Is the gambler's fallacy really a fallacy?

spirit-salamander

Philip Goff summarizes this supposed fallacy quite well:

"In the regular gambler’s fallacy, the gambler has been at the casino all night and has had a terrible run of bad luck. She thinks to herself, “My next roll of the dice is bound to be a good one, as it’s unlikely I’d roll badly all night!” This is a fallacy, because for any particular roll, the odds of, say, getting a double six are the same: 1/36. How many times the gambler has rolled that night has no bearing on whether the next roll will be a double six." (Philip Goff - Our Improbable Existence Is No Evidence for a Multiverse)

Now to my argument. First, let's assume an ideal dice, which is rolled ideally, whatever that means exactly. In this way, I may be establishing an analogy with the domain of quantum mechanical probabilities.
So if I roll our assumed ideal die, we are told that the probability of rolling a 6 is 1/6. So far, so good. The question that arises is what this 1/6 means philosophically or, if you like, mathematically.
What does the 1 mean here, what the 6 and what this symbol /?
It can only mean that out of 6 times rolling the dice, the 6 will occur one time, right? One must not forget that I presuppose an ideal framework of this thought experiment.
If I assume that after rolling the dice six times, a 6 came out each time, the probability in hindsight was 6/6, no? Maybe I'm on the wrong track with this retrospective view.
Therefore, before rolling the dice, I better say the probability would be 6/6, which should mean that out of 6 times rolling the dice always results in a 6.
I actually want to say 1/6 means that out of 6 rolls, 6 will come only once and that the rolled 6 stands in the context of 6 (possible) rolls.

How many times the player in Goff's example rolled the dice that night does, in an idealized context, affect whether the next roll will be a double six.

Let's assume 600 rolls. What would be the most probable result of an ideally rolled ideal dice? An absolutely even distribution? So 100 times the 1, 100 times the 2, 100 times the 3 and so on to 100 times the 6? Or 0 times the 1, 200 times the 2, 50 times the 3, 150 times the 4 and so on until the sum is 600?

I would say that the ideal dice must give an absolutely even distribution. If it were a little less ideal, you might have 99 times the 1, 101 times the 2.... . In other words, small unevenness. For example, the completely anti-ideal die would give 600 times 5, and all the others 0. The absolutely anti-ideal dice would have an infallible tendency to exclusively one number.

This means that probability is not based on complete chance, but on strict laws of probability. That is, if I have already rolled the 3 100 times in a row, the 3 should not occur again according to the ideal rolled ideal die. The probability in this case would be 0. The chance would consist in the fact that one cannot know which of the other numbers occur, but the probability law forbids that the 3 occurs again.

The same principle could be reduced to 6 rolls. According to the law of probability, each number can occur only once in the ideal dice roll. Exactly what the sequence of numbers will look like is left to chance.

Let's move from the ideal dice roll to the real one. In the real world of dice rolls, we may see odds that seem to be out of context.
If I roll 100 times and get a 6 every 100 times, that's obviously an outlier, but why couldn't I conclude with each subsequent roll of the dice that the probability of rolling a 6 is less than 1/6? The pattern so far has shown that it is extremely unlikely to continue, so couldn't we factor in the probability of that pattern to account for the next dice roll? If the pattern were to continue and last for years, there would obviously be a determining system behind it.

Someone could counter: In general, probability deals with uncertainty. Because of this randomness, a die would approach a even distribution as the number of rolls approaches infinity. For any finite value, there can be no guarantee that it will be perfectly even.

I would answer, infinity would be an ideal size, which would not simplify the matter.

The mention of infinity would also seem to be an excuse and evasion, since infinity cannot be completed according to its own definition.

One must conclude that the dice rolls, if there are several, somehow depend on each other, that is, mysteriously know about each other and about the intention of the dicer, how many times he intends to roll the dice.

How many times the player rolled the dice contra Goff has a bearing on how the next roll will turn out, even though the influence may be very marginal and very mysterious as well. Every roll is not unconditional, except for the first roll, of which, however, you can never know when it occurred. That is perhaps why there is the similarity to quantum physics (at least for ideal dice).

tim wood

Maybe this way. Imagine the sequence of 1s through 6s resulting from throwing a fair die a very large number of times. Within the sequence, for any arbitrary number n of throws, all the possible combinations of n 1s through 6s will occur many times.

