Submitted by Reason-Local t3_11de5ag in explainlikeimfive
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Submitted by Reason-Local t3_11de5ag in explainlikeimfive
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I remember arguing with someone about this regarding seeing a plane crash at an airport and then getting on another plane. Others were talking about how they wouldn't get on their own flight after seeing that. This person said that it would make them feel safer, because "what are the odds that two planes crash on the same day?" I said they weren't getting on two planes though, they were getting on one, and the odds of one plane crashing didn't change when another crashed (of course assuming that all plane flights are equal).
The irrational mind sees a plane crash and goes "Fuck that, I'm not getting on a plane now, it's likely to crash."
The rational mind says "what are the odds of two planes crashing?"
The reality is somewhere in between. The chance was always there, but it didn't go up or down simply because this event occured.
The odds probably went up slightly. Contaminated fuel, or bad weather, or technical problems with the ground systems, or terrorism, could cause multiple plane crashes. (Though I don't think that has ever happened accidentally... yet. Coordinated terrorist attacks have happened, such as 9/11).
Edit to add: There is also the risk of a design flaw on the aircraft. E.g. 737 Max. Or a consistent manufacturing flaw.
Or it could go down as everyone involved is now being more cautious.
I forget the exact scenario that inspired the conversation, but the person I was arguing with was definitely talking about the "odds" in the context of simple odds in the manner of "if a one in a million chance thing just happened, it's now probably closer to one in two million it happens again."
Odds only change when directly linked.
So if you are gambling on a dice roll and you bet that it will land on six, it's a one in six chance. If it does, and you bet on six again, the odds are still just one in six.
To change the odds, you would need to bet that it lands on six twice, because now you have linked the two rolls before they happen.
In the case of the plane crash, the odds of two planes crashing for independent reasons on the same day, involving the same airport, are much higher than a single plane crashing. Unfortunately for our hypothetical travellers, one plane already crashed, bringing the odds back down to the standard single-event odds.
The issue when it applies to the real world is that you rarely get truly independent events. And also probability is based on assumed knowledge (or lack thereof), so gaining information changes probabilities.
An example is you have a shuffled deck of cards, and you're betting on the odds of the top card being an Ace of Hearts. The probability at this point is 1/52. If you happened to notice a flash of red as the dealer was shuffling, nothing about the deck changed, but now your odds are 1/26 because of additional knowledge.
If the first place crash revealed some information that wasn't available before, that can change the probability calculation, even if nothing else changed about the situation.
Also in a theoretical situation you accept the circumstances presented as fact. In the real world, those assumptions could be wrong, or someone could be flat out lying to you. Like if someone flipped a could 25 times and it landed on heads every time, mathematically you'd say you still have 50/50 for the next toss. But you are assuming that it is indeed a fair coin and not a weighted or double-headed coin. If you want into someone on the street taking bets, you have to factor in not just the odds at face value, but also the odds that they're cheating.
Which is why they investigate aircraft crashes so thoroughly? So that a slight change in design, processes, procedures etc will further the body of knowledge about aircraft and the odds will decrease
It's the basis of all safety, I suppose.
Again, this was a discussion about theoretical simple odds. The real world isn't a coin flip, and like you pointed out the odds may be weighted in ways you don't know. But yeah, this was about him thinking that a die landing on six made it less likely to land on six again.
Right, the OP was talking simple odds, but the thread shifted to talking about plane crashes I think we exit the realm where simple odds apply. It makes sense when we're talking about dice.
I only brought it up because the guy I'm talking about was specifically talking about it in terms of simple odds, not arguing the fact that air travel is a very complex thing.
It was very much an application of the gambler's fallacy to a real world event, one that has extremely hard to quantify odds with countless variables to begin with.
Or “this airport has a shit mechanic/maintenance crew”
Here's a case where there might be a change in the odds. Usually, after a crash, there's a period of increased vigilance by ground crews to be on the look out for maintenance issues and speeding up checks of various systems. It's almost certainly safer to fly in the weeks following a high-profile crash.
Again, this discussion was very much about simple odds, not the complex odds of all the multiple factors. Dude even ironically used the words "gambler's fallacy" to describe what he thought I was doing by saying that one outcome does not effect the next.
In the real world, yes, there are connections between complex events like plane crashes, and the "odds" aren't static like dice.
Ok but what are the odds of 2 planes crashing from the same airport on the same day? That would would have a lower possibility right? If u saw a plane crash at the same airport u we’re taking off from hypothetically.
The odds of two planes crashing while flying in/out the same airport on the same day are far lower than the odds of a single plane crashing while flying in/out of that airport.
But.....
The odds of a plane crashing via that airport are not any lower after a plane has already crashed there, assuming in this hypothetical that plane crashes have specific odds and are not very complex events that are extremely hard to actually put "odds" on.
That's defective thinking as well.
Given that one plane crashed, it is much more likely that another plane of the same type will crash.
EG: a design mistake causes undue stress on a part, that part wears out faster that it should. That part breaks mid air and causes a plane crash. It is likely that the parts in all the similar planes has also been under stress, and might be breaking soon as well.
This is why the FAA tends to ground all planes of a certain kind until they figure out it wasn't the plane causing the issue.
See also the Boeing 737-800 MAX
See my other response.
