" or "the remaining door was correct". This is the definition of "the host chooses randomly".
> The host then peeks behind a door the door contestant didn’t choose (without showing anyone else what’s there) at random. If the prize isn’t there, the host reveals that there is no prize behind the door. Otherwise the host forgets they peeked and repeats the peek-and-choose step until they randomly get an empty door.
Picking random doors until an empty one is found absolutely the same thing as randomly choosing amongst the available doors that don't contain a prize. You can implement any function you like here. Either the host always opens a door that does not contain the prize, or the host has a 1/3 chance of revealing the prize and the game is restarted. There is simply no other possibility.
Not entirely. The issue is what, exactly, gets restarted.
Version A: contestant chooses a door. Then the host picks a door and, if the host finds the prize, restarts the whole game if they picked the door with the prize. There is a 1/3 chance the contestant finds the prize. There is a 1/3 chance the contestant does not find the prize and and host does not find the prize. And there is a 1/3 chance neither one finds the prize. So 1/3 of the time the game restarts and, conditioned on a completer run, the contestant’s first guess was right with 50% probability.
Version B: contestant chooses a door. Then the host picks a door and, if the host finds the prize, restarts just this step and picks another door. There is a 1/3 chance the contestant finds the prize. If this happens, the host won’t find the prize. If the contestant’s first guess was wrong, there is a 1/2 chance the host finds the prize on any given try, but the host keeps trying so the host will not find the prize when the loop is done. So, conditioned on a completed run, the contestant’s first guess is correct 1/3 of the time.
See the difference?
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".
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.
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.)
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?
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.
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?
Well, it's a restatement of the Gambler's Fallacy, but I'll taboo using that as an argument because it begs the question (but for reals!)
I thought about this for a while, and here's the best I can do for right now. It's like winning the MegaMillions lottery on Saturday, and then winning $1 on a scratch ticket on Sunday. You'd go running around screaming on Saturday, but you wouldn't on Sunday. (Would you?) Because the impossible already happened. Hitting $1 on a scratch-off is never remarkable (unless you're 8), including on the day after you won the MegaMillions. It's just a somewhat anti-climactic coincidence.
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).
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.
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 :[ ).
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.'
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.
> 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.
Your first example should say >=5 heads. (Also, your site says almost 23%, making it just over 45% for either via symmetry)
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.
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...
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."
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
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.
But if you see it run by carnies at the county fair where the wheel will be gone tomorrow?
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.
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.
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?
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.