Now consider your own sequence, whatever it is. Find all the instances of that same sequence in the very large and long sequence. List the the next number in sequence after your sequence. You will find that it's any of the of the 1 through 6 all with equal probability.

That is, your sequence of length m is always part of a sequence of length m+1, and you simply have no way of knowing which m+1 sequence yours is a part of, and hence no way of knowing what the result of your next throw will be.

TonesInDeepFreeze

[1/6] can only mean that out of 6 times rolling the dice, the 6 will occur one time, right? — spirit-salamander

Wrong. A probability is not a surety of what will happen. A probability of 1/6 of the occurrence of the six doesn't ensure that the six will occur within 6 rolls. And the rest of your argument is arbitrary confusion.

The probability of 6 rolls all turning up the six is 1/(6^6).

The conditional probability (from having 5 rolls all turn up the six) of all six rolls turning up the six is 1/6, which is also simply the probability of a roll turning up the six.

And empirically, if the gambler's fallacy weren't a fallacy, then in Las Vegas there would be a whole bunch of fabulously rich amateur dice players and the casinos all bankrupt instead of all the losers who leave town begging for spare change for a bus ticket and the casinos profiting by billions of dollars.

TheMadFool

First off, using a die instead of a coin complicates the matter (for me) but it really doesn't seem to be that much of a problem. I'll use a coin if it's all the same to you.

Let's begin with stating our assumptions:

1. The coin has two faces - heads & tails - and it's a fair coin.

2. The idea that the outcomes of a coin-toss will "balance" out i.e. you'll get an equal number of heads and tails is based on infinity. My memory of this is that as the number of trials approach infinity, the difference between empirical probability and theoretical probability approaches zero. The law of large numbers? I'm not sure.

Imagine now a gambler is tossing the coin. He gets, let's say, just to make things interesting, a million heads. Is he justified in thinking the next toss will be a tails? Unfortunately, no. The law of large numbers has infinitely many opportunities (beyond a million ad infinitum) to even things out i.e. the next million tosses might be all tails. You get the picture, right?

As you can see, a streak of heads doesn't necessarily imply that the next toss will yield a tail. Keep in mind that, my interpretation, possibly mathematical nonsense, concedes that the law of large numbers would, in a sense, cause the outcomes to be dependent on each other. In other words, yes the gambler is correct in expecting a tails after a streak of heads but what he can't know is when his streak will end. Hence, the fallacy, hence the countless lives destroyed by commiting this fallacy.

TonesInDeepFreeze

↪TheMadFool

We don't need infinitistic considerations to see that the gambler's fallacy is incorrect.

The conditional probability, no matter the results of any finite number of previous trials, is the same as the initial probability, which is 1/2.

SophistiCat

The question that arises is what this 1/6 means philosophically or, if you like, mathematically. — spirit-salamander

This is quite a profound question, but for better or for worse, it is irrelevant to the gambler's fallacy. (Your understanding of probability is way off though.)

The key to understanding the gambler's fallacy is in the last sentence of your quote:

How many times the gambler has rolled that night has no bearing on whether the next roll will be a double six." (Philip Goff - Our Improbable Existence Is No Evidence for a Multiverse) — spirit-salamander

This is not a result of a probabilistic calculation - this is a key assumption, a real-world knowledge on which probabilities are based. The assumption is that every roll of the dice is effectively independent. If you agree with this, then the fallacy in the Gambler's Fallacy should be readily apparent. If you don't agree, then all bets are off, as it were, and that 1/6 probability model can no longer be maintained.

spirit-salamander

A probability is not a surety of what will happen. A probability of 1/6 of the occurrence of the six doesn't ensure that the six will occur within 6 rolls. — TonesInDeepFreeze