But also think that there are all sorts of reasons planes crash, and all sorts of planes. A plane crashing at a large airport often does not significantly slow air traffic outside that directly impaired by the use of that runway or airport, depending on severity.
And again, see the part about the Boeing 737-800 MAX.
Sometimes its a problem with the whole class of plane.
That's the most recent example I can name off the top of my head.
There are many many examples of this.
You can keep going to see sometimes how changes in process/certification lead to the crashes.
And sometimes that first crash is the sign those problems are starting to come home to roost.
Yes, but that was not the conversation we were having.
And the grounding of whole fleets based on one event immediately after it has happened without reason to think it was a flaw in the design of the aircraft itself is not exactly common. When a plane overshoots the runway on landing due to pilot error, they don't ground entire fleets of that airframe.
Planes crashing is very complex thing and the "odds" are as well. It's not remotely static like dice.
That part was also part of the conversation I had with the guy in trying to explain the concept of odds to him.
Unless the crashing plane damaged the departing plane & compromised its ability to function properly. I'll see myself out.
But what are the chances that two different planes will crash at the same airport in one day?
Very low. Once one plane has crashed they’re likely to close the airport. If there are no planes taking off or landing, the only way for a second crash to occur is a plane falling out of the sky directly at the airport.
Nah but ignoring all that.
Like say it crashed but the after effects weren't taken into account.
Barring outside influences, and assuming the odds are static, like dice, exactly the same as one plane crashing there, once one already has.
But not before right?
I'm not arguing, this just always kind of messes up my head.
Like the changes of me winning two lottery being the same because the chances of winning the lottery are the same for each ticket.
But how likely is it that someone who plays the lottery all of their life wins, and how likely is it that they win twice?
The odds that they win twice in a lifetime is much more unlikely than the odds they win once.
But after they won once, the odds of winning again are exactly the same as if they had never won.
Think of the dice example. There is a one in six chance of rolling a specific number. Rolling again, there is still a one in six chance of rolling that same number. The number of sides hasn't changed, the number you chose didn't magically disappear.
The odds only change if you bet that you will roll two in a row before the first roll, because you are now betting on both events before they happen, not on a single event happening. The events themselves have no influence on each other.
Ah I get it, thanks a lot. The lottery chances "resetting" after the first win actually makes sense to me.
I think that's where my head gets confused. If I saw a plane crash I'd think of it as "well there's no chance of two of that thing happening today!" Rather than " it's still just as likely to happen today."
Glad I could help ya figure something out! It's definitely a bit counterintuitive at first, and I once had the same confusion. Cheers!
Do note. The guy seems to be talking about simple odds using real plane crashes for some reason.
Real world statistics do not work that way.
In the real world plane crashes are not independent events.
In the real world you knowledge of the odds of plane crashes is not complete.
A plane crash will cause the ground crew and flight crew to change their behavior, shifting the odds.
A plane crash is evidence that your assumptions on the safety of a given model of plane might be incorrect.
In the magical world of simple odds, safety audits and groundings wouldn't make a lick of sense. They do in the real world.
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> There's nothing connecting one roll to the next; they are completely independent.
...under the assumption of a completely fair die. (An assumption you are usually making in a statistics class.)
In practice, though, the fairness of the die may be in doubt in many real-world scenarios.
>>There's nothing connecting one roll to the next; they are completely independent.
> ...under the assumption of a completely fair die. (An assumption you are usually making in a statistics class.) > > > > In practice, though, the fairness of the die may be in doubt in many real-world scenarios.
You may have quoted the wrong passage there, because even without a fair die it still holds true. Whether the die is weighted towards a 6 or not, the individual rolls are still independent from each other, merely the probabilities of the outcomes are different.
> Whether the die is weighted towards a 6 or not, the individual rolls are still independent from each other, merely the probabilities of the outcomes are different.
The rolls are, but provided you have any uncertainty about the underlying probabilities, your beliefs about those rolls (and your expectations about the future rolls, which is the exact same thing) should be updating with each roll.
For a simple example, imagine I have two coins. One is loaded to always land heads, the other is fair. I pull one of the two from a box at random, and I do not know which I pulled. I want to estimate the probability of my next flip being heads. It's 75% in this case (50% to be loaded * 100% if it's loaded + 50% to be fair * 50% if it's fair).
I flip the coin, and it lands heads. This is evidence in favor of me having the biased coin. Specifically, I should update my probability that the coin is biased (using Bayes' rule) to:
P(loaded | heads) = P(loaded and heads) / P(heads) = 0.5 / 0.75 = 2/3.
Now I want to estimate the probability of the next flip. There is now a 2-in-3 chance (or more properly, that is my correct Bayesian estimation of that probability) that I am holding a biased coin, so the probability of the next flip being heads is 5/6 (it's 2/3 * 1 + 1/3 * 1/2 = 2/3 + 1/6 = 5/6). This is not equal to my original 3/4, even though the flips themselves are IID, because their underlying distribution depends on an unknown parameter about which I am gaining information.
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But if you roll a 6 ten times in a row, you could just have an unfair die. So in theory it's 1/6 always. But in the real world I would bet on it being a 6 the eleventh time 🤷
The probability of rolling a six on any fair die roll does not change. It will always be 1/6. The idea that the probability goes up if you haven't rolled a six in a while, or down if you just rolled a six, is known as the gambler's fallacy.