But probability cannot completely exclude surety. After all, the surety consists in a certain probability. Let's take the 1/6 probability of rolling a number like 4. The surety lies in the fact that 4 occurs at about 17,7 percent chance. If a probability of 1/6 for the occurrence of the six does not ensure that the six occurs within 6 rolls, then the 6 could never occur. That would be theoretically possible, wouldn't it? It would not be ensured that the six occurs in 60 throws, not in 600, not in 6000, and so on. But what is the point of using probability if it is not reliable? My examples referred to an ideal dice, which is ideally rolled. Thus, those odds of 17.6 percent would always have to be ideally fulfilled as a thought construct. Whether I roll 6, 60 or 60000 times.

spirit-salamander

The idea that the outcomes of a coin-toss will "balance" out i.e. you'll get an equal number of heads and tails is based on infinity. — TheMadFool

I find the concept of infinity problematic with the idea of probability. The idea is that at infinity all the numbers on a die have fallen equally. But the infinity knows no completion. It goes on and on. Therefore it would generate a bogus argument.

In other words, yes the gambler is correct in expecting a tails after a streak of heads but what he can't know is when his streak will end. — TheMadFool

The gambler's mistake, in my opinion, is that the probability of tails coming after a series of heads implies only a minimal increase in probability. He erroneously considers it to be very high. So if there is a long series of heads, the chance of tails is possibly only minimally increased. But nevertheless increased, in which the player is right. But for him not sufficiently increased to beat the bank.

TonesInDeepFreeze

If a probability of 1/6 for the occurrence of the six does not ensure that the six occurs within 6 rolls, then the 6 could never occur. — spirit-salamander

Probability does not ensure that the six will occur, but that does not entail that probability ensures that the six will not occur.

For any statement Q, from 'not ensured that Q' we do not infer 'ensured that not Q'.

It would not be ensured that the six occurs in 60 throws, not in 600, not in 6000, and so on. — spirit-salamander

Correct.

But what is the point of using probability if it is not reliable? — spirit-salamander

It's imperfect for making predications, but it's still good.

spirit-salamander

(Your understanding of probability is way off though. — SophistiCat

Let's assume 600 rolls. What would be the most probable result of an ideally rolled ideal dice? An absolutely even distribution? So 100 times the 1, 100 times the 2, 100 times the 3 and so on to 100 times the 6? Or 0 times the 1, 200 times the 2, 50 times the 3, 150 times the 4 and so on until the sum is 600?

I would say that the ideal dice must give an absolutely even distribution. If it were a little less ideal, you might have 99 times the 1, 101 times the 2.... . In other words, small unevenness. For example, the completely anti-ideal die would give 600 times 5, and all the others 0. The absolutely anti-ideal dice would have an infallible tendency to exclusively one number.

This means that probability is not based on complete chance, but on strict laws of probability. That is, if I have already rolled the 3 100 times in a row, the 3 should not occur again according to the ideal rolled ideal die. The probability in this case would be 0. The chance would consist in the fact that one cannot know which of the other numbers occur, but the probability law forbids that the 3 occurs again. — spirit-salamander

Is this way off? You can try it yourself at home. Roll the dice 600 times and write down the results. There will be an approximately even distribution. Now my argument was about a dice as a thought thing, the perfect dice rolled perfectly. The distribution should be perfectly even. If this were not the case, one would have to conclude that there was manipulation involved.

TonesInDeepFreeze

one would have to conclude that there was manipulation involved. — spirit-salamander

Less than a perfect distribution would have you infer cheating? That's crazy. And even if there were an astronomically long streak, one would very highly suspect that there was cheating, but it's not certain that there was cheating. There will be streaks. There is no upper limit on how long a streak can be.

Purely mathematical probability is not taken necessarily to be matched every time by real world outcomes.

Take the simplest example of a coin toss. The chance of heads is 1/2. But that does not entail that heads comes up exactly 50% in every experiment. Take even just an experiment of two flips. Easily it can happen that heads comes up twice - 100% heads rather than 50% heads. Indeed the probability of heads twice in a row is 25%. The probability of heads six times in a row is1/(2^6). So if you see it happen, then you happened to be there when that 1/(2^6) chance was realized. Just ask any mother and father who had six girls in a row.

TheMadFool

I find the concept of infinity problematic with the idea of probability. The idea is that at infinity all the numbers on a die have fallen equally. But the infinity knows no completion. It goes on and on. Therefore it would generate a bogus argument. — spirit-salamander

Consider infinity as it appears in the law of large numbers not as something completed/actual (not ok) but as something incomplete/potential (ok). It only means that one can continue to perform trials ad infinitum and perhaps also ad nauseum. There's no upper limit to how many trials can be conducted and thereby hangs a tale.

Allow me to illustrate my point. Suppose you play a coin-toss game in which heads you win, tails you lose. First toss, tails; second toss, tails; third toss, tails; fourth toss, tails. You're experiencing a losing streak - 4 tails in a row. What do you suppose the next toss (fifth) will be? Heads/tails? If you commit the gambler's fallacy, you'd say, "heads." Now consider the number of tosses are increased to 10 total tosses. Do you still think the fifth toss will be heads, breaking your losing streak? You can't because even if you get a tails again, chance (the law of large numbers) has 5 more tosses to ensure that the number of heads = the number of tails. Continuing along this trajectory, imagine the total number of tosses are increased to a 100. Would you feel confident about the fifth toss turning up heads, to break your run of losses? No! Your losing run could extend up to the fiftieth toss and chance still has 50 more tosses in which to break even so to speak. Imagine now that the number of tosses extends to infinity. No finite streak of losses, no matter how large/many, can give you good reason that the next toss will be different as there are an infinite number of tosses remaining for chance to "rectify" the situation.

TheMadFool

↪spirit-salamander

Here's something that has me puzzled...

Imagine you throw a six-sided die 5 times and all times you get a 6. The probability of this happening = (1/6)^5. Suppose also that getting a 6 is a loss.

What do you think is going to happen on the 6th throw?

Probability of six 6's in a row = S = (1/6)^6 [continuing the losing streak]

Probability of the 5 sixes and the 6th roll being something other than six = P = (1/6)^5 * (5/6) [breaking the losing streak]

P > S

Gambler's fallacy??

tim wood

↪TheMadFool

The trick, if it be that, to calculating odds and probabilities is to be very careful to be sure that you're calculating exactly what you think you're calculating. and sometimes that's not so easy to figure out.

TonesInDeepFreeze

Gambler's fallacy?? — TheMadFool

No. It's more probable not to roll a six than to roll a six, in any circumstance - no matter what was rolled before. That's not the gambler's fallacy.

TheMadFool

↪tim wood

↪TonesInDeepFreeze

Suppose you play a game with one die and the rules are getting a 6 means a loss. Getting any other number {1, 2, 3, 4, 5} is a win

You're very excited since the odds are in your favor: 5 to 1

What's the probability that,

1. You have a streak of losses (6's) on five consecutive throws and the sixth throw is also a 6?

This probability = L = (1/6)^6

2. You have a losing streak of five consecutive 6's and the sixth throw gives you any number from 1 to 5 (win and breaking the streak)?

This probability = W = (1/6)^5 * 5/6

W > L

:chin:

You should commit the Gambler's "fallacy."

TonesInDeepFreeze

↪TheMadFool

What should you commit? A bet? Or what? What is the payout? If the payout is proportionate to the odds, then mathematically it doesn't matter whether you bet on six or not-six (where 'not-six' means any number other than six wins) or whether you even play at all.

The gambler's fallacy would be to think that because there were 5 sixes then for the next throw the probability of a six is less than 1/6. Recognizing that the probability of a six is 1/6 is not fallacious. it doesn't matter what happened on the previous throws.

TheMadFool

If the payout is proportionate to the odds — TonesInDeepFreeze

This is what bothers me. What I know of probability is that this issue raised by you has to do with what is known as expected value (I'm in two minds about whether this is the relevant concept). Suppose a game of die is such that if you get a 1 or 2 you win, any other number and you lose. If you win, you get 12 cents and if you lose, you have to pay 6 cents.

The expected value for this game = (2/6) * 12 - (4/6) * 6 = 0. That is to say, the game is fair - your losses and your gains will balance each other out with no one (neither you nor the house) gaining the upperhand. Playing this game 6 times means you will, on average, win twice and lose four times.