This. The odds of rolling 6 6s in a row are pretty n low (like less than 1/1000). But once you've rolled 5 6s in a row the odds that the next one will be a 6 are 1/6.
Probability only looks forward
>Probability only looks forward
I like this phrase. I need to try and remember it.
Another way of saying this is that "dice don't have a memory"
It's not always true, events can be dependent
Sometimes, poker is a good example, assuming you're playing with a 52 card deck and there's 5 people playing and you're 4 flushing and waiting for the river to hit. You'd think that since you're 4 flushing your odds of hitting the flush are 1 in 4 since the river card can be 1 of 4 suits.
But it's actually slightly less because if you're 4 flushing then that means a disproportionate number of your suit were already played.
It doesn't throw it off MUCH (so if you're 4 flushing you should probably chase it, especially if you're pocket suited) but it is LESS than the 1 in 4
That's one example. It makes quite a difference tho, 6%. Also, whether you should call or not would depend on the bet size and the pot size. Getting slightly over 4 to 1 would be break even, not considering implied odds.
>Probability only looks forward
-ish. Bayes' theorem shows how probability of an expected scenario changes based on what's happened so far.
Your 6 6s scenario is a great, if somewhat trivial, example of this: the odds of rolling 6 6s is (1/6)^6, but the odds of rolling 6 6s given that you already rolled 5 6s is (1/6).
Things get slightly more complex when events are not fully independent, though the fundamentals of "probability of X given Y" is still basically the same.
>example of this: the odds of rolling 6 6s is (1/6)5, but the odds of rolling 6 6s given that you already rolled 5 6s is (1/6).
Your example supports the previous assertion, though. The probability of 1/6 is the probability of only that final roll. There's no difference between "the odds after 5 sixes" and "the odds after zero rolls" and "the odds after 6000000 sixes in a row." It's not looking back at all.
As I said, the probability of any single roll being a 6 is 1/6, because each roll is a fully independent event.
However, the probability of rolling 6 6s in a row varies based on the prior events. After 5 6s are rolled, the final roll is still a 1/6 event, but after, say, 3 6s, the probability is (1/6)^3. P(6 6s) ≠ P(6 6s given 5 6s). The given part is the piece that looks backwards.
>However, the probability of rolling 6 6s in a row varies based on the prior events.
I mean, sort of, but that's not what you're actually calculating.
>After 5 6s are rolled, the final roll is still a 1/6 event, but after, say, 3 6s, the probability is (1/6)3.
Yes, but no. That's still just the odds of rolling 3 in a row. That calculation does not care about or consider the previous 3 rolls at all. It's indistinguishable from just checking the probability of 3 rolls, because that's all it's doing.
You're not wrong in removing the previous 3 rolls from the calculation, but that's exactly what they mean by "only looking forward." You literally need to ignore the earlier rolls to calculate the future outcome. You don't look back.
>That's still just the odds of rolling 3 in a row.
Technically yes, but also technically no. It's similar to saying (2*2) is the same as 4. They are equivalent, but also technically different. P(3 6s) is equivalent to P(6 6s given 3 6s), but the latter has information about prior events that the former doesn't; it also is describing a different set of events.
You can only ignore the earlier rolls because each roll is independent of the rest. If the event in question is not fully independent, then it will alter the probability in a way that doesn't let you simply eliminate the prior events. This is what I meant by the dice example being trivial; because they're independent, calculating the future probability is trivial, regardless of the given priors.
This is literally Bayes' theorem. P(A|B) = (P(B|A)*P(A))/P(B)
. In words, the probability of A given B is equal to the probability of B given A times the independent probability of A all over the independent probability of B. If you plug in the dice example:
Again, trivial, but shows that prior events can be used in the calculations of future events (even if they are independent and don't actually impact the result).
It's trivial because they're independent. Bayes theorem applies to dependent events. Dice rolling isn't dependent. You're talking in circles to justify using incorrect logic.
"The 1998 Superbowl championship isn't related to how many Doritos I ate yesterday, but if it was here's how I'd twist this logic diagram to explain it."
That's what you're doing here.
First off, I have to correct an earlier mistake: a single die roll is fully independent from other rolls, but a sequence is not. If the ask is about the probability of rolling a sequence and you've already rolled some, then you can't ignore what's been rolled so far. It doesn't affect the next roll, but it does affect the probability of the sequence overall.
As for your nonsensical question, you could still apply Bayes' to independent events, but P(B|A) is just P(B), which cancels out with the denominator, leaving just P(A|B) = P(A) (which is to be expected, by definition).
Going back to the dice sequence: if dice sequences were fully independent, then P(3 6s given 6 6s) wouldn't be 1, but be (1/6)^3 , by definition. But that's silly, because we know the given here is that we already rolled them.
But we're now talking in circles. You insist on updating the expected scenario at each step (first, you ask about 6 6s, then about 5 6s, then 4 6s...) while I keep the information along the way (first, 6 6s, then 6 6s given 1 6, then 6 6s given 2 6s...). The answers were get are the same, but my notation preserves the original ask and the history of events.
The preservation of information between rounds is important, because it will impact the probablilty of arriving at that original end goal. The Monty Hall problem is a classic example of this: if you treat the second decision as independent of the first, then the odds for each door from your perspective becomes (1/2); but if you include the given information about prior events, then the odds of your first door being right becomes (1/3) and the other door becomes (2/3).
>If the ask is about the probability of rolling a sequence and you've already rolled some, then you can't ignore what's been rolled so far.
This is exactly what you do. If you want to know the probability of rolling 6 6s in a row, you calculate it for that. If you've rolled 3 out of 6 and want to know the probability of making it to 6 straight, you do the very basic 6 - 3 = 3, and then crunch the probability for only 3 rolls, because you are only actually calculating the probability of 3 rolls.
Your "preservation of information" is a simple subtraction. It's not some mystical connection influencing future rolls. All those future rolls are still entirely independent of the ones you've already made.
The probability of rolling the same on 6 dice, and the probability of rolling 3 of the same having already rolled 3 are identical, because you're still rolling 6 dice.
The probability of rolling 3 more of the same after any arbitrary amount is not the same question. This is where your confusion is coming from. Here, you only need the probability of 3 rolls, no matter how many you've rolled previously.
You didn't really answer the question imo. You are just repeating what they said, it doesn't go up or down. Not explaining why.
The probability of rolling a 6 on a die depends on the die; if it's perfectly manufactured then it's 1/6, if it's crooked then it's different than 1/6. The probability doesn't change because in theory the die does not change physically, when you roll it.
In practice, dice can get chipped / worn out, so the probability CAN change. But for the purpose of statistical problems, you're considering "perfectly manufactured" dice that are brand new.
Wow, you are saying if the dice are warped they no longer behave like perfect dice!? Incredible insight. /s
The insight here is that, provided you are not certain of the fairness of the die, each roll would give you information about its fairness (and therefore about future rolls). The rolls themselves are still independent even on an unfair die, but you will develop progressively better estimates of the actual underlying probabilities.
(Of course, in basic statistics we ignore this sort of thing and just assume the die is fair or loaded in some known way. But this is an important caveat in real-world statistics!)
Coin toss is an easier example. The coin always has a 50 percent probability that it will land heads or tails on any individual toss.
But when you observe how often a coin will land heads N number of times in a row, we see that as N increases, the probability of success decreases. This is because as you toss coins, each toss brings with it the collective probabilities from all previous tosses.
So while each toss always has a fifty/fifty chance of landing heads, trying to predict how often a particular number of heads in a row will occur depends on how many coins you intend to toss, and the probability that each toss will land on a particular side.
I've always understood it better with a coin toss. The odds of getting heads 9 times in a row then getting tales on the 10th flip is the exact same as getting heads 9 times in a row then getting heads again on the 10th flip.
In the scenario you gave, the reason that it doesn’t increase the probability is the because the other sides don’t go away after rolling. For example, if you roll a 10 on a sixteen sided die, that 10 still stays on the die for any subsequent roll attempts, so the amount of sides stays the same, and each number has a 1/16 chance of being rolled.
Probability can change in certain situations, for example a raffle. If 200 names are put in a hat, and one Is drawn, that name is now removed, and your name has now got a 1/199 chance of being drawn. Make sense?
Where you getting these d16s from?
Out of my ass lol just for the examples sake. I also realize it would’ve been easier to say a 6 sided die, but I just woke up so leave me alone 😅
Because the dice doesn't remember what's happened in the past. It's an inanimate object that doesn't undergo any significant change after being rolled.
Can you imagine if the opposite was true? If dice remembered the past and tried to correct for it? How would you know if you're holding a normal dice or dice that (unbeknownst to you) has recently rolled 6 100 times and is now determined to roll something different?
Probability (and science in general) hinges heavily on semantics. The chance to roll a 6 on a die never changes (1/6), but the chance of having 10 consecutive non-6 rolls is relatively low ([5/6]¹⁰=.162), so the chance of rolling exactly one 6 in a string of 10 rolls is 1-.162=0.838. So if you've had nine non-6s it feels like there is an ~84% chance of rolling a 6 on the next roll, but it is still just a ~17% chance.
0.838 is the probability of having at least one 6 among your ten rolls. You need to get into binomial distributions for the probability of a specific number of events. Exactly one six somewhere in the sequence is probability 0.323.
You are correct. I should have said at least 1 six. But the logic still holds for the types of arguments people try to make about 9 non-6s in a row, so the 10th must have a higher probability, but in-fact it doesn't.
Had to scroll way to far to get here.
You are mistaking this for the probability for a set of events. An event in this case is a roll on a dice - 1/6th chance to roll a number. The chance never changes. BUT the chance of rolling any other number decreases the more you add the events into the set. But that's the probability for the whole set.
Set of 1 roll is 5/6 (times 100 if you wanna see it in % ) which is roughly 83,33 %
But set of 10 rolls (any other than 6) is 5/6 ^ 10 which is roughly 16,15 %
Now from this it seems like the chance is getting bigger, but what's changing is the set you're tracking probability for. The rolls are still 1/6th for a number :)
edit> to give you maybe a better example considering my English and explanation skills :D
if you roll 9 times any other number than 6, then the next roll being any other number while considering the whole set is 16,15 % and the roll for the 6 thus increased to awesome 83,85 %! But that's ONLY because we make the probability based on the previous rolls. The chance the next roll will be any other number on its own is still 5/6 and the chance for the 6 is still 1/6.