What this appears to be is that fair games have been designed in a way that when played, a certain combination of outcomes (2 wins, 4 losses) are supposed/expected to occur. If not, no one would play. In other words, if I lose 4 times in a row [a streak], I should look forward to a change in my fortunes, for the better that is. In fact, that's the selling point of games of chance. This simply means that the gambler's fallacy isn't a fallacy since some games like the one above are designed in such a way that streaks of losses are common and, let's not forget to mention, these streaks are supposed to be overcome. Either that or the gambler's fallacy is a fallacy but in a different setting (one such would be games that have a 50/50 chance of winning/losing).

Frankly speaking, I'm more confused than when I began. :chin:

TonesInDeepFreeze

The way gambling against the house works is you give the house x amount of money. If you lose the bet then the house keeps your money. If you win the bet then the house gives you x+y amount of money, where y>0. But when you win, the x+y that you get is less than what a "fair" payout would be. For example, if the chances of winning are only 1/10, then a "fair" payout on a one dollar bet would be ten dollars (x=1 and y=9), but the house doesn't pay "fair" so you only get, say, nine dollars (x=1 and y=8). So the house has a percentage advantage (here it's a 10% advantage). Sucker games are those where the house has really great advantage.

If the game is "fair" (so there's no house making a profit) then there's no reason to expect you'll do better or worse, no matter how you bet. For example, if you bet a dollar ten times on an outcome that has a 1/10 chance, then you would expect to lose nine times, thus lose nine dollars, but you also expect to win once with a payout of ten dollars, which is a nine dollar profit on that particular bet (since you spent a dollar on that tenth bet). So expect that your losses will equal your profits.

The gambler's fallacy is thinking that the chance of an outcome is less because that outcome has recently occurred (or that the chance of an outcome is more because that outcome has not recently occurred). The fallacy is in not understanding that each outcome is independent, not affected by previous outcomes. But if the game is "fair" then, even though it is a fallacy to think the chance of an outcome is affected by previous outcomes, we would not expect that it would hurt to bet per the gambler's fallacy, since there's not an expectation of better advantages among bets anyway.

But if the game is against a house that does not pay "fair", then the gambler's fallacy is not a wise basis for placing bets. Why is it not wise? Because instead of placing bets where the house has not a very great percentage advantage, instead you're distracted by a fallacy into placing bets you ordinarily wouldn't place where the house does have a great percentage advantage. If the house takes a great percentage on a certain kind of bet, but you take that bet because you think your hoped for outcome has a better chance than it actually has, then we would expect that you're on a course to lose more money than you would if you played without the gambler's fallacy.

So irrespective of betting against a profit-taking house, or even irrespective of gambling at all, the gambler's fallacy is thinking that chances of a single event are affected by previous outcomes. And with regard to betting against a profit-taking house, the danger of the gambler's fallacy is this: Instead of making bets on games where the house doesn't have such a great percentage advantage, you think you can beat the house at games it does have a very great percentage advantage as you incorrectly expect that you will ameliorate the effect of that advantage because your chances on an outcome are 1/n when they're really 1/(n+k).

SophistiCat

Is this way off? — spirit-salamander

Yes. Like I said, history-independence is an assumption in common games of chance, such as dice. But if you start off assuming that the probability of each possible outcome in a single trial is 1/6 and then end up concluding that the probability of a particular outcome in the next trial is more than 1/6, then you have contradicted yourself.

SophistiCat

You can try it yourself at home. Roll the dice 600 times and write down the results. There will be an approximately even distribution. Now my argument was about a dice as a thought thing, the perfect dice rolled perfectly. The distribution should be perfectly even. If this were not the case, one would have to conclude that there was manipulation involved. — spirit-salamander

Here is a little script that implements this scenario using a pseudo-random number generator: https://groovyconsole.appspot.com/edit/5205481161228288 (click "Execute script" to roll a die 600 times and see the distribution).

But you can try a much simpler experiment at home. Take a coin and flip it twice. Repeat as many times as you please. According to your theory of perfect coin flips, you should be getting one head and one tail every time. In reality, you will get that result about half the time. So real coin flips are nowhere close to what you think perfect coin flips ought to be. In fact, perfect coin flips would have to be manipulated to produce an alternating sequence: heads, tails, heads, tails, etc. Anything else would violate your criteria of perfection.