I've understood the math on this for a long time, but can someone here help me understand....... if you rolled a 6 on 1,000 rolls simultaneously, aren't you forced at some point to grapple that the odds of that streak are incredibly, incredibly low, and the likelihood that it will be broken is a fair bet to take on that information alone?
Like, if someone said before the rolls "here, I'll give you a billion dollars if that dice rolls 6 1000 times", you inherently know that that is highly unlikely to happen.
I get it, each roll is independent, but the "gamblers fallacy" doesn't seem like complete fallacy in this scenario. Someone please help me close the gap on my understanding.
I'm guessing someone is going to answer, "You have the same chance regardless of 1,000 rolls". I know. But in a way you're betting now with historic information that 1,001 rolls in a row is likely enough to bet on it.
If 1,000 isn't enough to get us past the "independent roll" answer, let's go with one trillion. One quintillion rolls. I don't care. At some point you would question the validity of probabilites and all that, right?
If a die rolled the same 1000 times in a row, than practically it is far more likely that the die is rigged. But the way that probability theory is discussed wouldn't account for such a thing that isn't specifically mentioned in the problem presented. A real life situation has many more variables than a hypothetical one.
Because "not 6" is 5 times more likely to happen than 6 on a six sided die in a single throw. You're focusing on the one combination result that is 6 for all 1000 dice rolled and comparing it to the many combinations that are not that one single very specific result you want.
If the billionaire gave you a die to roll 1000 times to get a single specific order of numbers like "123545453621145..." you'd still have the same horrible odds of getting that one result.
Now if it were 1000 die to roll and the result must be all 6s except one single roll anywhere among the 1000 rolls, your odds of getting that goes up because there are more than 1 results that match that criteria.
Make it must get only one 6 anywhere out of 1000 throws/die while the rest are not 6 and your odds are much better because there are even more results that match that criteria.
Going back, if you had already thrown all 6s for 999 throws and you only have to throw the last die, your chances are 1/6 because the result you want is 1 out of 6 results now. The more 6s you got, the probability of getting the rest of the throws as 6s went up because there are less possible results available for the rest of the row of throws. Up until you achieve all 6s, at which point the probability is now 1 out of 1.
Indeed, you have repeated the part I already understand, but it feels like saying that, "The probability that I could transform into a duck are the same in this moment that they were 10 seconds ago according to quantum physics". At some point it seems that a certain number of throws in a row would force us to consider things differently. If I did indeed turn into a duck, folks would not shrug that off as a quantum possibility, even if highly improbable. Hypothetically, if you had one quintillion throws in a row, you would have a team of scientists on the scene and it would make international news. Nobody would ever shrug that off and say "ah well, the probability of the next roll doesn't change." At that point, all involved scientists and statisticians and any interested parties would effectively be falling victim to the gamblers fallacy, but it still seems to make sense that they would, right?
I feel bad for you because I 100% think the exact same way as you do about this and am second-hand frustrated that nobody has dealt with your actual point
Thanks for validating. I empathize with the math. When it comes to casinos or whatever, there are likely strings that will come out of any number of dice throws, but you bet your ass that at some point the pit bosses will start watching closely, never dismissing the lead up to this point. At some point the manager is called in from off duty. At some point the game is shut down. The gambler can't just get away with saying, "But each roll has an independent probability!" No, the casino crew has now "fallen victim to the gamblers fallacy". But inherently, we understand they haven't. They have made a reasonable decision that they can't afford that person throwing the dice one more time. But what was the point of shutting it down if the dice were fair and any number of gamblers could start doing the same thing on any other number of tables? Do all casinos just shut down forever, therefore falling victim to some version of the fallacy?
If the die is fair the odds don't change. The historical information is irrelevant as the die has no memory.
A long string of rolling the same number may indicate the die is not fair, in a non-hypothetical situation.
Would you personally brush off the one quintillion consecutive throws if the die was determined to be fair by a team of scientists? Then another quintillion? And if someone now gave you the chance to bet on the next outcome, which would you choose? I argue that if you are rational you would bet that the streak continues. But mathematically you shouldn't change your bet, and you should ignore the two quintillion consecutive throws up until this point, right? Do you see the problem here?
If the die is fair then it's fair, that's all she wrote. Rationally you should always follow the math. Constructing an absurdly unlikely scenario doesn't invalidate the math.
What does it being absurdly unlikely have to do with anything? And at what point do you consider it "absurdly unlikely"? 100 throws? 1000 throws? 1 million throws?
By even pointing out the absurdity seems to indicate that you have an inherent understanding that it does at some point begin to matter.
You're constructing an extremely unlikely scenario to rationalize thinking a die has memory. It does not.
The previous results don't affect future outcomes for a fair die, no matter what those previous results are or how unlikely that outcome was.
Let me toss that back at you: How many unlikely outcomes have to occur before it "begins to matter"? Is one enough to start ignoring the math? Ten? A thousand?
I never offered that the die has memory. I only offered a hypothetical in which a fair die rolled one quintillion times on the same number by what a mathematician would say is pure chance. And your suggestion that I believe that implies that you have some inherent belief that the math "breaks down" at some undefined, seemingly ridiculous point. Regardless of the number of rolls I pick, you will say it doesn't matter and I say at some point it does. That is the rub. You have absolute confidence that any limitless number I can think of wouldn't sway you into reconsidering which number would be most "rational" to pick next if this all occurred in front of your eyes.
I think what I'm really saying is that normally we'd expect, on average, a die that may roll the same number that can be explained with mathematical probabilities. And those probabilistic averages play out the same, all day every day in casinos everywhere, because we observe them, and the laws of physics appear fixed. Any gambler who thinks those laws of physics and probabilities will change based on their crude observations of a small number of rolls is, in fact, a fool.
Now, you suddenly have an outlier that outlies averages so far that the whole casino industry topples because of it. Although my scenario is absurdly unlikely, your math shows that it is equally possible, albeit unequally probable. Is the gambler who watches the seemingly supernatural phenomenon unfold in real time all that foolish if they were to bet on the next outcome to be the same as the prior quintillion?
I suppose this might be a question of philosophy and not math. And I'm not arguing with the defined math, but I firmly stand beside the point that eventually it is not irrational to assume the same number might be rolled one more time after observing it a quintillion times.
The “absurdly unlikely” part comes in to play in being able to view the events from two different perspectives.
One is that, if someone claimed at the outset, that they could predict the next quintillion rolls of the die (whatever values those might be), the probability of all of them being correct is vanishingly small - each of the 6^quintillion combinations almost certainly won’t show up, only one will, and you’re relying on picking that one sequence.
However, once you’ve correctly predicted the quintillion rolls in a row, if you then say “I’m going to roll a 6 next”, you aren’t any more or less likely to get it right than you were on the first roll.
The probability of being able to predict (N+1) correct dice rolls is N * 1/6.
1 roll: 1/6
2 rolls: 1/6 * 1/6 = 1/36
3 rolls: 1/36 * 1/6 = 1/216 Etc.
If you’ve already done the N dice rolls, you’ve already dealt with the probability of getting to where you are in the chain of rolls. The probability to advance to the next step in the chain is always the same though, even if the chances of you successfully getting to that point in the chain are infinitesimal. You’d still expect 5/6 to get it wrong at the next roll, 35/36 to get it wrong in the next 2 rolls, and 215/216 to get it wrong in the next 3 rolls.
You have done a good job of explaining the math. And thank you.
Now, you're sitting at that gambling table and someone gives you the opportunity to choose one number that will appear on the die for the next roll after one quintillion. Are you going to choose some number other than the one that came up one quintillion times or some other number? I imagine you would choose the same number instead of picking some other number at random.
Ultimately, it doesn’t matter which number I pick, I’ve still got the same chance of being right as with any other number.
People can use superstition to help them decide, but it doesn’t make them any more likely to be right. Some people will choose 6 because “that’s got to be right, it’s happened so often”. Others will choose their favourite number “because that’s lucky”. Others will choose anything but 6 because “they can’t be that lucky”. Any logic you try and apply to it to say “this outcome is more likely than any other” is just your brain tricking you.
Would you not sit up at a story on the news where someone rolled 7's at a craps table for a year straight, only stopping to eat and sleep? If you paid any attention to that story would you be a fool? They bring officials in and claim the game is still fair and allow it to continue. Are you a fool if you claim it is rigged? Now that same roller rolls for multiple decades. Do you still calmly say, "we are foolish to assume this person will roll 7's one more time just because of the past 50 years they have continued to roll 7's. Each roll is a new roll." ?
If you’re trying to construct an actual scenario, a casino wouldn’t let that happen. They’d kick the player out because “they believe them to be an advantaged player”, because they don’t like losing money. And eventually you reach a point where it’s simply more likely that there is a bias somewhere in the system that hasn’t been detected yet.
That means it would be a feature of the system (player / table / dice) rather than of the maths - the maths is based on perfectly controlled probabilities.
Practically, you can’t ensure it’s a 100% fair system, so the simple “each outcome is 1/6” breaks down. If you could guarantee that it was perfectly fair, then what I said earlier stands. In a Real-World situation, the assumptions change significantly - you can’t have perfect knowledge of everyone’s intentions, whether it could be a scam etc.
EDIT: however, most gamblers fallacies aren’t based on the idea “I have actual evidence that the system is rigged”. Things like “5 hasn’t come up on the roulette wheel, it must be overdue” aren’t based on an assumption of bias, they’re based on an assumption of fairness, which says that eventually all numbers will come up equally. However, they don’t have to come up equally before the heat death of the universe.
I'm not trying to construct an actual scenario. I am constructing a hypothetical scenario that says there is no chance that the system is rigged, and there are a quintillion throws that are all identical, which is entirely possible, but highly improbable. In real life we can say, "that would never happen", but the math says you are incorrect and it 100% could happen. Now, this situation, which is mathematically possible, plays out (hypothetically). Which bet are you going to make after the one quintillionth throw? And are you a fool if you use past information to say the next throw will remain the same as the past quintillion?
Again, if it’s guaranteed to be mathematically exactly fair, then by the maths I posted earlier, claiming you have better than 1/6 chance of getting the next one right is mathematically impossible, by definition.
To be clear: you’re defining a situation whereby you are guaranteed to only have a 1/6 chance of getting the next number correct, whichever you pick, and then saying “isn’t it better to stick with the number that came up before?”. Simply, no, it’s not, because of the way you defined the system.
Thanks for pointing out that commenter's moving goalposts.