Your misconception of probability is perfectly summarized here:

But what is the point of using probability if it is not reliable? — spirit-salamander

Probability is a measure of uncertainty. Where you can make a perfectly reliable prediction, you have no need of probability.

tim wood

This probability = W = (1/6)^5 * 5/6 — TheMadFool

No. The odds on any single throw are one in six - which you recognize. You're combining the odds on throwing six sixes with throwing the sixth six.

Another way. Let's suppose you're throwing a die and writing down the results. The question is what are the odds that on the tenth throw you will throw a six? Ans., one in six. Second question, how could any of the throws one through nine effect those odds? Ans., they don't.

TonesInDeepFreeze

↪TheMadFool

Following up on tim wood:

Given the assumption that there were 5 sixes, then what is the probability that the 6 roll sequence will be 6 sixes? That probability is 1/6, which is exactly the probability of any single roll having a six.

That is different from not having the assumption that there were 5 sixes, where the probability of 6 sixes is 1/(6^6).

If you think that, if you roll 5 sixes, then the payout on rolling a six on the next roll should be proportionate to 1/(6^6), then I'd love to play with you and bet on a six not being rolled on your 6th roll on a day when you happened first to roll 5 sixes.

spirit-salamander

There is no upper limit on how long a streak can be. — TonesInDeepFreeze

Any long streak of luck would always be thought of in the context of a larger sequence of rolls. Maybe I roll the 6 a thousand times in a row, but in that context, which is perhaps only a theoretical one, the 6 may not occur again for the next thousand rolls. However, it seems to me that one must always add an imaginary closed overall context to every throw, in which all numbers are evenly distributed.

Purely mathematical probability is not taken necessarily to be matched every time by real world outcomes. — TonesInDeepFreeze

I had tried to think about the ontology or metaphysics or reality behind this probability thing, which can be described clearly mathematically, about which you have a real understanding, as I see. One approach to make this 1/6 probability clear for each number in the roll of the dice would be to think that six possible worlds, or only theoretical worlds, are thought of in each roll. We just don't know which world we are in during the roll of the dice. In the world where the 1 appears, in the one with the 2 and so on? If finally the 6 comes, we know that we "are" in world with the 6. I could imagine it in such a way too. Because it would explain the "1/6" philosophically.

Take the simplest example of a coin toss. The chance of heads is 1/2. But that does not entail that heads comes up exactly 50% in every experiment. — TonesInDeepFreeze

What would actually be, if the experimental external conditions of the tossing machine and coin always remain exactly the same? Only one side of the coin would appear, let's say heads. The perfect coin toss would be one-sided. Let's find another setting of the machine, which only leads to always bring out the other side, number. Now we could set the machine so that it always alternates the conditions. First like this, then like that, and so on. Surely here we could say that there is a 50% probability? Since we are not gods, we would have to deal with probabilities in spite of the machine set up.

TonesInDeepFreeze

one must always add an imaginary closed overall context — spirit-salamander

Any given experiment would be a finite number of events. But there there are always longer and longer experiments - more events - than any given finite number of events.

possible worlds — spirit-salamander

That seems fine.

we could set the machine so that it always alternates the conditions. First like this, then like that, and so on. Surely here we could say that there is a 50% probability? — spirit-salamander

Of course, if we don't know which alternate the machine is about to produce, then we expect 1/2 probability of heads.

But in talking about probability in greatest generality, we don't have in mind a machine that alternates like that. We consider an abstract "framework" where results can't be predicted but only expectations can be given, based on the sheer mathematics that there are two equally likely possible outcomes and one of them is heads, so the probability of heads is 1/2. But probability does not ensure outcomes. Of course, there will be experiments in which heads comes up less than half among the flips and experiments in which heads comes up more than half among the flips.

spirit-salamander

But if you start off assuming that the probability of each possible outcome in a single trial is 1/6 and then end up concluding that the probability of a particular outcome in the next trial is more than 1/6, then you have contradicted yourself. — SophistiCat

Why? I actually wanted to say that the probabilities after the first throw change steadily and minimally. So if I rolled 6 on the first roll, the probability of a 6 appearing again on the second roll would be minimally lower. Lower in the sense of something like 0.0000000000000000000001. This is not meant to be mathematically correct.

So real coin flips are nowhere close to what you think perfect coin flips ought to be. In fact, perfect coin flips would have to be manipulated to produce an alternating sequence: heads, tails, heads, tails, etc. Anything else would violate your criteria of perfection. — SophistiCat

You are right, the perfect toss would always produce the same result.
This perfect throw could be accomplished also only by a presupposed God.
My imaginary god would arrange it in such a way that with two coin tosses in each toss case once head and once tail appears.

spirit-salamander

Probability is a measure of uncertainty. Where you can make a perfectly reliable prediction, you have no need of probability. — SophistiCat

How would you philosophically explain and describe the probability 1/6 in the dice rolls. What is the 1 here, what is the 6 and what / and how do they relate to the real world?
I have come to the conclusion that it is all very baffling and perplexing because you get to questions of chance and determination.

Albert Einstein's most famous quotes is, "God does not play dice". Is this to say that there are basically no probabilities in the world. And that our probability formula is empty and meaningless. Because as was said here, with the dice in infinite duration a 6 could never come.

TonesInDeepFreeze

if I rolled 6 on the first roll, the probability of a 6 appearing again on the second roll would be minimally lower. Lower in the sense of something like 0.0000000000000000000001. This is not meant to be mathematically correct. — spirit-salamander

It's not mathematically correct and there's no reason to think it's empirically correct.

In an empirical situation, if you suspect that one side of the coin has an advantage, then after a side comes up, one would expect that the probability of it coming up next is higher not lower.

What is the 1 here, what is the 6 and what / and how do they relate to the real world? — spirit-salamander

6 is the cardinality of the set of possible outcomes; that set is the event space. 1 is the particular outcome, which is one of the members of the event space. Division expresses the ratio of the particular outcome to the possible outcomes.

In actual rolls of the die, we would have an expectation, that in the long run, six comes up in close to 16 and 2/3 percent of the rolls, and we expect that it comes closer and closer to 16 and 2/3 percent as we do more and more rolls. That is an expectation but it is not ensured.

TonesInDeepFreeze

our probability formula is empty and meaningless [?] — spirit-salamander

If you thought probability is meaningless, wouldn't you be just as happy if the doctor told you that the chance of surgery survival is 1% as if he told you it was 90%?

spirit-salamander

It's not mathematically correct and there's no reason to think it's empirically correct. In an empirical situation, if you suspect that one side of the coin has an advantage, then after a side comes up, one would expect that the probability of it coming up next is higher not lower. — TonesInDeepFreeze

Okay, I've realized that I'm wrong here. Thank you very much for all your comments.

That is an expectation but it is not ensured. — TonesInDeepFreeze

This is undoubtedly true for humans, I have now come to realize that as well. But how would it like with Laplace's Demon? From his point of view?

If you thought probability is meaningless, wouldn't you be just as happy if the doctor told you that the chance of surgery survival is 1% as if he told you it was 90%? — TonesInDeepFreeze

Yes I would, nevertheless the connection of the mathematical probability calculation with reality still seems strange to me. At least we seem to assume some at least semi-fixed patterns of events in the world. From these patterns we then generate probabilities.

In the case of the dice, one would say that it is quite evenly shaped, without one side having more weight than another. And one would say that when you roll the dice, you always make a different roll. And if I now roll the dice 600 times, it is very likely that an approximate uniform distribution will result. This must be due to the laws of nature and the absence of any attempt to influence the dice. The 1/6 seem to be the mathematical expression for it (laws of nature and the absence of the manipulation).

TonesInDeepFreeze

But how would it like with Laplace's Demon? — spirit-salamander

I don't know.

In the case of the dice, one would say that it is quite evenly shaped, without one side having more weight than another. — spirit-salamander

Of course, in the physical world, dice are not perfect. We don't claim that the mathematical framework corresponds to every physical pair of dice. The mathematical framework is an "ideal" that is still useful even though such things as physical dice are not ideal.

The 1/6 seem to be the mathematical expression for it (laws of nature and the absence of the manipulation). — spirit-salamander

Perhaps some people think in terms of laws of nature in this regard. But personally I think of it just as I described it - an ideal event space and ideal events

Is the gambler's fallacy really a fallacy?

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