The poster is conflating two different and unrelated things. Determining whether a coin is fair, or putting a probability on it being fair after observing a number of tosses has very little to do with “if I just got three heads in a row with a fair coin, what’s the probability of getting a fourth?”.
To add to this, maybe what I'm touching on without knowing it is similar to the problems Einstein was trying to solve when euclidian geometry failed. Yes, space bends as does time when you get off the paper and into the real universe. Probablities that explain away the gamblers fallacy as a fallacy maybe break in the real world when pushed to the same brink as Einstein pushed things when he invented relativity. Maybe??
In a universe of infinite time and space, everything is probable.
No, there isn't. Every die roll is an independent event. Thus, the probability of rolling any number is the same no matter how many times you roll it. The likelihood of roll a 6 100 times in a row is very low, but it can happen. The next time you roll, the chances of any number coming up are the same as they were before.
The odds aren't per side, the odds are any side. Probability would only increase if after each roll, one side of the dice was eliminated. Other than that, the odds remain the same.
Would buying more lottery tickets increase your odds?
If they are all different (i.e. you choose your numbers and make them different) then yes you’ll increase your odds. Same for a raffle, except you don’t need to worry about the numbers as each ticket is unique. But if everyone buys more raffle tickets with the same multiple, then your odds don’t change. If everyone buys more lotto tickets, everyone’s chances of winning increase (but also the odds of having to share the jackpot).
The answer is simple, how would the dice remember what you rolled before? Each roll is an independent event, there's no memory of the last one that affects the next one.
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The problem with probabilities is always, imo, semantic. We call the unknown "probability", for lack of a better way to describe things. But nothing is random, in reality. That's where the issue lies.
If I roll a die, I don't really have 1/6 chance of rolling a 6. I have either 0% chance, or 100% chance - depends on how I roll it. If I roll it the same way every time, I'll get the same result every time. Rolling a dice isn't "random". We call it random, because we don't know HOW we rolled it.
As for why this thing we call probability doesn't change, that's the easy part. A die has 6 faces, and if we could toss it randomly (we can't, randomness doesn't exist), any of them can come up, and they'll come up the same number of times (as in, given a truly random toss, any of them can come up as much as any other). So we say if we roll a d6, each outcome happens 1/6 of the times. Doesn't matter if you rolled it once, or 1000 times - each roll can come up as any number 1/6 of the time.
Ask yourself this—does the die remember what it previously landed on? If not, how could the odds go up? It’s a physical thing that falls and bounces by the laws of physics. What physics laws would change because it was previously on a 5 instead of a 3?
First, I'll try to give you the intuition of why it doesn't work like that:
Imagine you have one regular die 1-2-3-4-5-6, and another die with colours instead Blue-Red-Yellow-Green-Orange-Pink.
You roll the first die and obtain 6, that was a 16%.
You then roll the second die. What would that mean to "have a smaller chance to roll a 6 again"? Which colour is "6"? Is that Pink because it was the last of my list? That's completely arbitrary. It doesn't make sense that my first roll with change the probability of obtaining a colour or another, so there is also a 16% chance of obtaining each colour.
Now, if I write some numbers on top of the colours, does that magically change the probability of obtaining the "Pink 6" just because I obtained a 6 before? Surely not.
And if instead I rolled the first die a second time, why would that be any different? It's not like objects have some "magical memory" that remember how often it rolled 6.
Now, here is why you did think it worked like that in the first place:
If you roll the first die, don't look at its result, then roll the second die, then yes, the probability of getting two 6s is quite low: 1/36 = 3%
Similarly, if you roll the first die, don't look at its result, then roll the second die, and then I look at both and say 'I promise you that there is at least one 6', then the probability of a double 6 remains quite low: 1/11 = 9%. What's the difference? Well, I didn't tell you whether the first or the second was a 6, so you have less information than in your example (where you knew that it was the first who was a 6). Just a slight change in information and the probability is different.
And lastly, in nature, there is a lot of things that look random but aren't. For example, there is a 50/50 chance of being night or day, but after enough night the day eventually come, because that's not actually random in the same way a die works.
But in the situation you described:
Let's say you roll 2 dice one at a time. There are 36 possible outcomes. You always had a 1 in 36 chance of rolling one of these combinations. That means that rolling a 6 then a 6 was always 1 in 36.
After you roll the first dice, you can no longer roll the other 5 numbers on the first dice, so as a result the chances of 30 of the combinations coming up is now 0%. The remaining 6 combinations are now 100%, up from 16.6% before the first roll.
The logical fallacy is that rolling a 6 and a 6 was always a 1 in 36 chance. But after you've rolled the first dice, you're not just asking if you're going to roll a 6 again. You're also asking what's the probability that I'm going to roll a double, which should be a lower probability right? The problem is that rolling 2 6's wasn't the only possible double before you rolled the first dice. You always had a 6 in 36 chance of rolling a double (1, 1 or 2, 2 etc). After rolling that first dice, you only have 1 chance to roll a 6 but there's no longer 36 possibilities, there's only 6.
The 1st time you roll a 6 sided die, you have a 1/6 chance to roll a 6.
The 100th time you roll, you still have a 1/6 chance because it's still a 6-sided die.
The 100000000th time you roll, you still have a 1/6 chance to roll a 6 because it's still a 6-sided die.
If you have rolled four 6s in a row, then on the 5th roll you have a 1/6 chance to roll a 6, because it's still a 6-sided die. Completely apart from that, the chance of rolling five 6s in a row is 1/7776, because there are 7775 other possible combinations for those five rolls.
All the results of dice throws are equivalent to each other... if you pay attention and take note of EVERY die number, that is. But if you're focusing on just the number 6, then the other numbers are all considered "not 6"s.
The "not 6"s are 5 times more likely to appear than 6 on a six sided dice. The probabilities are drastically different now.
The chances of a 1 being rolled after a 6 is the same as a 6 being rolled. But the chances of a "not 6" being rolled after a 6 is much greater.
A "13554663" result is just as likely as "66666666" to occur. But if you only note the 6s, then "########" (all "not 6"s) is much more likely than "66666666" to happen because "########" fits with multiple possible results while "66666666" fits with only one possible result.
Dice don't have a memory. They absolutely don't care what has happened already. Every time you roll one, you're starting from scratch, simple as that.
The dice have no way to remember the times you've rolled them before. Since they don't know you rolled a 6, they can't act differently on later rolls to make 6 less likely.
Because they are independent events.
Think of it this way - if you roll a die in Chicago and I roll a die in New York - they are not connected in any way. Agree?
Same is true if you roll the same die repeatedly in the same place. It has no memory.
It is possible to design a system that tries to balance outcomes explicitly so as to “load share” based on some memory but that is not how ordinary randomness works.
As many said: whatever you rolled before, the next roll will always have a 1/6 chance of becoming any specifuc number. There's 6 different outcomes, each is equally likely.
I'd like to add: When figuring out the probability of the sum of multiple rolls, you look at the number of combinations that make that sum.
There's only one way to make a 2 with 2 rolls (1 and 1).
There are two ways of making a total of 3: 1+2 or 2+1
There are three ways of making a total of 4: 1+3, 3+1, 2+2
5 = 1+4, 4+1, 2+3 or 3+2
And so on.
The probability of a certain sum (12 for two sixes) is the number of combinations that give you that sum (1) divided by the total number of combinations there are (6*6), so it would be 1/36.
If you wanted another specific outcome, say the first die must be a 3 and the secomd one must be a 1, that's also 1/36. If the sum just has to be 4 the probability is 3/36 or 1/12.
Dice have no memory. What happened in previous rolls in no way affects the outcome of future rolls. The odds of rolling a six don't change to "make up" for not rolling a six in a while.
If I say I am going to roll a die 4 times, the probability of getting 4 of the same number I picked in a row is 0.077%
Each time I roll the die, the probability of getting the number I picked is 16.67%
The probability of each roll achieving the result you want actually increases as the previous rolls are in your favour.
If you roll 2 of the same in a row, the odds of getting two more is 2.78%, and if you get 3 in a row, the odds of getting the final one is again 16.67%.
I think you're confusing two concepts:
The chance of any particular number for a roll of the die once (e.g. 1 in 6)
The chance of getting a series of specific results. For instance, the chance of not rolling a 6 if you rolled 30 times is quite low (something like .4% chance of happening).
However, the fact that a sequence of rolls is likely to occur is NOT predictive of any single roll. You know it's weird if you roll 30 times and never hit a 6, but if you roll a 31st time, your odds of any one number is still one in 6.
Don't confuse the probability of getting a sequence with the probability of any single roll. It's a little counter-intuitive for some people.
One thing I'm not seeing is that we tend to look at probabilities being 'fair' - If I roll a die 6 times, and it comes up 1, 2, 3, 4, 5 and 6, that feels 'fair' because each face has an equal probability to roll.
However, if I roll the die 12 times and it comes up with three 1s, three 2s, two 3s, two 4s, two 5s, and no 6s, that feels unfair - and it feels like we should roll two 6s to make it fair and match the probability.
However, probability doesn't really behave that nicely in the small scale. If I roll that die thousands and thousands of times, yes the results will be roughly equal per die face, but that doesn't mean that each set of 6 rolls maintains the large scale probability.
Every roll of the die is an completely independent, right? There's no way that your previous roll can effect your next roll. That means you might go 30 rolls without landing a 6, or you might roll a 6 ten times in a row. But averaged over many many rolls 1/6th of them should be 6s.
If you ask “what is the chance I roll a 6 on this one die roll?” That never changes at 16.7%.
But if you were to ask ‘what if I roll twice and get at least one 6?” That’s 30%
Chance on 3 rolls? 42%
6 rolls? 67%
But if you rolled 5 times and didn’t get a 6 yet, the chance your last roll is a 6 remains 16.7%.
The thing you might be looking for is the Law of large numbers. The theorem states that the average of the results obtained from a large number of trials should be close to the expected value and tends to become closer to the expected value as more trials are performed.
So if you roll the dice 1000 times, it is astronomically unlikely to never roll a 6, or that all 1000 will be 6s. It isn't impossible, but very unlikely. The distribution of how many times a number was rolled should approach 1/6 for each number.
FellowConspirator t1_ja83xfr wrote
Each time you roll the six-sided die, there's six possible outcomes: 1, 2, 3, 4, 5, 6. The probability of each one of those is 1/6 (16.66...%), and the total is 1 (100%). It doesn't change with each roll, because you aren't changing the die, or what you are doing. There's nothing connecting one roll to the next; they are completely independent.